2
\$\begingroup\$

This is a very basic question. I want to have a very simple BJT Signal Amplifier with variable voltage gain. Let's think about a common emitter with an emitter resistor like this:

enter image description here

The input signal is a sine wave and the signal amplitude is small enough to be considered as small signal (around 500mV pk-pk ) and RB,RE and RC are selected properly to work in good bias condition . According to the equation we expect the voltage gain( AV) to be approximated as AV=-RC/RE:

enter image description here

So I expect by turning the potentiometer, I should see different gains. This was on the paper but in reality it does not happen exactly: the gain changes well but some wave form distortion is seen in some frequencies for example in my case these distortions look like skewness to left in 2.5MHz, 7MHz ,.... What am I missing ?

\$\endgroup\$
  • \$\begingroup\$ When you say "in reality", do you mean simulation or a real circuit? \$\endgroup\$ – Vasiliy Oct 8 '13 at 14:44
  • \$\begingroup\$ @Vasiliy in a real circuit \$\endgroup\$ – Aug Oct 8 '13 at 14:46
  • 1
    \$\begingroup\$ Take a look at how much your bias point changes when you vary RC. Large excursion of bias point can lead to waveform distortion. \$\endgroup\$ – Michael Karas Oct 8 '13 at 14:54
  • 1
    \$\begingroup\$ A more typical approach is to design the transistor stage with fixed gain to meet the needs of the lowest level input signal. Then use a pot as a variable voltage divider at the input to provide attenuation of the input. \$\endgroup\$ – Michael Karas Oct 8 '13 at 14:57
  • 1
    \$\begingroup\$ The definition of small signal is a signal small enough not to cause distortion (aka nonlinear behavior). In your circuit you see some distortion. Therefore it was not a good assumption to think your input was small signal. There is no specific amplitude level that can be considered small-signal for all circuits, rather what is a small signal depends on the circuit you are analyzing (and even what node of the circuit you are looking at) and on what amount of distortion you are willing to consider negligible. \$\endgroup\$ – The Photon Oct 8 '13 at 16:21
3
\$\begingroup\$

This is not a good circuit for several reasons:

  1. The bias point is unpredictable, since it's quite dependent on the gain of the transistor. The effective resistance at the base is roughly RE times the gain. That means that the voltage divider of RB and this apparent base resistance varies strongly with the transistor gain, which in turn makes the bias point unpredictable since the gain of bipolar transistors is unpredictable. All you usually know is that it will be above some maximum level, not how high it could be. It's not at all unusual for actual devices to have several times higher gain than their minimum guaranteed gain.

  2. The gain adjustment also effects the bias point. Even worse, nothing is compensating for this change.

  3. The output impedance, and therefore the output rise time, change with the gain setting.

To fix these things, first create a circuit that has the highest gain you ever want and bias it predictably so that the output is close to the middle of its range with no signal in. Then put a potentiometer on the input of that amplifier to adjust overall gain. If the input signal has a DC offset, then AC couple the potentiometer both from the input signal and from the input to the amplifier. The potentiometer will therefore always operate with a DC level of 0, so changing the gain won't change any DC level anywhere.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

I'd like to add another answer from a different perspective.

There is a way to change the AC gain of the amplifier without disturbing the operating point and this is sometimes more desirable than a "volume" or "level" control in front of the amplifier.

For one example, use a fixed collector resistor \$R_C\$ and bypass the emitter resistor with a capacitor and rheostat:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, \$R_E \$ can be made large enough to provide a stable operating point while the AC emitter resistance is \$R_G || R_E \$.

When \$R_G\$ is very small, the AC gain is approximately \$-g_m R_C = -\dfrac{I_C}{V_T}R_C\$

For larger values of \$R_G\$, the AC gain is approximately \$-\dfrac{R_C}{R_G||R_E} \$

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.