How are the resistor values R1 and R2 calculated for a transistor amplifier that has a voltage divider bias

by Sâu   Last Updated September 11, 2019 21:25 PM

Good day friends

I am trying to analyse the working of transistor amplifiers with a prime goal of designing a good audio amplifier that does not produce much distortion in the output. I am restricting my analysis to the follow two general purpose transistors:

  1. BC107
  2. 2N3904

I started my analysis from the very basics. To calculate the value of resistors for the emitter and collector, I simply used the Kirchoff's voltage law for the output side and got the values as 1.2k and 2.2k respectively. Coming to the input side, I had a little bit of difficulty to properly deduce the necessary values. However I had studied once that the Base voltage should always lie in between that of the emitter and the collector. So I randomly chose the values as 22k and 68k for R1 and R2. However probability for a random guess to work is always less.

I constructed the circuit in Multisim to simulate the output. Here is the circuit:

enter image description here This circuit is highly inefficient and it attenuates the input signal rather than amplifying it. Here is the output from the oscilloscope:

enter image description here

The pulse like signal is the output that I got from the amplifier. Similar results were obtained when I used BC107 transistor. This result is very frustrating.

What I require is some guidance on properly calculating the resistor values in order to properly bias the transistor to the active region. Can anyone please provide a good systematic procedure with explanation on calculating the resistor values for this purpose?

Edit: I have changed the peak voltage to 2V. I am getting an amplification but the wave is pulse like. Also I constructed this circuit practically on a breadboard. The audio output intensity is much less than what I hear when I connect the audio source directly to speaker. What should I do? Also please provide a procedure to calculate the biasing resistor values.

enter image description here

Answers 2

What you see has nothing to do with the values of the resistors ! What you see is the behavior you get when you severely overdrive the input. You are applying way too much input signal to this circuit.

Is that 10 V peak you're putting in ? Geez, the supply is only 9 V. If you want low distortion then the output swing can only be 2 Vpeak or so.

Stay below 1 Vpeak-peak at the input and it should work properly.

If your goal is to design your own audio amp. you have a very long way to go. You will need to understand how this circuit works, understand why changing the resistors would not help and much much more. Designing a circuit is not putting something in a simulator and hoping for the best. Designing something required knowledge and practice.

Update: Oops, my mistake, forgot to consider that CE of 100 uF, this will increase the gain. IF THE CIRCUIT WAS DIMENSIONED PROPERLY which it is not. This amplifier is wrongly biased, RC is too high driving the transistor into saturation.

This explains why you get the distorted output !

You will need to study a bit more to learn how to bias a transistor properly.

January 15, 2016 14:58 PM

In an amplifier like this, your objective in selecting R1 and R2 is to bias the amplifier halfway between two extremes. The two extremes are:

  1. The transistor is fully on, and the collector current is limited only Rc and Re.
  2. The transistor is fully off, and the collector current is zero.

If you hit either of these extremes, the output is clipped. So if we can bias the amplifier to be halfway between these extremes, then we have maximized the input signal amplitude that can be amplified without clipping either the positive or negative side.

We can make a couple simplifying assumptions:

  1. Because the transistor has high current gain (β > 75, most likely), we can consider that the current into the collector is equal to the current out of the emitter. It also follows that the current through Rc must equal the current through Re.
  2. Because we are only interested in biasing at DC, we can ignore all capacitors as if they are open circuits.
  3. Because the saturation voltage for a BJT transistor is small (0.2V) relative to the supply voltage (9V), we can assume this saturation voltage is 0.
  4. This amplifier's output will be connected to a high impedance, so we consider this current to be zero. Notably, a speaker is not a high impedance (8Ω is typical). If you want to connect this circuit to a speaker, you need a buffer amplifier.

So the first question is this: in the first extreme, when the collector current is limited only by Rc and Re, what is that current?

Since Rc and Re are in series, we can add those resistances together and calculate the current with Ohm's law:

$$ I_c = {9\mathrm V \over R_c + R_e} = {9\mathrm V \over 2.2\:\mathrm{k\Omega} + 1.2\:\mathrm{k\Omega}} \approx 2.65\:\mathrm{mA} $$

Remember, this is the current through the collector of the transistor at one extreme of clipping. The other extreme is no current at all. So halfway between these points is just half of \$I_c\$, or about 1.32mA.

So what voltage needs to be at the base to make \$I_c = 1.32\:\mathrm{mA}\$ ?

The base emitter junction of a BJT transistor is effectively a diode. And we've already established that the current through Rc, the collector, and Re are equal (see simplifying assumption #1). So we can simplify this problem a bit:


simulate this circuit – Schematic created using CircuitLab

Given that we know the current through Re (1.32mA), we can calculate the voltage across it with Ohm's law:

$$ 1.32\:\mathrm{mA} \cdot 1.2\:\mathrm{k\Omega} = 1.58\:\mathrm V $$

We also know that the forward voltage of any silicon diode is about 0.6V. Added to the 1.58V above, that means if we want the current to be 1.32ma, then V1 will need to be:

$$ 1.58\:\mathrm V + 0.6\:\mathrm V = 2.18\:\mathrm V $$

So now you just need to come up with a pair of resistors for R1 and R2 that make a voltage divider with an output of 2.18V.

You must also keep in mind that the current into the transistor base will introduce some error into your voltage divider. We can estimate that this current will be the collector current, divided by the transistor's gain. If you then pick your voltage divider values such that the current through the divider is at least 10 times the base current, then this error will be negligibly small.

To keep the math simple it's reasonable to guess that transistor gain (β) is 100. So the base current will be something like 0.0265 mA. You want the current through the voltage divider to be at least 10 times this, or 0.265 mA. By Ohm's law:

$$ R_1 + R_2 < 34\mathrm{k\Omega} $$

Finally, you will want to adjust your simulator to input a much smaller amplitude signal. The output signal can't be possibly more than 9V peak-to-peak, and actually less than that because the transistor can't drive the output all the way to the supply rails. Since this is an amplifier, that means the signal will need to be very much less than 9V peak-to-peak, otherwise you will see clipping and attenuation.

Phil Frost
Phil Frost
January 15, 2016 18:08 PM

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