0
\$\begingroup\$

I'm trying to design a common-emitter amplifier with a gain of 25. I seem to have gotten the correct gain by guessing with my resistor values. Can anyone tell me how to calculate resistor values and verify that I am indeed getting a gain of magnitude 25? enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Are you looking at DC gain or AC gain? \$\endgroup\$ – The Photon Dec 5 '17 at 3:09
  • \$\begingroup\$ Since your local NFB (re) is signal and temperature dependent, you will need global NFB added to some larger circuit that this is only an element of, if you are going to use an AC-grounded emitter. \$\endgroup\$ – jonk Dec 5 '17 at 5:13
  • \$\begingroup\$ Also, are you serious about the \$5\:\textrm{k}\Omega\$ loading on the output, when using \$R_C=25\:\textrm{k}\Omega\$? Where is this design supposed to be applied? Is this a classroom demonstration lab? Or? \$\endgroup\$ – jonk Dec 5 '17 at 11:24
  • \$\begingroup\$ R1,R2 calculated by the requirement needed to bias the transistor, Rc, Re calculated from small signal model also you might want to consider what kind of load you are driving in the design process. \$\endgroup\$ – rsg1710 Dec 5 '17 at 15:55
1
\$\begingroup\$

The transistor is biased wrongly. Its base voltage is much too high for the 25k to 1k ratio of the collector and emitter resistors. Therefore this transistor is saturated and is not amplifying.

\$\endgroup\$
0
\$\begingroup\$

To a first order, especially if the emitter resistor is relatively large, the unloaded gain of a common-emitter stage is RC/RE (collector resistor over the emitter resistor.) This assumes the source impedance is << than the input impedance.

To get more accurate you should include the dynamic emitter impedance, which at room temperature is about 1/(40*IC) where IC is the collector bias current.

Over temperature re = VT/IC where VT=kTq K is Boltzmann's constant, q is the charge of an electron, T is temperature. Then your gain is Rc/(re+RE).

So making sure that your physical Re is >> re will make sure your gain is constant over temperature and shifts in bias due to signal. (Which can cause non-linear distortion.)

Of course if you bypass the emitter resistor then above the breakpoint your gain will increase to Rc/re in the limit, until other parasitics come into play.

The advantage of bypassing the emitter resistor is more gain, but with a more stable DC operating point.

\$\endgroup\$
  • \$\begingroup\$ I am changing the values of R1 and R2 and they seem to be affecting the gain. How exactly are they affecting it? \$\endgroup\$ – user166481 Dec 5 '17 at 3:06
  • \$\begingroup\$ Changing R1 and R2 changes the current of the transistor, thus changing the emmiter dynamic resistor, and since you have the emmiter resistor bypassed, the gain will be heavily dependent on the operating point of the transistor, try removing the 1uf capacitor, gain will drop but will be more predictable. \$\endgroup\$ – S.s. Dec 5 '17 at 3:13
  • \$\begingroup\$ @A.J. is correct- When I initially answered the question I didn't notice the emitter resistor bypass cap for some reason, probably because you said you were getting a gain of 25 and RC/RE was 25. I edited my answer to include that information. \$\endgroup\$ – John D Dec 5 '17 at 15:08
0
\$\begingroup\$

You can remove the 1uF bypass capacitor to get a precise and temperature Independent and more stable gain.

Because in your case(with the 1uF cap) the gain is just Rc/re that's because the 1uf cap shorts the emitter resistor Re at the signal frequency.

Since re depends on many parameters like temperature and collector current(which can change by varying R1 and R2) it's difficult to get the exact gain you desire.

But with the 1uf removed the gain is Rc/(re + Re).And because of Re>>re(hundreds Of ohms compared to tens of ohms) the gain can be simplified to just Rc/Re.

Now as you can see the gain is much less dependent on temperature or collector current by a factor equals re/(re + Re).

And because of Rc/Re = 25K/1K = 25 simply removing the 1uf will give the desired gain.

Another problem that rises due to the bypassing of Re is the distortion of the output signal when large swings about the Quiescent point occur(The amplifier is not perfectly linear) and this can be solved by removing the bypass capacitor.

\$\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.