7 Apr
2016
7 Apr
'16
1:41 p.m.
You should be just fine.
On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
If you have a circuit which is normally designed to
operate with an unregulated supply, through a regulator...
say unregulated +8 through a 7805 to a regulated +5 and
you want to test it independent of the +8 supply, if
you leave the unregulated rail unattached and put +5
switcher straight onto the regulated +5 rail, will you
damage the 7805? Clearly the VIN is open, but the ground
pin will still be attached. Would this push voltage
back through and screw things up?
Thanks,
Bill S.
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7 Apr
7 Apr
1:58 p.m.
Err.. unless the voltage of the switcher is identical to that of the 7805,
then one device will source current, and the other will sink it.
Like putting two 6V batteries in parallel, where one is fresh and the other
weak. Current will flow until the potentials are equalized. But with two
regulated circuits, I don't see how equality can be achieved.
Not saying it's going to smoke-out, but it does seem like a wonky thing to
do.
On Thu, Apr 7, 2016 at 3:41 PM, wulfman <wulfman at wulfman.com> wrote:
You should be just fine.
On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
If you have a circuit which is normally designed
to
operate with an unregulated supply, through a regulator...
say unregulated +8 through a 7805 to a regulated +5 and
you want to test it independent of the +8 supply, if
you leave the unregulated rail unattached and put +5
switcher straight onto the regulated +5 rail, will you
damage the 7805? Clearly the VIN is open, but the ground
pin will still be attached. Would this push voltage
back through and screw things up?
Thanks,
Bill S.
--
The contents of this e-mail and any attachments are intended solely for
the use of the named
addressee(s) and may contain confidential and/or privileged information.
Any unauthorized use,
copying, disclosure, or distribution of the contents of this e-mail is
strictly prohibited by
the sender and may be unlawful. If you are not the intended recipient,
please notify the sender
immediately and delete this e-mail.
2:03 p.m.
Per his description, the 7805's input will be open. It will not try to
source any current, as it will have none to give.
I suppose there might be a little leakage.
--
Will
On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com> wrote:
Err.. unless the voltage of the switcher is identical
to that of the 7805,
then one device will source current, and the other will sink it.
Like putting two 6V batteries in parallel, where one is fresh and the other
weak. Current will flow until the potentials are equalized. But with two
regulated circuits, I don't see how equality can be achieved.
Not saying it's going to smoke-out, but it does seem like a wonky thing to
do.
On Thu, Apr 7, 2016 at 3:41 PM, wulfman <wulfman at wulfman.com> wrote:
> You should be just fine.
>
> On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
> > If you have a circuit which is normally designed to
> > operate with an unregulated supply, through a regulator...
> > say unregulated +8 through a 7805 to a regulated +5 and
> > you want to test it independent of the +8 supply, if
> > you leave the unregulated rail unattached and put +5
> > switcher straight onto the regulated +5 rail, will you
> > damage the 7805? Clearly the VIN is open, but the ground
> > pin will still be attached. Would this push voltage
> > back through and screw things up?
> >
> > Thanks,
> > Bill S.
> >
> >
> >
>
>
> --
> The contents of this e-mail and any attachments are intended solely for
> the use of the named
> addressee(s) and may contain confidential and/or privileged information.
> Any unauthorized use,
> copying, disclosure, or distribution of the contents of this e-mail is
> strictly prohibited by
> the sender and may be unlawful. If you are not the intended recipient,
> please notify the sender
> immediately and delete this e-mail.
>
>
2:15 p.m.
On 04/07/2016 2:03 PM, William Donzelli wrote:
Per his description, the 7805's input will be
open. It will not try to
source any current, as it will have none to give.
I suppose there might be a little leakage.
--
Will
If his intention is to bypass the 7805 then it should have both input
and output legs cut and lifted.
John :-#)#
On Thu, Apr 7, 2016 at 4:58 PM, drlegendre .
<drlegendre at gmail.com> wrote:
> Err.. unless the voltage of the switcher is identical to that of the 7805,
> then one device will source current, and the other will sink it.
>
> Like putting two 6V batteries in parallel, where one is fresh and the other
> weak. Current will flow until the potentials are equalized. But with two
> regulated circuits, I don't see how equality can be achieved.
>
> Not saying it's going to smoke-out, but it does seem like a wonky thing to
> do.
>
> On Thu, Apr 7, 2016 at 3:41 PM, wulfman <wulfman at wulfman.com> wrote:
>
>> You should be just fine.
>>
>> On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
>>> If you have a circuit which is normally designed to
>>> operate with an unregulated supply, through a regulator...
>>> say unregulated +8 through a 7805 to a regulated +5 and
>>> you want to test it independent of the +8 supply, if
>>> you leave the unregulated rail unattached and put +5
>>> switcher straight onto the regulated +5 rail, will you
>>> damage the 7805? Clearly the VIN is open, but the ground
>>> pin will still be attached. Would this push voltage
>>> back through and screw things up?
>>>
>>> Thanks,
>>> Bill S.
>>>
>>>
>>>
>>
>> --
>> The contents of this e-mail and any attachments are intended solely for
>> the use of the named
>> addressee(s) and may contain confidential and/or privileged information.
>> Any unauthorized use,
>> copying, disclosure, or distribution of the contents of this e-mail is
>> strictly prohibited by
>> the sender and may be unlawful. If you are not the intended recipient,
>> please notify the sender
>> immediately and delete this e-mail.
>>
>>
--
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
Call (604)872-5757 or Fax 872-2010 (Pinballs, Jukes, VideoGames)
www.flippers.com
"Old pinballers never die, they just flip out"
2:23 p.m.
Maybe I'm overthinking this. If I just put
regulated +5 on the 7805 VIN will it work?
Bill S.
6:08 p.m.
"...if you leave the unregulated rail _unattached_ and put +5
switcher straight onto the regulated +5 rail..."
My error, I read that as "attached".
In any event, just lift both the 7805 IN and OUT pins, and then supply
known-solid +5DC between the OUT and GND pads on the board.
No, you can't feed the IN pin with +5V, for as others have mentioned,
the 7805 has a minimum dropout of 2V or so.
On Thu, Apr 7, 2016 at 4:03 PM, William Donzelli <wdonzelli at gmail.com>
wrote:
Per his description, the 7805's input will be
open. It will not try to
source any current, as it will have none to give.
I suppose there might be a little leakage.
--
Will
On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com> wrote:
Err.. unless the voltage of the switcher is
identical to that of the
7805,
then one device will source current, and the
other will sink it.
Like putting two 6V batteries in parallel, where one is fresh and the
other
weak. Current will flow until the potentials are
equalized. But with two
regulated circuits, I don't see how equality can be achieved.
Not saying it's going to smoke-out, but it does seem like a wonky thing
to
do.
On Thu, Apr 7, 2016 at 3:41 PM, wulfman <wulfman at wulfman.com> wrote:
> You should be just fine.
>
> On 4/7/2016 1:38 PM, Bill Sudbrink wrote:
> > If you have a circuit which is normally designed to
> > operate with an unregulated supply, through a regulator...
> > say unregulated +8 through a 7805 to a regulated +5 and
> > you want to test it independent of the +8 supply, if
> > you leave the unregulated rail unattached and put +5
> > switcher straight onto the regulated +5 rail, will you
> > damage the 7805? Clearly the VIN is open, but the ground
> > pin will still be attached. Would this push voltage
> > back through and screw things up?
> >
> > Thanks,
> > Bill S.
> >
> >
> >
>
>
> --
> The contents of this e-mail and any attachments are intended solely for
> the use of the named
> addressee(s) and may contain confidential and/or privileged information.
> Any unauthorized use,
> copying, disclosure, or distribution of the contents of this e-mail is
> strictly prohibited by
> the sender and may be unlawful. If you are not the intended recipient,
> please notify the sender
> immediately and delete this e-mail.
>
>
8 Apr
8 Apr
8:54 a.m.
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
http://www.ti.com/lit/ds/symlink/lm78l05.pdf ( page 8 gives you the
internals for a low power version )
Nothing in the data sheet saying you cant apply voltage to the output.
As per the low power version circuit diagram i cant see how you can do
any damage to it
I was not able to find an internal diagram for the higher power version but
i am sure its close to the same circuit just larger pass transistors.
I took a new regulator and measured the output pin to ground with my
trusty ohmmeter
on the diode setting and had no reading. If you think about it the
output on the regulator is a pass transistor
and the output to ground will be in effect a diode that will not allow
current to pass from output to ground.
My last comment still stands. i doubt you will cause any ill effects to
your regulator.
I personally have done this to some old arcade boards with no ill effects.
On 4/7/2016 6:08 PM, drlegendre . wrote:
"...if you leave the unregulated rail
_unattached_ and put +5
switcher straight onto the regulated +5 rail..."
My error, I read that as "attached".
In any event, just lift both the 7805 IN and OUT pins, and then supply
known-solid +5DC between the OUT and GND pads on the board.
No, you can't feed the IN pin with +5V, for as others have mentioned,
the 7805 has a minimum dropout of 2V or so.
On Thu, Apr 7, 2016 at 4:03 PM, William Donzelli <wdonzelli at gmail.com>
wrote:
> Per his description, the 7805's input will be open. It will not try to
> source any current, as it will have none to give.
>
> I suppose there might be a little leakage.
>
> --
> Will
>
> On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com> wrote:
>
--
The contents of this e-mail and any attachments are intended solely for the use of the
named
addressee(s) and may contain confidential and/or privileged information. Any unauthorized
use,
copying, disclosure, or distribution of the contents of this e-mail is strictly prohibited
by
the sender and may be unlawful. If you are not the intended recipient, please notify the
sender
immediately and delete this e-mail.
10:31 a.m.
On 04/08/2016 8:54 AM, wulfman wrote:
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
http://www.ti.com/lit/ds/symlink/lm78l05.pdf ( page 8 gives you the
internals for a low power version )
Nothing in the data sheet saying you cant apply voltage to the output.
As per the low power version circuit diagram i cant see how you can do
any damage to it
I was not able to find an internal diagram for the higher power version but
i am sure its close to the same circuit just larger pass transistors.
I took a new regulator and measured the output pin to ground with my
trusty ohmmeter
on the diode setting and had no reading. If you think about it the
output on the regulator is a pass transistor
and the output to ground will be in effect a diode that will not allow
current to pass from output to ground.
My last comment still stands. i doubt you will cause any ill effects to
your regulator.
I personally have done this to some old arcade boards with no ill effects.
On 4/7/2016 6:08 PM, drlegendre . wrote:
> "...if you leave the unregulated rail _unattached_ and put +5
> switcher straight onto the regulated +5 rail..."
>
> My error, I read that as "attached".
>
> In any event, just lift both the 7805 IN and OUT pins, and then supply
> known-solid +5DC between the OUT and GND pads on the board.
>
> No, you can't feed the IN pin with +5V, for as others have mentioned,
> the 7805 has a minimum dropout of 2V or so.
>
> On Thu, Apr 7, 2016 at 4:03 PM, William Donzelli <wdonzelli at gmail.com>
> wrote:
>
>> Per his description, the 7805's input will be open. It will not try to
>> source any current, as it will have none to give.
>>
>> I suppose there might be a little leakage.
>>
>> --
>> Will
>>
>> On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com>
wrote:
>>
I archive all data sheets I run across and I found this information on
both TI and Fairchild data sheets for the 78H05, LM340 and LM78XX
devices - it is usually under a heading called Application Hints and for
some reason Fairchild have removed these Hints from the sheets you quote.
In ALL cases the manufacturers caution that putting voltage to the
output when the input is not powered CAN DAMAGE THE CHIP.
The sheets I have appear to be a bit hard to find so here are my
archived copies:
http://www.flippers.com/pdfs/LM340_LM78XX_National_2003.pdf (page 11)
http://www.flippers.com/pdfs/LM78H05A_Fairchild.pdf (page 4)
So, yes, you can get away with it for a short time, but you are
stressing the regulator outside its design limits and that will shorten
its life.
John :-#(#
--
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
Call (604)872-5757 or Fax 872-2010 (Pinballs, Jukes, VideoGames)
www.flippers.com
"Old pinballers never die, they just flip out"
11:16 a.m.
Keep in mind that "not powered" and "open" are very different things.
What do the data sheets say?
I will admit that this is all careening towards the academic at this point...
--
Will
On Fri, Apr 8, 2016 at 1:31 PM, John Robertson <jrr at flippers.com> wrote:
On 04/08/2016 8:54 AM, wulfman wrote:
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
http://www.ti.com/lit/ds/symlink/lm78l05.pdf ( page 8 gives you the
internals for a low power version )
Nothing in the data sheet saying you cant apply voltage to the output.
As per the low power version circuit diagram i cant see how you can do
any damage to it
I was not able to find an internal diagram for the higher power version
but
i am sure its close to the same circuit just larger pass transistors.
I took a new regulator and measured the output pin to ground with my
trusty ohmmeter
on the diode setting and had no reading. If you think about it the
output on the regulator is a pass transistor
and the output to ground will be in effect a diode that will not allow
current to pass from output to ground.
My last comment still stands. i doubt you will cause any ill effects to
your regulator.
I personally have done this to some old arcade boards with no ill effects.
On 4/7/2016 6:08 PM, drlegendre . wrote:
>
> "...if you leave the unregulated rail _unattached_ and put +5
> switcher straight onto the regulated +5 rail..."
>
> My error, I read that as "attached".
>
> In any event, just lift both the 7805 IN and OUT pins, and then supply
> known-solid +5DC between the OUT and GND pads on the board.
>
> No, you can't feed the IN pin with +5V, for as others have mentioned,
> the 7805 has a minimum dropout of 2V or so.
>
> On Thu, Apr 7, 2016 at 4:03 PM, William Donzelli <wdonzelli at gmail.com>
> wrote:
>
>> Per his description, the 7805's input will be open. It will not try to
>> source any current, as it will have none to give.
>>
>> I suppose there might be a little leakage.
>>
>> --
>> Will
>>
>> On Thu, Apr 7, 2016 at 4:58 PM, drlegendre . <drlegendre at gmail.com>
>> wrote:
>>
I archive all data sheets I run across and I found this information on
both
TI and Fairchild data sheets for the 78H05, LM340 and LM78XX devices - it is
usually under a heading called Application Hints and for some reason
Fairchild have removed these Hints from the sheets you quote.
In ALL cases the manufacturers caution that putting voltage to the output
when the input is not powered CAN DAMAGE THE CHIP.
The sheets I have appear to be a bit hard to find so here are my archived
copies:
http://www.flippers.com/pdfs/LM340_LM78XX_National_2003.pdf (page 11)
http://www.flippers.com/pdfs/LM78H05A_Fairchild.pdf (page 4)
So, yes, you can get away with it for a short time, but you are stressing
the regulator outside its design limits and that will shorten its life.
John :-#(#
--
John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
Call (604)872-5757 or Fax 872-2010 (Pinballs, Jukes, VideoGames)
www.flippers.com
"Old pinballers never die, they just flip out"
11:18 a.m.
First, a few quick "whys":
1) The 7805 is actually a Motorola MC7805CP, date
code 7308 with gold leads. Very hard to exactly
replace.
2) As per the design of the unit, the 7805 is soldered
to the foil side of a single sided, copper clad
circuit board with no through hole plating. Lifting
the traces seems like a significant possibility that
I don't want to risk.
3) I'm still waiting for the components to build the
8 volt power supply. Hopefully they will arrive
Monday.
4) No, one thing missing from my bench, that I've
never really needed up to now, is a good adjustable
DC supply. My have to ask for one for Christmas.
Three thoughts have come to me since my original post.
1) I have a 12 volt DC supply. 12 volts seems to be
within the VIN range for the 7805s whose data
sheets I've now read. Can I simply apply 12 volts?
2) Could I place a resistor in series between the 12V
supply and the 7805 to drop the voltage at the 7805
to somewhere around 8?
3) If I was to "tack on" a jumper between VIN and VOUT,
would that protect the 7805 and allow me to power the
circuit with 5 volts?
Thank You,
Bill S.
11:27 a.m.
I archive all data sheets I run across and I found
this information
on both TI and Fairchild data sheets for the 78H05, LM340 and LM78XX
devices - it is usually under a heading called Application Hints and
for some reason Fairchild have removed these Hints from the sheets
you quote.
In ALL cases the manufacturers caution that putting
voltage to the
output when the input is not powered CAN DAMAGE THE CHIP.
I have the datasheet PDF for the On Semiconductor version, and I have
been unable to find anything of the sort in it.
Whether On was sloppy when writing their documetnation or whether some
manufacturer's devices are safe or whether I just missed something, I
don't know. But, when dealing with a device of unknown provenance, it
does seem clear you're better off not powering the output side if the
input side has a low-impedance path to ground. (Powering the output
with the input disconnected appears to be a completely different thing;
I have trouble imagining a design that would be damaged by that.)
/~\ The ASCII Mouse
\ / Ribbon Campaign
X Against HTML mouse at rodents-montreal.org
/ \ Email! 7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B
11:58 a.m.
I don't know why I bothering to be coy about it...
My unit here:
http://wsudbrink.dyndns.org:8080/images/cyclops-latest/P4060005.JPG
http://wsudbrink.dyndns.org:8080/images/cyclops-latest/P4060006.JPG
Actually, I'm further along than that, but I
don't have more recent photos.
Original unit here:
http://www.swtpc.com/mholley/PopularElectronics/Feb1975/PE_Feb_1975_pg30.jpg
Schematic here:
http://www.swtpc.com/mholley/PopularElectronics/Feb1975/PE_Feb_1975_pg28.jpg
Bill
3:11 p.m.
On 2016-Apr-08, at 11:27 AM, Mouse wrote:
http://www.ti.com/lit/ds/symlink/lm340-n.pdf
page 11
I suspect it would be OK if VIN and VOUT were shorted during the procedure (and VIN was
otherwise open or has just a small cap to gnd) as that likely takes care of reverse
current and reverse bias situations being set up on internal junctions, but one
doesn't know for sure unless you examine the internal schematic adequately or test the
device.
I archive all
data sheets I run across and I found this information
on both TI and Fairchild data sheets for the 78H05, LM340 and LM78XX
devices - it is usually under a heading called Application Hints and
for some reason Fairchild have removed these Hints from the sheets
you quote.
In ALL cases the manufacturers caution that
putting voltage to the
output when the input is not powered CAN DAMAGE THE CHIP.
I have the datasheet PDF for the On Semiconductor version, and I have
been unable to find anything of the sort in it.
7 Apr
7 Apr
2:18 p.m.
drlegendre wrote:
Not saying it's going to smoke-out, but it does
seem like a wonky thing to do.
I disagree about "wonky" let me try with more
diagram and less English:
(+8)
|
VIN| ceramic cap
|-----][-----
___|____ |
| 7805 |----------GROUND
-------- |
| |
VOUT|(+5) |
| |
___|____ |
| LOAD |-------|
--------
+8 is not currently available (no pun intended).
I would like to test LOAD without removing 7805
as it is soldered in place. Is damage to 7805
likely if alternative regulated current is applied
at (+5) and (+8) is left open?
Bill S.
3180
days inactive
3181
days old
14 comments
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cclist＠sydex.com -
drlegendre＠gmail.com -
hilpert＠cs.ubc.ca -
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wdonzelli＠gmail.com -
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