Will a battery act like a wire once it has lost charge?

[dougman0988]

I'm building an amplifier (should end up being around 100 watts peak) and in order to help prevent crossover distortion in the output stage, I want a 1.5V offset by use of a AA battery.  Once the battery loses charge, will it act like a short and therefore the amp will still work (probably just with slightly more distortion) or is a dead battery more like an open circuit?  Thanks for the help!

[GibsonGM]

hmmm, interesting, man.  Well, a battery is made mostly of carbon, so my guess is that it'd act like an open - or a resistor! You could find a dead one and check it out...or just set up a resistive voltage divider and eliminate the battery bias altogether...
Anyone else?

[dougman0988]

A voltage divider was considered, but even my preamp voltage is fairly large compared to 1.5V and I'd need a fairly high-wattage resistor network to break it down...I figured a battery would be the simplest and cheapest way to get the 1.5V offset.  But obviously I don't want to go down that route if the battery is going to prevent the amp from working when it dies and have to replace the battery.

I will certainly try this out on my own but I was just wondering if anyone has ever dealt with this before. 

[cpm]

i guessing two things:
1. as voltage goes off it acts more lka a big resistor
2. i wouldnt put a battery V+ where it is susceptible to sink current

[PRR]

In a 100W amp, there is probably a more clever bias than a battery. We normally bleed current through diodes to get 0.6V, 1.2V, etc. An added advantage (sometimes) is that the Vd tracks the Vbe with temperature.

A dry cell will probably stay low-resistance for several years after it goes "flat". However if it doesn't, how big a fire will your 100W amp make? If "nah!", then how much cash can you lose if you don't play the whole night? If "Nah!", then do it. Otherwise, I'd put the thinking-cap back on.

[dougman0988]

PRR,

Thanks for the response.  I suppose I thought that a diode would be difficult to forward bias properly, but simply having 2 of them in series with the double-Vbe loop would do the job, would it not?

[dougman0988]

Wow, nevermind I'm an idiot lol.  I completely forgot about the Vbe multiplier.  Basically does the same thing as the diodes but they provide thermal tracking to help prevent thermal runaway.  This is the way to go.

[PRR]

> I suppose I thought that a diode would be difficult to forward bias properly

It can be as easy as your battery, right voltage near-enough, and cheaper.

Look at Rod Elliot's #3. The stage before your power-buffer is usually a class-A scheme flowing several mA rail-to-rail. He sticks two 7-penny diodes in there, put 0.22 ohm resistors under the output emitters, and made money with the thing.

Then down below "Update" he discusses how this can sometimes run a little hot. He suggests changes. Power-amp design is very much a "learning process". In fact he discovered so many changes that he says "Don't Use This Circuit!" and points to a later design. But I wanted you to follow the history, not jump to the answer, for your own leanring.

But don't use a battery. Silicon Vbe changes 2mV per degree C. Your power devices may heat 50 degrees C. If your desired current needs 600mV cold, then when hot they only need 500mV. If you stick with 600mV bias, the idle current soars 30 times higher! Emitter resistors reduce the effect, but you still want a bias source which varies with temperature similar to your power devices.

Oh- in Rod's topology the bias actually should track his improved-version Q5 Q6 Vbe. If you use the simpler Darlington output, you need four diode-drops and should track the whole darlington (easy if you use pre-made Darlingtons).