A little overwhelmed with BJT amplifier design


Joined Dec 26, 2010
All true, but what I was trying to get across was the relatively high value of the supply needed in order to get the power below 50mW.
Since there is no output level specified in the OPs problem, why would a large supply voltage be required to obtain a solution at a lower power dissipation than 50mW?

If we accept the OPs requirement for a gain of 25 when working into the rather odd load impedance of 16.81kΩ, we require the transistor to provide a gm of about 25/16.81kΩ or about 1.5 milli-Siemens. Even if we allow for parallel loading from a reasonable DC collector load of say 10kΩ, reducing the total load to 6.27kΩ* this does not rise all that much to about 25/6.27kΩ, about 4mS. *Ignoring the transistor's own output impedance.

We can estimate the minimum possible collector current by assuming gm≈Ie/26mV, so Ie = 26mV*gm = 26mV*0.004mS = 0.1mA approximately. This is a really small current, and would only develop 1V in a 10kΩ load, so potentially the whole thing could be run at a power supply voltage of something like 3V, consuming maybe 0.35mW.

Let's not be too unreasonable though, the signal level obtainable would be pretty low this way, especially since the transistor is working with no feedback and so will distort very noticeably. If the collector current is made say four times bigger to 400μA, and the supply voltage is raised to say 10V, get more gain available, we can afford to have a resistor in the emitter circuit. How much? We need the total emitter impedance to be 6270/25 = 251, and re = 26/0.4 = 65, so the external emitter impedance is 251-65 = 186Ω

The picture shows what this ended up with: I don't propose to flog through describing every resistor value (the OP needs to do that), but it illustrates the point that a high voltage is not needed. I have kept the huge coupling cap values apparently specified in the OPs problem, although there is normally little point in the input and output capacitors being as big as the emitter decoupling. Power consumed should be about 4.3mW.

Edit: But where is the 186Ω resistance? It's left for the student to find.


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Joined Feb 19, 2009
Correct, my bad. I read it as one resistor needs to be over 500k (other thread), and all caps needed to be equal to 1uF, so I was solving a different problem and that's why it looks silly.