I'm getting stuck at the problem 2.16 in Art of Electronics.

"Design a tuned common emitter amplifier stage to operate at 100kHz. Use a bypassed emitter resistor, and set the quiescent current at 1mA. Assume Vcc is 15V, L = 1mH, and put a 6.2k resistor across LC to set Q = 10. Use capacitive input coupling."


simulate this circuit – Schematic created using CircuitLab

Is this a correct design? I found two way to find C:

1) f = 1/(2*pi*sqrt(LC)) ( and I know f = 100kHz, L = 1mH)

2) Q = w0 RC ( Q is 10, R is 6.2k, and w is based on f = 100kHz)

And they give different results. How could I solve this problem?

  • 1
    \$\begingroup\$ You're missing the bypassed part of 'bypassed emitter resistor'... \$\endgroup\$
    – brhans
    Commented Jan 25, 2016 at 2:52
  • \$\begingroup\$ Yes, I forgot that. But how do I solve the capacitor problem? \$\endgroup\$ Commented Jan 25, 2016 at 2:55
  • \$\begingroup\$ Parallel RLC Circuit - Wikipedia What more of an explanation did you need? \$\endgroup\$
    – Dave
    Commented Jan 25, 2016 at 5:25
  • 2
    \$\begingroup\$ Your "different results" should not be very far apart. They can be accounted for by the fact that L = 1 mH and R = 6.2k are only approximately correct for 100 kHz and Q = 10. If L = 1 mH exactly, then R needs to be 6283 ohms (C = 2533 pF), and if R = 6200 exactly, then L needs to be 1.08 mH (C = 2341 pF). \$\endgroup\$
    – Dave Tweed
    Commented Jan 25, 2016 at 13:15

1 Answer 1


Your resistor needs to be 6.283K, then both methods will give you a value for C1 = 2,533 pf. It is only a matter of doing the calculations with higher precision components.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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