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I know that in common emitter amplifier, \$R_C\$ has effect on output swing and also on gain. and the two resistors \$R_1\$ and \$R_2\$ stabilize from \$h_{fe}\$ change.

schematic

simulate this circuit – Schematic created using CircuitLab

But the question for me is how to choose the best resistor for \$R_E\$. As it has effect on \$V_{CE}\$. When we increase \$R_E\$ the \$V_{CE}\$ decreases. And also it has no effect on voltage gain.

At the other hand we can have constant \$I_c\$ by choose appropriate \$R_E\$ and different \$R_1\$ and \$R_2\$.

What limitations should be considered in order to choose best \$R_E\$?

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    \$\begingroup\$ R1 and R2 do not stabilize against HFE changes; this is thanks to RE. R1 and R2 cannot precisely program a specific collector current without RE. Moreover the, R1 and R2 bias is susceptible to VCC varations; RE helps with that too. More VCC, means more current through the transistor, means higher voltage at the top of RE, means transistor fights the effect of more VCC. \$\endgroup\$
    – Kaz
    Nov 21, 2013 at 0:27

2 Answers 2

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No, Re does effect voltage gain. To first approximation, the voltage gain of \$V_o/V_i\$ is \$-R_c/R_e\$ at DC. Due to \$C_1\$, the gain increases with frequency near and above the rolloff frequency of the filter formed by \$R_e\$ and \$C_1\$.

To chose \$R_e\$, make it \$R_c/G\$, with \$G\$ being the desired voltage gain of the circuit. G should be a few times less than the current gain of the transistor, else the gain won't be predictable and will have significant dependence on the transistor gain. Being dependent on the the transistor gain is not good for repeatability since transistor gains vary quite a lot even within the same model. That is why datasheets usually only tell you the minimum guaranteed gain at a few operating points. The actual gain of any one part can be significantly more, like 5-10x the minimum. When predictable gain matters, you design the circuit so that any transistor with a gain from the guaranteed minimum to infinity will still let the circuit operate correctly. A emitter resistor setting a smaller gain is one way of achieving this.

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  • \$\begingroup\$ Excuse me, But I can't understand the exact meaning of the gains. Is it the Vo/Vi or h_fe or something else? And also why should it be few times less than current gain of transistor? And is h_fe the current gain of transistor? \$\endgroup\$ Nov 20, 2013 at 20:01
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    \$\begingroup\$ @MohammadEtemaddar: Generally, the voltage gain of a circuit is the number of millivolts by which the output will change for each millivolt of change on the input. The current gain of a transistor is essentially hFE, though hFE is just a number and actual gain may vary with operating conditions. \$\endgroup\$
    – supercat
    Nov 20, 2013 at 21:17
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    \$\begingroup\$ @MohammadEtemaddar: I think it may be helpful to imagine the circuit behavior given an "ideal infinite gain" transistor where Ice will be zero when Vbe is below 0.7 volts or Vce is below 0.2 volts, and infinite if neither voltage is above its indicated threshold; Ibe will be zero unless Vbe is over 0.7 volts or more and Vce is less than 0.2 volts. One could then model the circuit behavior using some linear inequalities which would then resolve into piecewise-linear equations. \$\endgroup\$
    – supercat
    Nov 20, 2013 at 21:21
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And also it has no effect on voltage gain.

Assuming \$C_1\$ is "large enough", then for the frequencies of interest, the emitter is at AC common and thus the phrase common emitter.

I'll assume that the intent of \$C_1\$ is to bypass AC signals around \$R_E\$. In that case, the small-signal voltage gain is approximately:

$$\dfrac{v_o}{v_i} \approx -g_m R_C$$

where

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

Since the DC collector current \$I_C\$ will depend on the value of \$R_E\$, it is necessarily the case that \$R_E\$ affects the voltage gain

Now, it is straightforward to show that the equation for the DC collector current is:

$$I_C = \dfrac{V_{CC}\frac{R_2}{R_1+R_2} - V_{BE}}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}} $$

Examining the denominator of the right hand side, note that if you want \$I_C\$ to be relatively stable against variations of \$\beta\$, the value of \$R_E\$ should be chosen such that:

$$R_E \gg \dfrac{R_1||R_2}{\beta_{min} + 1}$$

For example, assume \$R_1||R_2 = 10k\Omega\$ and \$\beta_{min} = 99\$ then

$$R_E \gg \dfrac{10k\Omega}{100} = 100 \Omega$$

In this case, you want \$R_E \ge 1k\Omega\$ for good AC gain stability. But this is only a constraint. There may be other constraints on \$R_E\$ that require you find a compromise.

Also, you can add another degree of freedom by adding a resistor \$R_3\$ in series with \$C_1\$.

Then, the AC voltage gain is approximately:

$$\dfrac{v_o}{v_i} \approx - \dfrac{R_C}{R_E || R_3} $$

assuming \$R_E || R_3 \$ is "large enough".


The OP, in comments below, disputes the correctness of the \$I_C\$ equation so I will derive it here.

Find the Thevenin equivalent of the circuit connected to the base to be:

$$V_{BB} = V_{CC} \dfrac{R_2}{R_1 + R_2}$$

$$R_{BB} = R_1 || R_2 $$

Now, apply KVL around the base-emitter loop:

enter image description here

$$V_{BB} = I_BR_{BB} + V_{BE} + I_ER_E = \dfrac{I_C}{\beta}R_{BB} + V_{BE} + \dfrac{I_C}{\alpha}R_E$$

Collect terms and rearrange:

$$I_C = \dfrac{V_{BB} - V_{BE}}{\frac{R_{BB}}{\beta} + \frac{R_E}{\alpha}} = \dfrac{V_{CC}\frac{R_2}{R_1+R_2} - V_{BE}}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}}$$

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  • \$\begingroup\$ The \$V_{CC}\$ term in the equation for \$I_C\$ should be instead \$V_{CC} \frac{R_2}{R_1 + R_2}\$ but for some reason, I continue to get an error submitting the edit. \$\endgroup\$ Nov 20, 2013 at 22:25
  • \$\begingroup\$ I am disagree with I_c = (V_B - V_BE)/(R_B/beta + R_E/alpha) \$\endgroup\$ Nov 21, 2013 at 6:49
  • \$\begingroup\$ @MohammadEtemaddar, the equation you wrote is not the equation I wrote. The equation I wrote is correct. It's just a straightforward application of Thevenin's theorem and KVL. \$\endgroup\$ Nov 21, 2013 at 11:59
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    \$\begingroup\$ @MohammadEtemaddar, my equation is correct. You're not interpreting it correctly. The numerator is not \$V_E\$. \$\endgroup\$ Nov 21, 2013 at 13:10
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    \$\begingroup\$ @MohammadEtemaddar, I have added the derivation of the equation to my answer. \$\endgroup\$ Nov 21, 2013 at 13:20

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