# how to choose resistors' value for common emitter amplifier? [closed] I want to choose the best values of R1 , R2 , R3 and R4 to get the maximum gain (high performance).

I design the circuits practically, so I don't need to do a lot of calculations. I want the simplest way to calculate the four resistors.

This is the datasheet of the transistor, I think it may help : http://www.onsemi.com/pub_link/Collateral/2N5401-D.PDF

• You're going to need to add a lot more information. When you say maximum gain, is that voltage gain or current gain? Also, gain and performance are very different things. Also, the load you want to drive is a critical factor in resistor choice, so you cannot really chose the resistors in isolation. Current gain will be ultimately limited by the transistor's hFE, and you will have a AC highpass effect from the two capacitors. – Connor Wolf Sep 2 '14 at 8:14
• Building circuits without knowing the equations is pointless. Go through the charts in datasheet and do some pen-and-paper analysis. – venny Sep 2 '14 at 10:06
• This question appears to be off-topic because there is no evidence of any research. – Leon Heller Sep 2 '14 at 10:23
• This question makes no sense, and sounds a lot like gimme da codz anyway. You say you want the maximum gain, but then say you want "high performance". Performance measured by what criteria? How high is "high". You say you don't need a lot of calculations, but specifically ask how to calculate something in the very next sentence. Closing this mess since we do engineering here, not handwaving. Also, we are happy to help people understand electronics, but not just give them answers, especially whey they don't seem to want to understand how the circuit works. No thanks. – Olin Lathrop Sep 2 '14 at 12:25

Its not really practical to design any circuit without calculations and Alfred's answer gives a comprehensive approach (so +1 from me). However, there are other 'rule of thumb' type" approaches that might suit a 'practical' designer to get started. The thinking behind this approach is very much based in the theory developed by scientists and engineers over decades and if you are to advance in any design work I would recommend that you try to develop a solid mathematical approach. (Note: I still need to use Ohm's law V= IxR and an understanding of percentages/fractions with this approach!)

The circuit is pretty much standard, well understood and lends itself to a simple analysis.

Let's start with the value of emitter resistor (R4). The voltage across it should be between 10 and 20% of the supply (rule of thumb) to give an 80 - 90% voltage swing at the collector.

Decision 1. Let's take Ve as 10% of the supply (= 0.5V for a 5V supply).

Now choose the current you want through the transistor. 100uA, 1mA, 10mA? This will depend on your application. Do you need to minimise current used or do you need to drive a high current into the next stage?

Decision 2. Let's choose a typical value, say 1mA

Now we need a little bit of (simple) maths (using V=IR).

For a 0.5V drop @ 1mA we need a 500R resistor - BUT this is not a 'standard' E24 or 5% value so choose a nearest preferred value (n.p.v.) - either 470R or 510R

Decision 3. I choose the 510R

This value represents 10% supply drop leaving 90% to be divided between the transistor (c-e) and the collector resistor. Allowing for a minimum saturation voltage (say 0.2V) we need about 40% of the supply dropped across R2. (to give maximum output swing at collector). As the emitter resistor represents 10% supply drop we can simply calculate the collector resistor at about 4 X emitter resistor. (= 4 X 510 = 2040R). Once again not a preferred value so we can choose from 2k0 or 2k2. ( the collector voltage will be set around 3V and have a +/- swing of 2V)

Decision 4. 2k0

The size of the decoupling capacitor is next. This basically sets the low frequency response of the stage. The bigger the value the lower the cut off frequency. Let's assume you are designing an audio stage. A 100uF capacitor will have a reactance of about 80 ohms @ 20Hz. A 10uF will have about 800 ohms. Doubling the capacitance will half the reactance. A 22uF will be about 400 ohms which is about the size of the emitter resistor. Choosing a higher or lower value will decrease or increase the cut off frequency of the amplifier stage.

Decision 5. I choose 22uF (a preferred value capacitor)

The final parts of the design is choosing suitable values for R1,R3 and Cin.

R1 and R3 form a voltage divider. We 'know' that the voltage at the emitter is about 0.5V and that there will be a 0.6V drop from the base to the emitter (assuming its a silicon transistor). So we know the voltage at the base will be 0.5 + 0.6 = 1.1V.

For a small signal transistor the current gain will be at least 100. This means that the current going into the base will be a maximum of 1/100 th of the emitter current. This current will be taken from the voltage divider circuit. The rule of thumb is to have at least 10 x this base current going through the bias resistors so that we can 'ignore' this loading effect in the calculation. This means that we need about 1/10th of the emitter current through R1 and R3. this is 1mA/10 = 0.1mA.

R3 can now be calculated because we know it needs 1.1V across it and a current of about 0.1mA through it. (V=IR). This gives R3 = (1.1/0.1) * 1000 = 11k (a preferred value).

R1 has a voltage drop of 5V - 1.1V = 3.9V for the same current. This gives a value of 39k (a preferred value). If the calculations don't give preferred values then adjust to the n.p.v. You can always tweak the values later by measuring the voltage at the base but generally they will be of the correct magnitude.

Decision 6. R1 = 39k and R3 = 11k

These values set the input impedance of the stage. Very roughly R3 will be reduced by about 10% due to current taken through the base of the transistor making it about 10k. This will effectively be in parallel with R1. It unlikely you require a precise answer, so rounding the 39k to 40k we have approx. 10k // 40k. Without calculating anything we can say the input impedance just under the 10k. (lowest resistance will dominate).

This value is about 20x the size of R4 and as we chose 22uF for the by-pass capacitor we need something about 20x smaller at the input to give the same cut off frequency - say 1u0 (preferred value) - which by coincidence matches the value shown.

Decision 7. Cin = 1u0

What if you want a different current say Ic = 100uA or 10mA? In the words of the great Alexandre Orloff - SIMPLES!

Just scale values by ratio.

for Ic = 100uA (= 1/10th), Resistors = 10 x values, capacitors = 1/10th values for Ic = 10mA (= x 10) , Resistors = 1/10 x values, capacitors = 10 x values

Since there are 4 resistor values to choose, one needs 4 (realizable) constraints to uniquely choose the 4 values.

Some examples of specifications that impose constraints are

(1) input impedance

(2) output impedance

(3) open-circuit voltage gain

(4) operating point

(5) positive and negative clipping levels

(6) idle power dissipation

You've only mentioned one specification, the gain, and have not actually specified it except that it should be, in some sense, high performance. So, there is no unique 'best' set of 4 resistor values.

Below are some of the design equations one might use to find resistor values given some of the above are specified.

Assuming the impedance of the capacitor is insignificant, the open-circuit voltage gain of this circuit is given by

$$|A_{voc}| \approx g_m R_2||r_o$$

where

$$g_m = \frac{I_C}{V_T}$$

and

$$r_o = \frac{V_A + V_{CE}}{I_C}$$

The quiescent collector current is given by

$$I_C = \frac{5V \frac{R_3}{R_1 + R_3} - V_{BE}}{\frac{R_1||R_3}{\beta} + \frac{R_4}{\alpha}}$$

The quiescent collector-emitter voltage is given by

$$V_{CE} = 5V - I_C\left(R_2 + R_4 \right)$$

The input impedance is given by

$$r_{in} \approx R_1 || R_3 || r_{\pi}$$

where

$$r_{\pi} = \beta \frac{V_T}{I_C}$$

The output impedance is given by

$$r_{out} \approx R_2 || r_o$$