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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."

schematic

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?

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    \$\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
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    \$\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

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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.

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