# How to calculate the base/ collector resistors.

#### R!f@@

Joined Apr 2, 2009
9,750
May be they just throw in some values using trial and error.
Might also be dependent on the mic R and it's gain.

#### Dodgydave

Joined Jun 22, 2012
10,050
I've seen several of these circuits, and they all seem to use 100K to 1M base resistor, and 10-100K collector resistor but no answer as to why!

#### bertus

Joined Apr 5, 2008
21,532
• Dodgydave

#### #12

Joined Nov 30, 2010
18,223
@Dodgydave
Have a look at the section "Collector Feedback Biasing a Transistor
That's a little tighter math than the PM I did with The Dodger because I neglected the Ib term in the collector current...because it is less than 1% of Ic. Another habit from designing 1% meters. I neglect things that are less than 1/2% until I'm doing the error budget calculations. In this circuit, the errors are a lot more about diminished Hfe at low Vc than it is about the base current being part of the collector resistor current.

• Dodgydave

#### MrAl

Joined Jun 17, 2014
8,475
View attachment 123697

Trying to find out how to calculate the resistors R2, R3, and seen this type of circuit used several times in electret mic preamps, and why they use it?...
Hi,

The short answer is that you take the transistor to be a perfect current amplifier with infinite gain and Vbe=0v, and then calculate the circuit current gain, and from there you can calculate the voltage gain, or just 100k/Zin.

The circuit input current is based on the total input impedance but if we estimate that to be 10k/2 then the voltage gain is 100k/5k which is 20.
This is a rough approximation that shows how the basic circuit works, which basically turns it into an ideal current mode op amp with some resistors. Of course for this to be an approximation the circuit gain cant be too high. If you want to you can place more restrictions on the transistor gain.

The gain above will be higher than really in the circuit, so it is a very rough calculation. If you want to get more exact, you have to include the gain of the transistor and then the analysis gets a little more complicated. I am not sure how deep you want to get into it. Once you do the calculation above you can test and then change the feedback resistor if needed as long as the transistor always stays in the active region.
For your circuit the current gain calculation comes out to about 20, and so the output voltage is 20v with 1v input. In a real circuit where the gain of the transistor is limited to maybe 100, the gain will be less like maybe around 15.

To make this really short, if you calculate the initial gain as Gx=R2/R1 (which is really what we did above) then that's the very rough estimate, and the correction is then:
G=Gx*B*R3/(B*R3+R3+Gx*R1)

where G is the actual gain and B is the transistor Beta.

In the limit as B gets very high which is the initial rough assumption, G=Gx.

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• Dodgydave

#### Dodgydave

Joined Jun 22, 2012
10,050
Hi,

To make this really short, if you calculate the initial gain as Gx=R3/R2 (which is really what we did above) then that's the very rough estimate, and the correction is then:
G=Gx*B*R3/(B*R3+R3+Gx*R1)

where G is the actual gain and B is the transistor Beta.

In the limit as B gets very high which is the initial rough assumption, G=Gx.
Do you mean Gx=R2/R3??
This would be 100K /10K =10...

#### MrAl

Joined Jun 17, 2014
8,475
Do you mean Gx=R2/R3??
This would be 100K /10K =10...

Hi,

Im sorry i meant to type Gx=R2/R1. I corrected it in the original post.
That means 100k/5k=20. So the initial estimate for the gain is 20.

There is another approximation though that is almost as good as the more exact, and that is:
G=(Gx*B)/(R2/R3+B)

That holds approximately when B is maybe 50 or above provided the other values are normal too.

This is also estimating the input resistance is half of the 10k which is 5k and the input cap passes all frequencies of interest without adding much to the input impedance. This is because of the attempt to keep the defining equations simpler.

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#### MrChips

Joined Oct 2, 2009
24,178
Here is how I would tackle the problem.

I begin with the load resistor R3. This is in effect the output impedance of the circuit.
The choice will depend on the input impedance of the amplifier stage to follow. You want this to be lower than the input impedance of the amplifier.
R3 = 10kΩ seems a bit high. I would have chosen something more like 2k-5k. No matter, 10kΩ will do for now.

Next, I select the DC operating point or Q-point. Since this is class-A amplifier, I choose the collector voltage to half the supply voltage.
Let's set Vc to 4V for an 8V supply.
Now I can calculate the collector current Ic = (Vs/2)/R3 = 4V/10k = 0.4mA

I want the transistor biased in the linear region. Hence the base-emitter voltage, Vbe ≅ 0.6V
As someone else rightly pointed out, I can ignore this and use Vbe = 0 for a rough calculation.
I have no idea what is the current gain β. I will assume β = 100.
Now I can estimate Ib = Ic/β = 400μA/100 = 4μA
So now I can calculate R2 = (Vc-Vb)/Ib = 4V/4μA = 1MΩ
If I had used Vb = 0.6V, R2 = 3.4/4μ = 850kΩ

That is a starting point. I would then breadboard the circuit and look at the output signal at the collector (before the output capacitor) on an oscilloscope. I would tweak the value of R2 and aim for a signal with the least amount of distortion with the maximum dynamic range.

More base current (lower R2) will drive Vc lower. The bottom of the wave will clip earlier.
Less base current (higher R2) will drive Vc higher. The top of the wave will clip earlier.

Hope this helps.

• Dodgydave

#### Bordodynov

Joined May 20, 2015
2,906
• Dodgydave

#### Dodgydave

Joined Jun 22, 2012
10,050
Thanks, can you do a model for R2 from 470K to 2M2, so i can see what happens...

#### Bordodynov

Joined May 20, 2015
2,906
• Dodgydave