Designing Bias Supply (Vbias or Vb) Networks for Effects

Copyright 2002-2004 R.G. Keen. All rights reserved. No permission for local copies or serving from pages other than

 A question that comes up pretty frequently is:

I've seen values ranging from 10K to 47K in the bias divider that makes the bias voltage in effects. Sometimes the values are not given. How do you decide what value to use for the two resistors in the voltage dividers that bias your opamps at 4.5V? 

At the risk of sounding overly simplistic, it's done with Ohm's law.

That actually is a complete answer, but I suspect that some background is needed. 

The intent of providing a bias voltage is to keep the reference at some middle voltage. If you do this with resistors, any current you suck out of the middle point (or push into it!) will move the reference voltage around. How much depends on the resistor values.

The trick is to size the resistors so that the amount of current you pull out or push in will not move the reference voltage around more than some allowable amount.

Which moves the original question from the single question "what resistor values do I use?" to "what is the allowable variation in the reference voltage?" and "What amount of current does the circuit source/sink from the reference?"

The first is a judgement question, the second is a design question. Engineering judgement almost always uses the rule that says "10% is almost always insignificant": if what you do changes things less than 10%, you can ignore it as insignificant, as long as you're not near some boundary (don't want to fall off a cliff). So the very first reaction is - the allowable variation in a split 9V supply of 4.5Vdc should be less than .45Vdc.

The design question then demands that you either know with some accuracy, or more likely, can bound the current draw on the reference point. For instance - what's the bias current for those two opamps? Don't know? Go look up the data sheets on the internet. Too lazy to look it up? No problem. There are 100K resistors between the reference voltage and the inputs. Even if the inputs were solidly grounded, those 100K bias resistors would keep the current down to 4.5V/100K = 45uA per input. The bias resistor are only 10K? That makes it 450uA per input.

By looking at everything connected to the bias point, you can estimate or bound the amount of current going into or out of the bias point. With that number in hand, you can get back to the resistors. Remember the rule of the insignificance of 10%? If we have an estimate of something we want to make insignificant, we can just multiply by ten.

Let's say that the current drained is 120uA. To make that insignificant, it needs to be less than 10% of the current already running in our bias resistors. So the bias resistors need to be carrying 1.2ma. Now Georg Ohm rides back to our rescue: The resistors in the bias chain need to be no larger than 9V/1.2ma to get that current to flow. That's 7500 ohms. We know that each resistor needs to be half that (**Thank you, Dr. Ohm!!!**), so each resistor is 3750 ohms?

All done.

Well, not quite. 3750 is not a standard 5% value. But 3.3K is. So we use two 3.3K resistors, 9V/6600 = 1.36ma flows, and all is well. Thanks to Ohm's law, we could go back and calculate exactly how much the reference voltage is likely to be moved around.

This is a pain, right?

There is another way to do it.

Many commercial opamp circuits with more than a few IC's need a more solid reference than the sloppy 10% we assumed. (Worse, actually. Did you notice that I didn't even calculate the variation in the reference voltage that happens because the reference resistors have a +/-5% tolerance, let alone end of life and temperature drift?)

In that case, they will make a resistor divider for ONE opamp, tie it to the + input, hook the output to the - input, and use the output for the reference voltage. Now, the opamp sources and sinks current to the rest of the circuit. The resistor reference has a low and constant load, and things have the potential to be much more accurate.

Recently, an even simpler way popped up in my head. The LM386 power amplifier chip comes in an 8 pin DIP package, is quite cheap, and self biases to half the power supply. With no input signal, its output sits at half the power supply and sources/sinks current to keep the output nice and quiet at that voltage. You don't even need biasing resistors or caps, since its inputs are self-biased at ground.

But where great accuracy doesn't matter, the two resistor divider rules. And that's how you calculate it.

Have fun!