# Using diodes to drop voltage

Discussion in 'Electrical Engineering' started by Skenny, Jun 9, 2006.

1. ### SkennyGuest

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

2. ### RobyGuest

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

3. ### SkennyGuest

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

4. ### BeachcomberGuest

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

5. ### repatchGuest

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

6. ### SkennyGuest

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?)

7. ### repatchGuest

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

8. ### BeachcomberGuest

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

9. ### SkennyGuest

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!

10. ### JohnR66Guest

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

11. ### SkennyGuest

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"

12. ### ehsjrGuest

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

Ed

13. ### PanHandlerGuest

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.

14. ### SkennyGuest

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

15. ### Paul Hovnanian P.E.Guest

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.

16. ### BeachcomberGuest

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

17. ### Paul Hovnanian P.E.Guest

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.

18. ### SkennyGuest

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..