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..
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.
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.
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.
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.
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.
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!
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.
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.pdfIt will give you 9 volts, regardless of what current your amplifier draws.
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.
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.
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.