Electrical Engineering - Using diodes to drop voltage

Author

Using diodes to drop voltage

Skenny

2006-06-09, 1:21 am

I have a small computer speaker amp that uses a 9 volt wall wart rated at
400MA.
Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
to series connect 5 diodes (1 amp each) between the amp and a 12 volt
computer power supply?
Shouldnt that drop the voltage to 9 VDC?
5 * 0.6= 3.0
12 VDC - 3 VDC = 9 VDC.
Anyone see any problems with doing it this way?
Thanx..
--Skenny

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Roby

2006-06-09, 10:21 am

Skenny wrote:

> I have a small computer speaker amp that uses a 9 volt wall wart rated at
> 400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC?
> 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way?
> Thanx..
> --Skenny

The string of diodes will reduce the voltage just as you describe. There
is one difference in powering the amplifier this way: the 12 volt return
is connected to the computer power common. With a wall transformer, the
amplifier supply voltage is isolated from the computer supplies.

That's *probably* not a problem and seems worth trying. Perhaps another
reader has done it and will jump in here.

I am often tempted to pitch out the multitude of wall warts nested behind
my computer (speakers, modem, router, etc) and tap the computer supply or
build a one-supply-for-all box.

Roby

Skenny

2006-06-09, 11:21 am

Thanks Roby,
This is for an aracde cabinet that Im building, since the speakers and amp
wont be coming into contact with any other parts, except the power supply, I
dont think I will have any trouble with isolation.
The idea about a common power supply for everything on a PC sounds like I
good idea. Please keep us posted if you pursue this.
--Ken

"Roby" <roby@no-address.net> wrote in message
news:KNadnXetUZ516RTZnZ2dneKdnZydnZ2d@adelphia.com...
> Skenny wrote:
>
>
> The string of diodes will reduce the voltage just as you describe. There
> is one difference in powering the amplifier this way: the 12 volt return
> is connected to the computer power common. With a wall transformer, the
> amplifier supply voltage is isolated from the computer supplies.
>
> That's *probably* not a problem and seems worth trying. Perhaps another
> reader has done it and will jump in here.
>
> I am often tempted to pitch out the multitude of wall warts nested behind
> my computer (speakers, modem, router, etc) and tap the computer supply or
> build a one-supply-for-all box.
>
> Roby

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Beachcomber

2006-06-09, 1:21 pm

On Fri, 9 Jun 2006 08:53:50 -0500, "Skenny" <skenny@high_streamDOTnet>
wrote:

>Thanks Roby,
>This is for an aracde cabinet that Im building, since the speakers and amp
>wont be coming into contact with any other parts, except the power supply, I
>dont think I will have any trouble with isolation.
>The idea about a common power supply for everything on a PC sounds like I
>good idea. Please keep us posted if you pursue this.
>--Ken
>

A better solution is to buy a 9V. regulator from some place like
Jameco Electronics. This will take 12VDC in and give a constant 9VDC
out no matter what the load, up to the capacity of the regulator.

The problem with series diodes is keeping the voltage drop across each
equal and balanced. Some power supplies do this with resistors.

Beachcomber

chuck

2006-06-09, 1:21 pm

Skenny wrote:
> I have a small computer speaker amp that uses a 9 volt wall wart rated at
> 400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC?
> 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way?
> Thanx..
> --Skenny
>
>
>
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Skenny, it may work ok. Remember that the voltage
drop across a semiconductor diode is not constant,
but is a non-linear function of current and
temperature. Often, we don't care much about a
diode's exact forward voltage drop, but in this
application it is the main performance factor.

Download a spec sheet for the diode you are
considering and see what the voltage drop is
likely to be at around 400 mA.

The spec sheet will also help you to see what the
effects of temperature might be.

Good luck.

Chuck

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repatch

2006-06-09, 4:21 pm

On Thu, 08 Jun 2006 22:30:42 -0500, Skenny wrote:

> I have a small computer speaker amp that uses a 9 volt wall wart rated at
> 400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC? 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way? Thanx..
> --Skenny

The voltage drop of a forward biased diode is NOT constant.

It has a non linear relationship to the current going through the diode.
0.6V is a "rough" number based on a typical amount of current (usually in
the 10-30mA range in my experience). The ACTUAL drop may be much more,
especially when nearing the upper end of the current spec. I've seen drops
of over 1V when dealing with currents of > 1A.

So, with all that said, 0.6V is certainly a "low" estimate (when dealing
with power), and as such is generally safe. Chances are you speakers will
see less then 9V, that's usually OK, you'll just get less volume.

If sound is important to you then this is a bad idea since you will be
introducing distortion (the larger input signals will cause a larger dip
in the power supply), but since the word computer is in there twice I
doubt 100% audio is a concern.

TTYL

Skenny

2006-06-09, 5:21 pm

Thanks everyone for your input.
I didnt know that the voltage drop across a diode is linear to the current.
So I agree, this may not work, since the current wont be constant (due to
sounds changing), the supply voltage will not be constant, so I could see
where distortion could result.
Actually, I didnt give all the data either. The wall wart output is 9VAC.
The power leads to the amp board go directly to a full wave rectifier (4
small diodes on the board), so I figured 9VDC should be OK. I hooked it up
to a 9v transistor type battery and it seems to work good, but I didnt crank
up the volume much.
I dont really want to use a zener or a regulator chip.
Do you think it would matter much if I used the full 12 volts on the amp?
(Might let out the smoke?)

"repatch" <repatch42@yahoo.com> wrote in message
news:pan.2006.06.09.18.47.54.201687@yahoo.com...
> On Thu, 08 Jun 2006 22:30:42 -0500, Skenny wrote:
>
>
> The voltage drop of a forward biased diode is NOT constant.
>
> It has a non linear relationship to the current going through the diode.
> 0.6V is a "rough" number based on a typical amount of current (usually in
> the 10-30mA range in my experience). The ACTUAL drop may be much more,
> especially when nearing the upper end of the current spec. I've seen drops
> of over 1V when dealing with currents of > 1A.
>
> So, with all that said, 0.6V is certainly a "low" estimate (when dealing
> with power), and as such is generally safe. Chances are you speakers will
> see less then 9V, that's usually OK, you'll just get less volume.
>
> If sound is important to you then this is a bad idea since you will be
> introducing distortion (the larger input signals will cause a larger dip
> in the power supply), but since the word computer is in there twice I
> doubt 100% audio is a concern.
>
> TTYL

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repatch

2006-06-09, 5:21 pm

On Fri, 09 Jun 2006 14:27:46 -0500, Skenny wrote:

> Thanks everyone for your input.
> I didnt know that the voltage drop across a diode is linear to the
> current.

It's not linear, in any way, it's actually closer to exponential.

> So I agree, this may not work, since the current wont be constant
> (due to sounds changing), the supply voltage will not be constant, so I
> could see where distortion could result.
> Actually, I didnt give all the data either. The wall wart output is 9VAC.
> The power leads to the amp board go directly to a full wave rectifier (4
> small diodes on the board), so I figured 9VDC should be OK. I hooked it up
> to a 9v transistor type battery and it seems to work good, but I didnt
> crank up the volume much.
> I dont really want to use a zener or a regulator chip. Do you think it
> would matter much if I used the full 12 volts on the amp? (Might let out
> the smoke?)

Well, this is an interesting question. In reality, the wall wart you have
WILL measure probably more then 12V when unloaded. The 9V rating is at the
rated current. OTOH the 12V rail in a computer usually is close to 12V. So
you've got a situation where the electronics in the speaker can
likely handle 12V+ at low currents, but at high currents expect to be fed
closer to 9V.

Therefore, when the speakers aren't producing any sound everything is
fine, when the volume gets turned up things will start getting
interesting. Chances are if you don't blast it (i.e. never let the
speakers clip) you'll be fine, but it's also possible the amp will
dissipate more power then it's designed for and burn out.

Only way to tell would be to try it. Generally computer speakers are
designed in such a way that they are usually OK with being driven by a
higher voltage, but that's no guarantee.

TTYL

Beachcomber

2006-06-09, 6:21 pm

>
>Well, this is an interesting question. In reality, the wall wart you have
>WILL measure probably more then 12V when unloaded. The 9V rating is at the
>rated current. OTOH the 12V rail in a computer usually is close to 12V. So
>you've got a situation where the electronics in the speaker can
>likely handle 12V+ at low currents, but at high currents expect to be fed
>closer to 9V.
>

Is there any particular reason you don't want to use an external 9 VDC
regulator? It's cost is probably pennies and it will do the job.
At 400 ma, you probably don't even need to worry about a heat sink.

DC output wall wart transformers come in both regulated and
non-regulated varieties. When you need a fairly precise output
voltage or you don't know if putting too many volts will harm your
device, its best to go with the regulated version.

Beachcomber

Skenny

2006-06-09, 8:21 pm

I think using the regulator is what I will do.
Thanks to everyone for the valuable input.
It's good to have forums like this where you can get different approaches to
a problem.
BTW, Im not an engineer. I am an electrician, have been for nearly 30 years.
Most experience has been industrial, so I have done some "fly by night"
engineering too. LOL
Currently I work in an aluminum rolling mill. We maintain several AC & DC
drives, PLC's, HMI's, etc.
Ok, Im off the soap box now, once again, thanks for the help!

"Beachcomber" <invalid@notreal.none> wrote in message
news:4489de48.10358078@news.verizon.net...
>
>
> Is there any particular reason you don't want to use an external 9 VDC
> regulator? It's cost is probably pennies and it will do the job.
> At 400 ma, you probably don't even need to worry about a heat sink.
>
> DC output wall wart transformers come in both regulated and
> non-regulated varieties. When you need a fairly precise output
> voltage or you don't know if putting too many volts will harm your
> device, its best to go with the regulated version.
>
> Beachcomber
>
>

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JohnR66

2006-06-10, 1:21 am

"Skenny" <skenny@high_streamDOTnet> wrote in message
news:1149823421_15493@sp6iad.superfeed.net...
>I have a small computer speaker amp that uses a 9 volt wall wart rated at
>400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC?
> 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way?
> Thanx..
> --Skenny
>
>
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Yes it will work fine. The 12v computer supply is pretty well regulated, so
you get pretty close. Small audio amp ICs that are likely used in you
speaker don't need tight regulation anyway.
John

Skenny

2006-06-10, 8:21 am

Thanks.
Im not sure I care for the voltage difference between speaker ground and
computer audio out ground.
Do you think it would be enough to hurt the computer?
The speakers are cheap, I think I paid around 7 bucks for them at Big Lots,
but the computer wasnt quite that cheap.

"Alfredo E. Torrejon"
<alfredo@atorrejon.dsl.DONT-SPAM-ME.pdx.MR_SPAM_MAN.spiretech.com> wrote in
message news:Pine.LNX.4.20.0606100151190.638-100000@hendrie.1610am.lan...
> On Fri, 9 Jun 2006, Skenny wrote:
>
>
> Wait--before you get off the soap box, you said that the wall wart that
> powers these speakers is rated for 9 Volts *AC* and that there are
> rectifier diodes in the speakers. That does make a difference.
>
> The 9 Volt AC rating of the wall wart is most likely RMS, meaning that its
> peak voltage hits about 12.7 volts. The filtercapacitor in the speaker's
> power supply will thus charge to something like 11.5 volts during normal
> operation. You might want to verify this by powering the speakers and
> measuring the voltage across this capacitor.
>
> In short, you will probably be OK with feeding the 12 volts directly
> without regulation because the voltage seen by the speaker's electronics
> will be about the same as that seen when they are connected to the wall
> wart. However, be aware that the speaker's internal ground will be about
> 0.6 volts positive with respect to the computer's metal chassis because
> there will be a diode drop in series (assuming that a full-wave rectifier
> was used).
>
> Alfredo
>
>

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ehsjr

2006-06-11, 2:21 am

Skenny wrote:
> I have a small computer speaker amp that uses a 9 volt wall wart rated at
> 400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC?
> 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way?
> Thanx..
> --Skenny
>

It will work fine.
The ~.6 volt drop is at very low current (10 mA) for a
1N4001 diode. At 40 mA that diode will drop ~.7, at 100
mA it will drop ~.8, at 200 mA it will drop ~.85 and at
400 mA it will drop ~.9 So guessing that your amp will
draw an average of 50 mA, 5 diodes would give you a 3.5
drop.

Those cheap computer amplified speakers will work fine
at anything from 9 down to 7 volts (and maybe even less).

You could also use a LM7809 regulator.
See the datasheet, page 6, for a 3 component circuit:
http://www.ortodoxism.ro/datasheets/unisonic/LM7824.pdf
It will give you 9 volts, regardless of what current
your amplifier draws.

Ed

Lost'n Found

2006-06-12, 3:21 am

What I would do is make a voltage divider to get 9.6v, and connect the amp
to NPN transistor in Emitter follower configuration where the base gets to
the 9.6V, and the amp represents the emitter resistance.
The voltage will drop by about 0.6 to get 9v at the emitter. The amp will
get current as needed.

I usually put a large resistance in the emitter, but I see no reason why you
can not replace the whole emitter resistance with the amp itself.

Goodluck on that

"Skenny" <skenny@high_streamDOTnet> wrote in message
news:1149823421_15493@sp6iad.superfeed.net...
>I have a small computer speaker amp that uses a 9 volt wall wart rated at
>400MA.
> Since the voltage drop across a diode is approx. 0.6 volts, is it feasible
> to series connect 5 diodes (1 amp each) between the amp and a 12 volt
> computer power supply?
> Shouldnt that drop the voltage to 9 VDC?
> 5 * 0.6= 3.0
> 12 VDC - 3 VDC = 9 VDC.
> Anyone see any problems with doing it this way?
> Thanx..
> --Skenny
>
>
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> Newsgroups
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> =----

PanHandler

2006-06-12, 6:21 pm

"Skenny" <skenny@high_streamDOTnet> wrote in message
news:1149880843_22787@sp6iad.superfeed.net...
> Do you think it would matter much if I used the full 12 volts on the amp?
> (Might let out the smoke?)

Every electronic component I've ever owned RUNS on smoke. When the smoke
comes out, it quits. The only variable is the the more expensive the
equipment, the thicker and smellier the smoke.

Skenny

2006-06-12, 9:21 pm

Ive had a few cars that ran on smoke too, but they never seemed to run out,
just kept on smoking.

"PanHandler" <panhandler@emptyhat.net> wrote in message
news:cjkjg.6098$y%3.478@bignews1.bellsouth.net...
>
> "Skenny" <skenny@high_streamDOTnet> wrote in message
> news:1149880843_22787@sp6iad.superfeed.net...
>
> Every electronic component I've ever owned RUNS on smoke. When the smoke
> comes out, it quits. The only variable is the the more expensive the
> equipment, the thicker and smellier the smoke.
>
>

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Paul Hovnanian P.E.

2006-06-12, 10:21 pm

Any reason you can't just use a 3 terminal 9V regulator?
An LM7809 or LM7909 (depending on the load current) should probably work
just fine.

--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
Programmers don't die, they just GOSUB without RETURN.

Beachcomber

2006-06-13, 4:21 am

On Mon, 12 Jun 2006 17:54:32 -0700, "Paul Hovnanian P.E."
<paul@hovnanian.com> wrote:

>Any reason you can't just use a 3 terminal 9V regulator?
>An LM7809 or LM7909 (depending on the load current) should probably work
>just fine.
>
>--
>Paul Hovnanian mailto:Paul@Hovnanian.com
>------------------------------------------------------------------
>Programmers don't die, they just GOSUB without RETURN.

I think some are not familiar with "regulators" (through no fault of
their own of course). It's sounds difficult to use something you've
never used before. But an LM7809 or LM7909 is as simple as can be.

Back in my high school days, I had a shop/electronics teacher who only
taught tube theory because he didn't understand transistors.

Beachcomber

Paul Hovnanian P.E.

2006-06-13, 3:21 pm

Beachcomber wrote:
>
> On Mon, 12 Jun 2006 17:54:32 -0700, "Paul Hovnanian P.E."
> <paul@hovnanian.com> wrote:
>
>
> I think some are not familiar with "regulators" (through no fault of
> their own of course). It's sounds difficult to use something you've
> never used before. But an LM7809 or LM7909 is as simple as can be.
>
> Back in my high school days, I had a shop/electronics teacher who only
> taught tube theory because he didn't understand transistors.
>
> Beachcomber

This is true. When the OP presented his problem, some try to educate him
on the particulars of the Ebers-Moll model, which is a noble pursuit.
But some people just need a practical solution. The existence of a
device such as a 3 terminal regulator may be of much more value in the
short term.

--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
IRS: We've got what it takes to take what you've got.

Skenny

2006-06-13, 9:21 pm

Thanks guys, but I do know about the small regulator chips, I have built
several circuits with them.
I just wanted to know how everyone else feels about the diode idea.
Again, thanks..

"Paul Hovnanian P.E." <paul@hovnanian.com> wrote in message
news:448EFAB9.BF174933@hovnanian.com...
> Beachcomber wrote:
>
> This is true. When the OP presented his problem, some try to educate him
> on the particulars of the Ebers-Moll model, which is a noble pursuit.
> But some people just need a practical solution. The existence of a
> device such as a 3 terminal regulator may be of much more value in the
> short term.
>
>
> --
> Paul Hovnanian mailto:Paul@Hovnanian.com
> ------------------------------------------------------------------
> IRS: We've got what it takes to take what you've got.
>

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