17 May
2009
17 May
'09
4:51 p.m.
Hi! Does anyone know of an available transformer suitable for making an
S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
outputs.
The best I solution I can find is two separate 80VA transformers; one
parallel 10VAC output and another 36VCT output which can be configured for
+/-18VAC outputs.
I would prefer a single transformer rather than dual transformers because
they are the most expensive part of a linear power supply.
I know Northstar and Vector Graphic used similar dual output transformers
but I cannot find anything like them after searching the usual vendors like
Digikey, Mouser, Jameco, etc. Even transformer manufacturers don't seem to
carry a stock product of this configuration although you can order custom
units.
Thanks in advance for any help. Your ideas greatly appreciated.
Thanks and have a nice day!
Andrew Lynch
17 May
17 May
9:06 p.m.
On 17 May 2009 at 16:51, Andrew Lynch wrote:
Hi! Does anyone know of an available transformer
suitable for making
an S-100 power supply? I need 120VAC input and 10VAC output and +/-
18VAC outputs.
Hi Andrew,
I'm not certain if the S100 linear PSU transformers were eve a
standard item. The voltages are strange enough not to be common.
The transformer in my Integrand box is certainly a custom job. MITS
certainly used 2 transformers.
FWIW, the Stancor P-6138 transformer used in the MITS Altair 8800 was
a 7.5VCT unit and the 133P5 was something like 12.6VRMS. Less diode
drop, the voltage on a capacitor-input filter, will approach
1.4xVRMS, so even with a 12.6VRMS transformer, you've got plenty of
headroom for the regulators. If your supply voltage is too high,
you'll pay for it in excess heat in the series (analogue) regulators.
Cheers,
Chuck
9:15 p.m.
----------------------------------------
From: lynchaj at yahoo.com
To: cctalk at classiccmp.org
Subject: S-100 power supply transformer
Date: Sun, 17 May 2009 16:51:44 -0400
Hi! Does anyone know of an available transformer suitable for making an
S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
outputs.
The best I solution I can find is two separate 80VA transformers; one
parallel 10VAC output and another 36VCT output which can be configured for
+/-18VAC outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
_________________________________________________________________
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11:55 p.m.
----------------------------------------
From: dkelvey at hotmail.com
To: cctalk at classiccmp.org
Subject: RE: S-100 power supply transformer
Date: Sun, 17 May 2009 18:15:27 -0700
----------------------------------------
Hi
I forgot to mention. You need to derate the transformer
as well. The problem is that the current flows when the
voltage is the highest ( or close to it ). The math is a little
complicated but it means you need to increase the rating of
the transformer by at least 1.414.
If you can't find a transformer with the right windings,
I recommend looking at toroidal transformers. These can
easily have turns added to increase or buck the voltage
of the secondary.
Dwight
_________________________________________________________________
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From: lynchaj at yahoo.com
To: cctalk at classiccmp.org
Subject: S-100 power supply transformer
Date: Sun, 17 May 2009 16:51:44 -0400
Hi! Does anyone know of an available transformer suitable for making an
S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
outputs.
The best I solution I can find is two separate 80VA transformers; one
parallel 10VAC output and another 36VCT output which can be configured for
+/-18VAC outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
19 May
19 May
2:36 a.m.
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a full
wave BRIDGE topology.
reference: The holy book " The art of electronics " page 47
Best regards, Steven
Hi
I forgot to mention. You need to derate the transformer
as well. The problem is that the current flows when the
voltage is the highest ( or close to it ). The math is a little
complicated but it means you need to increase the rating of
the transformer by at least 1.414.
If you can't find a transformer with the right windings,
I recommend looking at toroidal transformers. These can
easily have turns added to increase or buck the voltage
of the secondary.
Dwight
----------------------------------------
> From: lynchaj at yahoo.com
> To: cctalk at classiccmp.org
> Subject: S-100 power supply transformer
> Date: Sun, 17 May 2009 16:51:44 -0400
>
> Hi! Does anyone know of an available transformer suitable for making an
> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
> outputs.
>
> The best I solution I can find is two separate 80VA transformers; one
> parallel 10VAC output and another 36VCT output which can be configured
for
+/-18VAC
outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
10:56 a.m.
Hi
This is the result of when the current actually flows.
It only flows when the voltage is near the peak. This
changes the average power dissipated in the secondary winding
( the VA rating ).
Altough, the 1.414 is the worst value and the real value
is somewhere between 1 and 1.414.
It make no difference what type of rectification is used,
fullwave bridge or center tap. Watts are watts and there is
no way around it. It is true that the load current rating
is halved for the secondary turns for the center tapped
fullwave rectification but the VA rating of the transformer
remains the same.
I'm not sure what the book you reference says. I do
know how power is calculated and it is always the vectored
volts times amps. The current flows out of the transformer
when the output voltage is close to the peak. This means
that the rms current of the secondary needs to be compensated
to account for when the current actually flows.
If your transformer has a secondary rated at 10 amps that
can only feed a rectifier and capacitor filter of 7.07 amps.
That would change if feeding a inductive filter because of
when the current flows.
If it was a center tapped with two diodes, it would be
14.14 amps instead of 20 amps. Still, the VA rating of the
transformer does change.
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctech at classiccmp.org; cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Mon, 18 May 2009 23:36:52 -0700
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a full
wave BRIDGE topology.
reference: The holy book " The art of electronics " page 47
Best regards, Steven
Hi
I forgot to mention. You need to derate the transformer
as well. The problem is that the current flows when the
voltage is the highest ( or close to it ). The math is a little
complicated but it means you need to increase the rating of
the transformer by at least 1.414.
If you can't find a transformer with the right windings,
I recommend looking at toroidal transformers. These can
easily have turns added to increase or buck the voltage
of the secondary.
Dwight
_________________________________________________________________
Windows Live?: Keep your life in sync.
http://windowslive.com/explore?ocid=TXT_TAGLM_BR_life_in_synch_052009
----------------------------------------
> From: lynchaj at yahoo.com
> To: cctalk at classiccmp.org
> Subject: S-100 power supply transformer
> Date: Sun, 17 May 2009 16:51:44 -0400
>
> Hi! Does anyone know of an available transformer suitable for making an
> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
> outputs.
>
> The best I solution I can find is two separate 80VA transformers; one
> parallel 10VAC output and another 36VCT output which can be configured
for
+/-18VAC
outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
2:15 p.m.
On 19 May 2009 at 7:56, dwight elvey wrote:
It make no difference what type of rectification is
used,
fullwave bridge or center tap. Watts are watts and there is
no way around it.
What surprises me is to see that a transformer with 2 secondary
windings would be operated with the windings in parallel to feed a
bridge rectifier rather than in series to feed a full-wave center-
tapped arrangement. Why incur the heat loss of an extra diode pair
with a bridge, when a simple 2-diode fullwave will do? With a low-
voltage high-current supply a substantial amount of power is
dissipated by those diodes.
Another thing that might be considered is a simple switching
regulator for the output to allow one to adjust the voltage for
optimimum on-board linear regulator operation. High-current FETs are
cheap and the circuit can be driven by a wide variety of choices of
IC. I realize that it wouldn't be "vintage", however.
--Chuck
5:34 p.m.
I respectfully disagree. ? In a center-tapped full-wave rectifier circuit
the output voltage is half what you get if you use a bridge rectifier. It is
not the most efficient circuit in terms of transformer design, because each
half of the secondary is used only half the time. Thus the current through
the winding during that time is twice what it would be for a true full-wave
circuit. Heating in the windings, using Ohm?s law, is I^2R, so you have four
times the heating half the time, or twice the average heating of an
equivalent full-wave bridge rectifier circuit. You would have to choose a
transformer with a current rating 1.4 times as large as compared with the
( better ) bridge circuit; besides costing more, the resulting supply would
be bulkier and heavier ?. from the good book ? ? The art of Electronics ?
The disadvantage of a bridge is two diode voltage drops. Diodes are a lot
cheaper than transformers though.
Best regards, Steven
-------------------------------------------------------------------------
Hi
This is the result of when the current actually flows.
It only flows when the voltage is near the peak. This
changes the average power dissipated in the secondary winding
( the VA rating ).
Altough, the 1.414 is the worst value and the real value
is somewhere between 1 and 1.414.
It make no difference what type of rectification is used,
fullwave bridge or center tap. Watts are watts and there is
no way around it. It is true that the load current rating
is halved for the secondary turns for the center tapped
fullwave rectification but the VA rating of the transformer
remains the same.
I'm not sure what the book you reference says. I do
know how power is calculated and it is always the vectored
volts times amps. The current flows out of the transformer
when the output voltage is close to the peak. This means
that the rms current of the secondary needs to be compensated
to account for when the current actually flows.
If your transformer has a secondary rated at 10 amps that
can only feed a rectifier and capacitor filter of 7.07 amps.
That would change if feeding a inductive filter because of
when the current flows.
If it was a center tapped with two diodes, it would be
14.14 amps instead of 20 amps. Still, the VA rating of the
transformer does change.
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctech at classiccmp.org; cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Mon, 18 May 2009 23:36:52 -0700
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a
full
wave BRIDGE topology.
reference: The holy book " The art of electronics " page 47
Best regards, Steven
Hi
I forgot to mention. You need to derate the transformer
as well. The problem is that the current flows when the
voltage is the highest ( or close to it ). The math is a little
complicated but it means you need to increase the rating of
the transformer by at least 1.414.
If you can't find a transformer with the right windings,
I recommend looking at toroidal transformers. These can
easily have turns added to increase or buck the voltage
of the secondary.
Dwight
----------------------------------------
> From: lynchaj at yahoo.com
> To: cctalk at classiccmp.org
> Subject: S-100 power supply transformer
> Date: Sun, 17 May 2009 16:51:44 -0400
>
> Hi! Does anyone know of an available transformer suitable for making an
> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
> outputs.
>
> The best I solution I can find is two separate 80VA transformers; one
> parallel 10VAC output and another 36VCT output which can be configured
for
+/-18VAC
outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
20 May
20 May
9:52 a.m.
Hi
Wow, it just goes to show that you can't believe everything
you read. What a bunch of bull. While I'm sure most of the
book is quite useful, this part if flat wrong.
Ever wonder why every high power switcher I've ever seen
used a center taped transformer? I guess those engineers
never read this book.
Lets take things one at a time. First, it is correct that
the voltage is half but that is just a matter of doubling
the turns on each winding.
Now the next part. Each winding carries current for only
half the time( so far ok ).
Now for the point of deviation. Why would the current
be double? It takes two pulses of current during each
cycle to charge the capacitor. Does it make any difference
were that current came from? Even a more important question
is where did double the current go? Seems like there is
a rule for current that says what goes in must come out
( Nortons rule as I recall ).
For the bridge circuit, there are two pulse of current
per cycle. Each one of these pulse is split into the
two halves of the center tapped. Each pulse is exactly
the same current except one is on each leg of the center
tapped transformer ( again Norton rule demands it ).
This means that each half of the winding has half the power
of the single winding ( assuming that the windings are the
same resistance ).
The number he states as 1.4 is completely bogus.
Please do think it through. To have the same loss in the
secondary, the winding have to be exactly 2 times as much
wire but being the low voltage windings in our case, they are just
not that much bigger on the transformer.
All I can say is WOW, what a blupper!!!
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Tue, 19 May 2009 14:34:12 -0700
I respectfully disagree. ? In a center-tapped full-wave rectifier circuit
the output voltage is half what you get if you use a bridge rectifier. It is
not the most efficient circuit in terms of transformer design, because each
half of the secondary is used only half the time. Thus the current through
the winding during that time is twice what it would be for a true full-wave
circuit. Heating in the windings, using Ohm?s law, is I^2R, so you have four
times the heating half the time, or twice the average heating of an
equivalent full-wave bridge rectifier circuit. You would have to choose a
transformer with a current rating 1.4 times as large as compared with the
( better ) bridge circuit; besides costing more, the resulting supply would
be bulkier and heavier ?. from the good book ? ? The art of Electronics ?
The disadvantage of a bridge is two diode voltage drops. Diodes are a lot
cheaper than transformers though.
Best regards, Steven
-------------------------------------------------------------------------
Hi
This is the result of when the current actually flows.
It only flows when the voltage is near the peak. This
changes the average power dissipated in the secondary winding
( the VA rating ).
Altough, the 1.414 is the worst value and the real value
is somewhere between 1 and 1.414.
It make no difference what type of rectification is used,
fullwave bridge or center tap. Watts are watts and there is
no way around it. It is true that the load current rating
is halved for the secondary turns for the center tapped
fullwave rectification but the VA rating of the transformer
remains the same.
I'm not sure what the book you reference says. I do
know how power is calculated and it is always the vectored
volts times amps. The current flows out of the transformer
when the output voltage is close to the peak. This means
that the rms current of the secondary needs to be compensated
to account for when the current actually flows.
If your transformer has a secondary rated at 10 amps that
can only feed a rectifier and capacitor filter of 7.07 amps.
That would change if feeding a inductive filter because of
when the current flows.
If it was a center tapped with two diodes, it would be
14.14 amps instead of 20 amps. Still, the VA rating of the
transformer does change.
Dwight
----------------------------------------
_________________________________________________________________
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From: steven.alan.canning at verizon.net
To: cctech at classiccmp.org; cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Mon, 18 May 2009 23:36:52 -0700
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a
full
wave BRIDGE topology.
reference: The holy book " The art of electronics " page 47
Best regards, Steven
Hi
I forgot to mention. You need to derate the transformer
as well. The problem is that the current flows when the
voltage is the highest ( or close to it ). The math is a little
complicated but it means you need to increase the rating of
the transformer by at least 1.414.
If you can't find a transformer with the right windings,
I recommend looking at toroidal transformers. These can
easily have turns added to increase or buck the voltage
of the secondary.
Dwight
----------------------------------------
> From: lynchaj at yahoo.com
> To: cctalk at classiccmp.org
> Subject: S-100 power supply transformer
> Date: Sun, 17 May 2009 16:51:44 -0400
>
> Hi! Does anyone know of an available transformer suitable for making an
> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
> outputs.
>
> The best I solution I can find is two separate 80VA transformers; one
> parallel 10VAC output and another 36VCT output which can be configured
for
+/-18VAC
outputs.
---snip---
Hi
What you want is an 8 volt and a 13.5 volt AC. You forget
that the DC voltage is created from the peaks not the RMS
average. You have to account for diode drop as well.
That is how I got the numbers. I assumed full wave rectification
and .6v per diode.
Dwight
23 May
23 May
9:51 p.m.
Taken from Stancor Transformer Design Guidelines Page 7
The formula for the relation between secondary RMS current (IAC)
which the transformer has to deliver and the D.C. output current taken
from the rectifier (IDC) is: IAC = KFF x IDC where KFF
is the form factor.
The factor for each circuit and type of load is as follows:
Capacitor Load
Form Factor Rectifier Circuit
1.2 FWCT ( Full Wave Center Tap )
1.8 FWB ( Full Wave Bridge )
How can you say the rectifier topology doesn?t matter ?
Best regards, Steven
----- Original Message -----
From: "dwight elvey" <dkelvey at hotmail.com>
To: "General Discussion: On-Topic and Off-Topic Posts"
<cctalk at classiccmp.org>
Sent: Wednesday, May 20, 2009 6:52 AM
Subject: RE: S-100 power supply transformer
Hi
Wow, it just goes to show that you can't believe everything
you read. What a bunch of bull. While I'm sure most of the
book is quite useful, this part if flat wrong.
Ever wonder why every high power switcher I've ever seen
used a center taped transformer? I guess those engineers
never read this book.
Lets take things one at a time. First, it is correct that
the voltage is half but that is just a matter of doubling
the turns on each winding.
Now the next part. Each winding carries current for only
half the time( so far ok ).
Now for the point of deviation. Why would the current
be double? It takes two pulses of current during each
cycle to charge the capacitor. Does it make any difference
were that current came from? Even a more important question
is where did double the current go? Seems like there is
a rule for current that says what goes in must come out
( Nortons rule as I recall ).
For the bridge circuit, there are two pulse of current
per cycle. Each one of these pulse is split into the
two halves of the center tapped. Each pulse is exactly
the same current except one is on each leg of the center
tapped transformer ( again Norton rule demands it ).
This means that each half of the winding has half the power
of the single winding ( assuming that the windings are the
same resistance ).
The number he states as 1.4 is completely bogus.
Please do think it through. To have the same loss in the
secondary, the winding have to be exactly 2 times as much
wire but being the low voltage windings in our case, they are just
not that much bigger on the transformer.
All I can say is WOW, what a blupper!!!
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Tue, 19 May 2009 14:34:12 -0700
I respectfully disagree. ? In a center-tapped full-wave rectifier circuit
the output voltage is half what you get if you use a bridge rectifier. It
is
not the most efficient circuit in terms of transformer
design, because
each
half of the secondary is used only half the time. Thus
the current through
the winding during that time is twice what it would be for a true
full-wave
circuit. Heating in the windings, using Ohm?s law, is
I^2R, so you have
four
times the heating half the time, or twice the average
heating of an
equivalent full-wave bridge rectifier circuit. You would have to choose a
transformer with a current rating 1.4 times as large as compared with the
( better ) bridge circuit; besides costing more, the resulting supply
would
be bulkier and heavier ?. from the good book ? ? The
art of Electronics ?
The disadvantage of a bridge is two diode voltage drops. Diodes are a lot
cheaper than transformers though.
Best regards, Steven
-------------------------------------------------------------------------
Hi
This is the result of when the current actually flows.
It only flows when the voltage is near the peak. This
changes the average power dissipated in the secondary winding
( the VA rating ).
Altough, the 1.414 is the worst value and the real value
is somewhere between 1 and 1.414.
It make no difference what type of rectification is used,
fullwave bridge or center tap. Watts are watts and there is
no way around it. It is true that the load current rating
is halved for the secondary turns for the center tapped
fullwave rectification but the VA rating of the transformer
remains the same.
I'm not sure what the book you reference says. I do
know how power is calculated and it is always the vectored
volts times amps. The current flows out of the transformer
when the output voltage is close to the peak. This means
that the rms current of the secondary needs to be compensated
to account for when the current actually flows.
If your transformer has a secondary rated at 10 amps that
can only feed a rectifier and capacitor filter of 7.07 amps.
That would change if feeding a inductive filter because of
when the current flows.
If it was a center tapped with two diodes, it would be
14.14 amps instead of 20 amps. Still, the VA rating of the
transformer does change.
Dwight
----------------------------------------
From: steven.alan.canning at verizon.net
To: cctech at classiccmp.org; cctalk at classiccmp.org
Subject: Re: S-100 power supply transformer
Date: Mon, 18 May 2009 23:36:52 -0700
Dwight,
I think a little explanation is in order, otherwise your info borders on
incorrect. You need to increase the CURRENT rating of the transformer by
1.4, but this only applies to a full wave rectifier topology and not a
full
> wave BRIDGE topology.
>
> reference: The holy book " The art of electronics " page 47
>
> Best regards, Steven
>
>
>> ----------------------------------------
>>> From: lynchaj at yahoo.com
>>> To: cctalk at classiccmp.org
>>> Subject: S-100 power supply transformer
>>> Date: Sun, 17 May 2009 16:51:44 -0400
>>>
>>> Hi! Does anyone know of an available transformer suitable for making an
>>> S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
>>> outputs.
>>>
>>> The best I solution I can find is two separate 80VA transformers; one
>>> parallel 10VAC output and another 36VCT output which can be configured
> for
>>> +/-18VAC outputs.
>>>
>> ---snip---
>>
>> Hi
>> What you want is an 8 volt and a 13.5 volt AC. You forget
>> that the DC voltage is created from the peaks not the RMS
>> average. You have to account for diode drop as well.
>> That is how I got the numbers. I assumed full wave rectification
>> and .6v per diode.
>> Dwight
>>
>
> Hi
> I forgot to mention. You need to derate the transformer
> as well. The problem is that the current flows when the
> voltage is the highest ( or close to it ). The math is a little
> complicated but it means you need to increase the rating of
> the transformer by at least 1.414.
> If you can't find a transformer with the right windings,
> I recommend looking at toroidal transformers. These can
> easily have turns added to increase or buck the voltage
> of the secondary.
> Dwight 
20 May
20 May
1:49 p.m.
On 19/05/2009 22:34, Scanning wrote:
I respectfully disagree. ? In a center-tapped full-wave rectifier circuit
the output voltage is half what you get if you use a bridge rectifier. It is
not the most efficient circuit in terms of transformer design, because each
half of the secondary is used only half the time.
Well I agree so far, and I find the same text on page 47 of my copy. But...
Thus the current through
the winding during that time is twice what it would be for a true full-wave
circuit.
Why?
One each half-cycle, the transformer is delivering the required current
to the rectifier circuit. One one half-cycle, that's coming from only
one half of the winding, but it's still the same current. On the other
half-cycle, that half of the winding is producing no current, but the
other half is. That's the same current, just in the other half of the
winding. The argument seems to only make sense for a half-wave
rectifier circuit.
Unless they mean that, because you only get half the voltage from a CT
full-wave rectifier circuit compared to the voltage from a bridge
connected to the ends of the same transformer secondary, you need twice
the current to get the same power output. But that's not how it works,
nor how CT circuits are designed. If that's what's meant, it's
comparing apples with oranges. You don't compare a 9-0-9 volt
transformer which gives 9VDC using a CT full-wave circuit to an 18V
transformer driving a bridge rectifier; you compare it to a 9V
transformer driving a bridge.
Heating in the windings, using Ohm?s law, is I^2R, so
you have four
times the heating half the time, or twice the average heating of an
equivalent full-wave bridge rectifier circuit.
Assuming what I wrote above, I disagree. It's the same current so the
same I^2R, so the same heating, but for half the time in each part of
the winding. Half times two equals one.
Hence the following conclusion is incorrect.
You would have to choose a
transformer with a current rating 1.4 times as large as compared with the
( better ) bridge circuit; besides costing more, the resulting supply would
be bulkier and heavier ?. from the good book ? ? The art of Electronics ?
The transformer does end up bulkier, but only because each half of the
winding has to satisfy the voltage requirement (as opposed to a whole
winding supplying that voltage). In fact given that circumstance, you
only need 0.7 times the current rating.
The disadvantage of a bridge is two diode voltage
drops. Diodes are a lot
cheaper than transformers though.
Can't argue with that :-)
--
Pete Peter Turnbull
Network Manager
University of York
6:07 p.m.
A slightly different take on this transformer requirements issue..
Hammond used to (perhaps they still do) provide a piece of paper with every
transformer, listing part numbers and specs. Included on the paper was a table
for voltage and current factors for basic rectifier/filter circuits: (extract)
Rectifier Filter Voltage Current VA
------------------- ------ ------------------ ---------------- -----------------
Half-wave cap in DCV = ACV * 1.16 DCA = ACA * 0.38 DCVA = ACVA*0.44
Full-wave center tap cap in DCV = ACV * 1.25/2 DCA = ACA * 1 DCVA = ACVA*0.63
Full-wave bridge cap in DCV = ACV * 1.25 DCA = ACA * 0.56 DCVA = ACVA*0.7
AC values are RMS; VA figures I added in.
They don't say precisely how these figures were arrived at, but reality is a
little different than theory (e.g. 1.25 vs 1.414 for the voltage output from
the filter, don't know whether that's reduced due to R losses or an average
with ripple under load).
I take it these values were provided as guidelines, as complete design takes
into account C size, tolerable ripple, etc.
--
TI published "The Voltage Regulator Handbook" in the 70's, which covers a
lot
of design issues for linear power supplies. Even it references a design
procedure from 1943 (Schade graphical techniques) for the rectifier/filter design.
11:58 p.m.
----------------------------------------
Date: Wed, 20 May 2009 15:07:57 -0700
From: hilpert at cs.ubc.ca
To: General at invalid.domain
Subject: Re: S-100 power supply transformer
A slightly different take on this transformer requirements issue..
Hammond used to (perhaps they still do) provide a piece of paper with every
transformer, listing part numbers and specs. Included on the paper was a table
for voltage and current factors for basic rectifier/filter circuits: (extract)
Rectifier Filter Voltage Current VA
------------------- ------ ------------------ ---------------- -----------------
Half-wave cap in DCV = ACV * 1.16 DCA = ACA * 0.38 DCVA = ACVA*0.44
Full-wave center tap cap in DCV = ACV * 1.25/2 DCA = ACA * 1 DCVA = ACVA*0.63
Full-wave bridge cap in DCV = ACV * 1.25 DCA = ACA * 0.56 DCVA = ACVA*0.7
AC values are RMS; VA figures I added in.
They don't say precisely how these figures were arrived at, but reality is a
little different than theory (e.g. 1.25 vs 1.414 for the voltage output from
the filter, don't know whether that's reduced due to R losses or an average
with ripple under load).
I take it these values were provided as guidelines, as complete design takes
into account C size, tolerable ripple, etc.
--
TI published "The Voltage Regulator Handbook" in the 70's, which covers a
lot
of design issues for linear power supplies. Even it references a design
procedure from 1943 (Schade graphical techniques) for the rectifier/filter design.
What voltages were they talking about for these types
of numbers?
The 1.414 number is the best case. allowing some for the diode
and as you save, expected ripple.
Dwight
_________________________________________________________________
Hotmail? has ever-growing storage! Don?t worry about storage limits.
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21 May
21 May
2:30 a.m.
dwight elvey wrote:
----------------------------------------
They came in the box with Hammond's low-voltage transformers (3 to 100VAC):
165,166,167 series.
It may be that there is some specificity to Hammond's design characteristics. I
mention them as an example of real-world relationships, and that there is some
derating of the VA capability of the transformer when used in these circuits.
Date: Wed, 20 May 2009 15:07:57 -0700
From: hilpert at cs.ubc.ca
To: General at invalid.domain
Subject: Re: S-100 power supply transformer
A slightly different take on this transformer requirements issue..
Hammond used to (perhaps they still do) provide a piece of paper with every
transformer, listing part numbers and specs. Included on the paper was a table
for voltage and current factors for basic rectifier/filter circuits: (extract)
Rectifier Filter Voltage Current VA
------------------- ------ ------------------ ---------------- -----------------
Half-wave cap in DCV = ACV * 1.16 DCA = ACA * 0.38 DCVA = ACVA*0.44
Full-wave center tap cap in DCV = ACV * 1.25/2 DCA = ACA * 1 DCVA = ACVA*0.63
Full-wave bridge cap in DCV = ACV * 1.25 DCA = ACA * 0.56 DCVA = ACVA*0.7
AC values are RMS; VA figures I added in.
They don't say precisely how these figures were arrived at, but reality is a
little different than theory (e.g. 1.25 vs 1.414 for the voltage output from
the filter, don't know whether that's reduced due to R losses or an average
with ripple under load).
I take it these values were provided as guidelines, as complete design takes
into account C size, tolerable ripple, etc.
--
TI published "The Voltage Regulator Handbook" in the 70's, which covers a
lot
of design issues for linear power supplies. Even it references a design
procedure from 1943 (Schade graphical techniques) for the rectifier/filter design.
What voltages were they talking about for these types
of numbers?
The 1.414 number is the best case. allowing some for the diode
and as you save, expected ripple.
Dwight
9:50 a.m.
New subject: sdk 2920
Hi
I generally don't post about things on ebay but
this is a rare one.
Intel SDK-2920
It is #270390988255
As I recall the chip was Intel's first and
last attempt at a DSP chip. It was EPROM and
processor in one.
Dwight
_________________________________________________________________
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22 May
22 May
1:59 a.m.
New subject: sdk 2920
Thanks for the tip. The guy was delusional if he thought he could get $300
for that setup. It ( 2920 ) was one ( of many ) of Intel's big screw-ups. As
a DSP chip, all you could do was download your "stuff" into the EPROM and
wiggle the input with an analog input and observe the analog output. No way
to see inside and what was happening inside the chip. Made development and
troubleshooting a real bitch ( which probably speaks volumes for its success
in the market ).
----- Original Message -----
From: "dwight elvey" <dkelvey at hotmail.com>
To: "General Discussion: On-Topic and Off-Topic Posts"
<cctalk at classiccmp.org>
Sent: Thursday, May 21, 2009 6:50 AM
Subject: sdk 2920
Hi
I generally don't post about things on ebay but
this is a rare one.
Intel SDK-2920
It is #270390988255
As I recall the chip was Intel's first and
last attempt at a DSP chip. It was EPROM and
processor in one.
Dwight
2:46 a.m.
New subject: sdk 2920
Scanning wrote:
Thanks for the tip. The guy was delusional if he
thought he could get $300
for that setup. It ( 2920 ) was one ( of many ) of Intel's big screw-ups. As
a DSP chip, all you could do was download your "stuff" into the EPROM and
wiggle the input with an analog input and observe the analog output. No way
to see inside and what was happening inside the chip. Made development and
troubleshooting a real bitch ( which probably speaks volumes for its success
in the market ).
I disagree about that being the problem, as basically all
microcontrollers were that way at the time. In fact, many
microcontrollers didn't even *have* EPROM versions, so development for
those required an expensive in-circuit emulator. The 2920 gave you the
choice: do many iterations with the EPROM programmer and eraser, *or*
buy the ICE.
In my not so humble opinion, the reason that the 2920 didn't catch on
but the TI TMS32010 did is that the 2920 didn't have hardware multiply
and accumulate, so it had very low performance for common DSP tasks such
as digital filters and FFT.
Another problem Intel has had from 1980 to the present is that while
they want to play in multiple markets, it is almost always the case that
they make the most money per wafer by making mainstream processors.
That results in few wafers being devoted to other products, which is why
they seem to have such on-again-off-again support for embedded
processors. On the other hand, that worked to their advantage when DRAM
prices dropped through the floor in the mid-1980s, but the managers
didn't have the balls to drop the DRAM product line. Their normal
method of allocating wafer starts naturally resulted in few wafers for
DRAM, minimizing their losses on it until management had the sense to
drop it entirely.
Eric
17 May
17 May
9:32 p.m.
Hi,
You might contact:
http://www.antekinc.com/
for transformers.
As Chuck mentioned, a full-wave bridge feeding a cap will give 1.4x the
secondary output voltage.
You can parallel similar output windings in series or parallel as long
as you watch phase.
Consider that back in the 70s linear regs like LM309 and 7805 needed
fair overhead on the rail.
Newer LDOs by def need less.
If you want to scrounge, you might look in an older high current battery
charger with a "6V" tap.
At the risk of incurring purist wrath you might consider jumpering out
the regs on some boards and using one of the many almost free linear or
switching supplies to be found.
Some boards had multiple regs at same rail IIRC; might be a decoupling prob.
Haven't looked at S-100 since 1979, but believe you might have enough
pins and copper.
GL,
Steve
Andrew Lynch wrote:
Hi! Does anyone know of an available transformer
suitable for making an
S-100 power supply? I need 120VAC input and 10VAC output and +/- 18VAC
outputs.
The best I solution I can find is two separate 80VA transformers; one
parallel 10VAC output and another 36VCT output which can be configured for
+/-18VAC outputs.
I would prefer a single transformer rather than dual transformers because
they are the most expensive part of a linear power supply.
I know Northstar and Vector Graphic used similar dual output transformers
but I cannot find anything like them after searching the usual vendors like
Digikey, Mouser, Jameco, etc. Even transformer manufacturers don't seem to
carry a stock product of this configuration although you can order custom
units.
Thanks in advance for any help. Your ideas greatly appreciated.
Thanks and have a nice day!
Andrew Lynch
10:51 p.m.
On 17 May 2009 at 21:32, Steve Stutman wrote:
Consider that back in the 70s linear regs like LM309
and 7805 needed
fair overhead on the rail.
Newer LDOs by def need less.
While I can understand that using a 78xx/79xx vintage regulator might
be important for the sake of accuracy, I'd be sort tempted to use
modern monolithic switching regulators if one wanted to preserve the
S100 "on board" regulation.
No more "scorched PCB" or big honking heatsinks--and less conversion
of amps into heat.
--Chuck
11:02 p.m.
Chuck Guzis wrote:
On 17 May 2009 at 21:32, Steve Stutman wrote:
The problem with the onboard regulation is that many of those original
regulators, though they might be working, are actually no longer any
good. Not sure if it's just age, or it's a lifetime of certain ones
running too near max, but in all the ones I've replaced, I've cut the
amperage draw of the board way down, sometimes in half.
If they're going to be replaced *anyway*, then makes sense just to
remove them and their associated, equally suspect tantalums/aecs
altogether. In my IMSAI, I feed each board power from a dedicated,
pluggable line running to it... ground is still done through the connector.
jS
Consider that back in the 70s linear regs like
LM309 and 7805 needed
fair overhead on the rail.
Newer LDOs by def need less.
While I can understand that using a 78xx/79xx vintage regulator might
be important for the sake of accuracy, I'd be sort tempted to use
modern monolithic switching regulators if one wanted to preserve the
S100 "on board" regulation.
No more "scorched PCB" or big honking heatsinks--and less conversion
of amps into heat.
--Chuck 
11:21 p.m.
On Sun, 17 May 2009, js at cimmeri.com wrote:
In my IMSAI, I feed each board power from a dedicated,
pluggable line
running to it... ground is still done through the connector.
What did you do with the power feeds on the backplane? Got any pics?
--
David Griffith
dgriffi at cs.csubak.edu
A: Because it fouls the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing in e-mail?
18 May
18 May
11:13 a.m.
In my IMSAI, I
feed each board power from a dedicated, pluggable line
running to it... ground is still done through the connector.
David Griffith wrote: What did you do with the power feeds on the
backplane? Got any pics? Alexandre Souza wrote: The problem is compatibility.
Imagine that
each S-100 board you would connect on your system has to be modified.
And every board that will be sold, lended or given away, will ALSO
have to be modified.
Why not just change the regulator for a newer one? They are so
cheap nowadays :o)
I get around the compatibility issue by leaving the original
backplane
power in place. Btw, this regulator failure problem is only showing up
on the 5V line. I've not had any issue with the 12V regulators. So,
I run a separate unregulated 8V wire from the primary PSU over to my
regulator board. This board has the new 5V regulated rail. Wires
from cards that have been converted plug in here. I
don't have a lot
of cards, so conversion is not a big deal... but I find
converting the
card easier than replacing its regulator + associated caps.
You can just replace the regulator... but the job is not just the
regulator... it's also all the supporting caps. If the design
overstressed the regulator to begin with, then you are back where you
started with a high heat situation.
jS
17 May
17 May
11:59 p.m.
On 17 May 2009 at 22:02, js at cimmeri.com wrote:
The problem with the onboard regulation is that many
of those original
regulators, though they might be working, are actually no longer any
good. Not sure if it's just age, or it's a lifetime of certain ones
running too near max, but in all the ones I've replaced, I've cut the
amperage draw of the board way down, sometimes in half.
I suspect that it's a combination of high-power use and the quality
of the TO-220 packaging back then. I recall that on some MITS
boards, (was it the 4K DRAM board?) bridging the regulator with a
power resistor was done to keep the temperature down to that of a
slow bonfire.
The early 2102-based SRAM boards were pretty power hungry too.
Cheers,
Chuck
18 May
18 May
3:20 a.m.
I suspect that it's a combination of high-power
use and the quality
of the TO-220 packaging back then. I recall that on some MITS
boards, (was it the 4K DRAM board?) bridging the regulator with a
power resistor was done to keep the temperature down to that of a
slow bonfire.
Maybe old boards can be retrofitted with low-power chips? :o)
6:01 a.m.
New subject: Problem with an old Tandy CM-4 monitor
A while ago I got a Tandy 1000 system with a CM-4 monitor. The system worked
fine for 20 minutes and then the monitor died. So far I havn't seen anything
that looks burnt on the inside, but the tube doesn't light up when you
provide power and there is a crackling noise going on. Is there a common
issue with these old CGA monitors, and is this fixable without gutting a
working one?
6:24 a.m.
New subject: Problem with an old Tandy CM-4 monitor
Teo,
First check the solder joints for bad soldering like very good at all power
consuming components because due to temp./length variations the solders are
getting lose.
If the CRT isn't glowing you should also check the powersupply and the
hor.line-circuit.
Most time the switching transistor (you'll measure a shortcut between the
legs after you desoldered it) or a capacitor(elco) dies.
And please be carefull not all old monitors have a discharge resistor placed
over the high-voltage elco's.
-Rik
-----Oorspronkelijk bericht-----
Van: cctalk-bounces at classiccmp.org
[mailto:cctalk-bounces at classiccmp.org] Namens Teo Zenios
Verzonden: maandag 18 mei 2009 12:02
Aan: General Discussion: On-Topic and Off-Topic Posts
Onderwerp: Problem with an old Tandy CM-4 monitor
A while ago I got a Tandy 1000 system with a CM-4 monitor.
The system worked fine for 20 minutes and then the monitor
died. So far I havn't seen anything that looks burnt on the
inside, but the tube doesn't light up when you provide power
and there is a crackling noise going on. Is there a common
issue with these old CGA monitors, and is this fixable
without gutting a working one?
1:47 p.m.
New subject: Problem with an old Tandy CM-4 monitor
A while ago I got a Tandy 1000 system with a CM-4 monitor. The system worked
fine for 20 minutes and then the monitor died. So far I havn't seen anything
that looks burnt on the inside, but the tube doesn't light up when you
provide power and there is a crackling noise going on. Is there a common
When you say 'The tube doesn't light up' do you mean that the screen is
dark, or that there's (also) no glow from the heater in the electron gun
at the back of the CRT?
And is this crackling noise there all the time the monitor is turned on?
issue with these old CGA monitors, and is this fixable
without gutting a
working one?
The most common problem that's _not_ fixable without a 'spares unit' is a
breakdown of the horizontal output transformer (line output transformer
over here :-)). This often does result in a continuous crackling (small
sparks from the EHT supply to chassis, etc). It'll result in a totally
dark screen, it may also remove the heater supply.
But it could be a lot of other things. Many faults result in a totally
dark screen. 'Crackling noises' may come from ther sections of the
monitor nosies are remarkable hard to describe.
What I would do is pull the case and check the CRT electrode voltages
(watch out for the focus pin which may have around 5kV on it!). Check the
EHT at the anode connector if you have a suitable meter (expect around
25kV here on a colour CRT). And work back from there.
-tony
3:12 a.m.
The problem with the onboard regulation is that many
of those original
regulators, though they might be working, are actually no longer any
good. Not sure if it's just age, or it's a lifetime of certain ones
running too near max, but in all the ones I've replaced, I've cut the
amperage draw of the board way down, sometimes in half.
If they're going to be replaced *anyway*, then makes sense just to
remove them and their associated, equally suspect tantalums/aecs
altogether. In my IMSAI, I feed each board power from a dedicated,
pluggable line running to it... ground is still done through the
connector.
The problem is compatibility. Imagine that each S-100 board you would
connect on your system has to be modified. And every board that will be
sold, lended or given away, will ALSO have to be modified.
Why not just change the regulator for a newer one? They are so cheap
nowadays :o)
7:19 a.m.
You might also consider a 7.5V and a couple 15V switching supplies
adjusted up to +8 and +16. This is what Grant Stockley does in his
Altair 680b replicas. The 7.5V units are little more rare but you start
to see them at 100W and higher even at places like Marlin P Jones.
eg,
http://www.mpja.com/prodinfo.asp?number=16012+PS
--
Chris Elmquist
5519
days inactive
5526
days old
28 comments
15 participants
participants (15)
-
alexandre-listas＠e-secure.com.br -
ard＠p850ug1.demon.co.uk -
cclist＠sydex.com -
chrise＠pobox.com -
dgriffi＠cs.csubak.edu -
dkelvey＠hotmail.com -
eric＠brouhaha.com -
hilpert＠cs.ubc.ca -
hp-fix＠xs4all.nl -
js＠cimmeri.com -
lynchaj＠yahoo.com -
pete＠dunnington.plus.com -
steve＠radiorobots.com -
steven.alan.canning＠verizon.net -
teoz＠neo.rr.com