4 Dec
2002
4 Dec
'02
4:43 p.m.
Hi all,
I'm currently in a very good mood because I've found a Jupiter Ace in
near-mint condition, complete with 16k RAMPack, intro tape and a copy of the
book "Jupiter Ace FORTH Programming" by Steven Vickers. All this, including
the polystyrene box for the rampack (and rampack instructions) cost me the
rather feeble sum of two British pounds.
Now, there is a catch. The machine didn't come with a power supply. So,
can someone with an Ace, working PSU and a multimeter please tell me what
the polarity and voltage of the PSU's output is? I'm hesitant to power up
the Ace in case I end up fouling up the settings, with fairly predictable
results.
Thanks.
--
Phil.
philpem(a)dsl.pipex.com
http://www.philpem.dsl.pipex.com/
4 Dec
4 Dec
7:35 p.m.
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5 Dec
5 Dec
2:01 a.m.
Tony Duell wrote:
Well, it certainly seems that way - from the incomplete and poor-quality
schematics I've got, it uses a 7805 1A linear regulator which needs around
8V to 15V to run properly. The usual supply voltage is 9V. IIRC...
In other words, it's a ZX80 workalike
that runs FORTH instead of BASIC...
Now, there
is a catch. The machine didn't come with a power
supply. So, can someone with an Ace, working PSU and a multimeter
please tell me what the polarity and voltage of the PSU's output is?
I'm hesitant to power up
I am darn sure it's 9V DC (probably about 1A) and
tip +ve. But you can check it for yourself. Open up the Ace
(the easiest way
to do this is to push the centre pins of the 5 fixing studs in, so
that they fall out inside the cae. Remove the studs and the case top,
then recover the pins from inside the machine. When reassembing, put
the top on, fit the studs, and push the pins in from the outside).
Well, that was
easy enough, except my Ace only had three fixing studs. I
guess the studs made it a little quicker to assemble an Ace in the factory,
but they're a real pain to remove when the case needs opening...
The PSU is a conventional 7805-based design (so the 9V
1A input
should be fine). One side of the input jack socket (I think it's the
sleeve) goes to the 0V line.
That would seem logical - ring ground, tip power. Same
as when a jack plug
is used as an audio connector.
the Ace in
case I end up fouling up the settings, with fairly
predictable results.
There is nothing totally irreplaceable in the Ace. A ROM dump
should
be available if you need to re-burn the EPROMs. And the other chips
are standard -- there's no ULA... In any case, most of the time when you get the
polarity wrong, the
7805 dies, but the rest of the machine survives. Don't try this for
the fun of it, though ;-)
Strange, I've never killed a 7805 by reversing the
polarity. Mainly because
I usually power them up on current-limited power supplies...
I don't suppose you have a copy of the Jupiter Ace schematics, do you?
Thanks.
--
Phil.
philpem(a)dsl.pipex.com
http://www.philpem.dsl.pipex.com/
6:21 p.m.
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6 Dec
6 Dec
2:43 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be
under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
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2:53 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
You could just put a silicon diode that is forward biased in series with the
6 volts. It will drop close to 0.7 volts. Make sure it has the current
handling capacity that you need. In other words, do not use a signal diode.
At 0.5 amps the diode would dissipate 0.35 watts. Regards - Mike
Mike B. Feher, N4FS
89 Arnold Blvd.
Howell NJ, 07731
(732) 901-9193
----- Original Message -----
From: "Ethan Dicks" <erd_6502(a)yahoo.com>
To: <cctalk(a)classiccmp.org>
Sent: Friday, December 06, 2002 3:44 PM
Subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good
find!!!)
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
3:19 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
From: "Ethan Dicks"
<erd_6502(a)yahoo.com>
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
The 7805 is an _old_ design. A modern LDO (Low Drop Out) regulator like an
LM2940 can maintain 5v out down to around 5.5v in.
Of course with many types of batteries you have the problem of voltage sag
as they discharge, and then you would need some kind of buck-boost (DC-DC)
converter to maintain 5v.
4:03 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
You can use a switcher chip, but that's little bit of a pain to build by
hand, since the majority are surface mount, and you need a good ground plane
(not really hard, just not as easy as a 7805 and a bridge rectifier).
Basically, the problem is that a diode drop is .6 volts, so you need a
minimum of 5.6 volts of input, and that's with a low drop-out regulator.
That doesn't leave much room for your battery to die. A normal 7805 really
wants 7V minimum to regulate reliably.
And depending on the current load, 15V on a 7805 is a Bad Idea, because
it's got to dump that 10V drop as heat. That's 10 watts at 1 amp, which is
a damn lotta heat in a TO-220 case. Ideally, a 7805 should be run at 7V or
8V to keep heat to a minimum.
If you do want to build a switcher, just about everybody has a switcher
chip that will work about .2V to .4V above the regulated output voltage.
Maxim is the first place I'd look. Get some double sided copper clad board
for a ground place, use an Xacto or a Dremel tool to route traces. Some of
the companies also offer free or very low cost eval boards. For instance,
National Semi has a switcher eval that takes 4V to 30V and produces 3.3V out
at 1 amp. They're $8. It has two caps on it that cost me $2 each, plus
having boards turned, the switcher itself, and a few other incidental parts.
It's cheaper for me to buy them than it is produce my own (I use them a lot
in prototyping projects).
When trying to get 5V from 6V, especially on batteries, linear regulators
aren't really the way to go. You'll dump a lot of the battery's energy as
heat, which is wasteful. Same for a Zener type regulator. You could use a
a couple diodes in series, but your output voltage will sink as your battery
voltage drops, since the drop across the diode is the same.
--John
-----Original Message-----
From: cctalk-admin(a)classiccmp.org [mailto:cctalk-admin@classiccmp.org]On
Behalf Of Ethan Dicks
Sent: Friday, December 06, 2002 15:45
To: cctalk(a)classiccmp.org
Subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first
good find!!!)
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be
under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
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4:20 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
There are low-dropout regulators that can do this, but
they are more expensive.
Another thing I've done in thes3
--- Ethan Dicks <erd_6502(a)yahoo.com> wrote:
__________________________________________________
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Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
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[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this
6V battery
pack that is intended to clamp to the back of a ZIP
drive, allowing
it to be used away from power, say with a laptop.
I've wanted
to use it for a portable source of juice for hobby
projects (it
has, among other features, a built-in 110V mains
plug and recharger,
making it handy to recharge), but I haven't expected
to be able to
feed a 7805 at +6VDC and get +5VDC out the other
side reliably.
How can I take +6VDC of battery power and get +5VDC
regulated power
from it? If it matters, the currents involved will
be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC
converter, but I'm not
skilled enough to design one, just skilled enough to
build it.
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up
now.
http://mailplus.yahoo.com 
4:31 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
I think there are some forms of the 7805 that only need .6V (one diode drop) of
overhead so you should be able to run them on as little as 5.6V and still get 5V output.
Also you may look inside your battery pack and find that it uses a higher voltage
internally and then regulates it down to 6VDC. If so you can take power off ahead of the
intenal regulator.
Joe
At 12:44 PM 12/6/02 -0800, you wrote:
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
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Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
5:43 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
Ethan Dicks wrote:
You don't have to build one. They are available as sealed
"modules" with pins for gazinta and gazouta; very easy to
use. An added advantage, when you're running off battery
power, is very high efficiency for maximum operating time
off your batteries. Just be sure to figure out what your
battery is putting out when it's just about to be exhausted,
and try to get a converter that can operate on this input
voltage to get all the capacity out. Check out Jameco,
Digi-Key's website to see if there's anything that fits the
bill.
I have a huge bin full of the venerable old 7805s. They
work and they're simple, but they're also terribly
inefficient and tend to run quite hot unless extensively
heat-sinked, especially towards the upper end of the voltage
input range and/or close to rated current flowing to the 5
volt load.
I fondly remember a home-made KIM-1 power supply made with a
7812 and 7805. Despite a relatively huge heat-sinking area,
the damn thing was running nearly hot enough to cause
burns. I think the input voltage was around 8-10 volts
somewhere. After the "MacDonald's Coffee-in-Lap" lawsuit, I
wonder if I coulda sued TI for burning my fingers. ;)
-- Ross
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
6:23 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
P.S. - I
don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
You don't have to build one. They are available as sealed
"modules" with pins for gazinta and gazouta; very easy to
use.
7:12 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
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9:51 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
I'll bet it used a NiCd, which gives ~4.8 volts for
the vast percentage of a charge - which is "kinda" in
tolerance of most +5 logic. Those things were so
cheap, do you think they'd use a step-up DC-DC
converter? Nah.
--- Tony Duell <ard(a)p850ug1.demon.co.uk> wrote:
__________________________________________________
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How can I take
+6VDC of battery power and get
+5VDC regulated power
from it? If it matters, the currents involved
will be under 1000mA,
but probably over 200mA.
Take a look at 'low dropout regulators'. National
Semiconductor make
(made?) them -- LM2940 series I think. They will
work down to about 0.6V
difference between input and output (so for a 5V
regulator, you need at
least 5.6V in). These are similar to the 7805 -- 3
terminals, and you
need to put a couple of decoupling caps near the
chip.
That probably won't let you use all the capacity of
your '6V' battery,
but it should let you use some of it.
Incidentally, I assuem the Zip drive produced 5V
internally from this
battery pack. Any ideas what it used?
-tony
7 Dec
7 Dec
6:50 a.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
Loboyko Steve wrote:
I'll bet it used a NiCd, which gives ~4.8 volts
for
the vast percentage of a charge - which is "kinda" in
tolerance of most +5 logic.
I have a Microwriter that used four NiCd cells in series
to get "+5V". What's more, it didn't bother with the extra
expense of a regulator, it just used the battery. Oh, what
a crap design -- now I have four dead NiCds (a funny sized
cell, too) and a non-functional Microwriter. Pics:
http://www.gifford.co.uk/~coredump/org.htm
--
John Honniball
coredump(a)gifford.co.uk
1:11 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first
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4:16 a.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
Hi Ethan,
Most modern DC-DC converters are designed with simplicity in mind. It
doesn't take a slide rule, or a PDP-8 to figure out component values,
they are usually supplied as part of an app-note with the data sheet.
If you go to www.digikey.com, you can execute a parametric search for
'DC converter', select a part that has reasonable characteristics
(voltage in, current/voltage out), and download the data sheet. At
worst you will have to calculate a couple resistors to set the voltage.
Good Luck!
Clint
On Fri, 6 Dec 2002, Ethan Dicks wrote:
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
5:49 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
Try a PowerTrends PT6322N DC->DC converter; they're small, cheap, and work
well.
On Fri, 6 Dec 2002, Ethan Dicks wrote:
[ACE PSU]
...a 7805 1A linear regulator which needs around
8V to 15V to run properly.
So I was thinking about this recently... I have this 6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing it to be used away from power, say with a laptop.
I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
-ethan
P.S. - I don't mind the concept of a DC-DC converter, but I'm not
skilled enough to design one, just skilled enough to build it.
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
16 Dec
16 Dec
6:56 p.m.
New subject: Driving a 7805 or how else to get +5VDC reg (was Re: My first good find!!!)
On Friday, December 6, 2002, at 03:44 PM, Ethan Dicks wrote:
So I was thinking about this recently... I have this
6V battery
pack that is intended to clamp to the back of a ZIP drive, allowing
it to be used away from power, say with a laptop. I've wanted
to use it for a portable source of juice for hobby projects (it
has, among other features, a built-in 110V mains plug and recharger,
making it handy to recharge), but I haven't expected to be able to
feed a 7805 at +6VDC and get +5VDC out the other side reliably.
How can I take +6VDC of battery power and get +5VDC regulated power
from it? If it matters, the currents involved will be under 1000mA,
but probably over 200mA.
You could use a low-dropout linear regulator like the LM2940. These
are readily available and easy to use, but not quite as cheap as the
7805.
-Dave
--
Dave McGuire "Why would a brothel need
St. Petersburg, FL a streaming media server?" -Kevin
7 Dec
7 Dec
5:24 p.m.
-----Original Message-----
From: cctech-admin(a)classiccmp.org [mailto:cctech-admin@classiccmp.org]On
Behalf Of Philip Pemberton
Sent: 04 December 2002 22:46
To: cctalk(a)classiccmp.org
Subject: My first good find!!!
I'm currently in a very good mood because I've found a Jupiter Ace in
near-mint condition, complete with 16k RAMPack, intro tape and a
copy of the book "Jupiter Ace FORTH Programming" by Steven Vickers. All
this,
including the polystyrene box for the rampack (and
rampack instructions)
cost me the
rather feeble sum of two British pounds.
Nice one! I wonder if that book is online somewhere since it's the only
thing I haven't got.....
Now, there is a catch. The machine didn't come
with a power
supply. So,
can someone with an Ace, working PSU and a multimeter please tell me what
the polarity and voltage of the PSU's output is? I'm hesitant to power up
the Ace in case I end up fouling up the settings, with fairly predictable
results.
Tony's not far wrong with his guess (as usual :). The PSU I have in my
sweaty mitts here says 9V 800ma and I've tested centre (tip) positive. I was
going to say you could get away with a ZX81 PSU but they're 700ma and the
ZX80 is 600ma. Oh, the voltage I got was ~14V so it requires a load to bring
it down to 9V.
cheers
--
adrian/witchy
www.binarydinosaurs.co.uk - the online computer museum
www.snakebiteandblack.co.uk - monthly gothic shenanigans
8059
days inactive
8071
days old
19 comments
15 participants
participants (15)
-
archer＠topnow.com -
ard＠p850ug1.demon.co.uk -
bill＠timeguy.com -
coredump＠gifford.co.uk -
erd_6502＠yahoo.com -
eric＠brouhaha.com -
jcwren＠jcwren.com -
jim＠jkearney.com -
mcguire＠neurotica.com -
n4fs＠monmouth.com -
philpem＠dsl.pipex.com -
rigdonj＠cfl.rr.com -
sloboyko＠yahoo.com -
vaxman＠earthlink.net -
witchy＠binarydinosaurs.co.uk