I have some more questions in my ongoing saga re CE amplifiers.
(1) My Heathkit course manual calculates C1 And C2 using the following:
C1 = 3.18 / (f1 * Rin) = 3.18 / (100 Hz * 3.94K) = 8.07 uF
C2 = 3.18 / (f1 * RL) = 3.18 / (100 Hz * 5.6K) = 5.68 uF
Why wouldn't they use 0.159 rather than 3.18 for C1 and C2?
C3 = 0.159 / (f1 * RNF) = 0.159 / (100 Hz * 5 Ohms) = 318.47 uF
(2) I came up with a formula for computing the value of RNF:
RE = 500 ohms
RL = 5.6K
RC = 2.5K
re' = 37mV / 3mA = 12.3 ohms
Av = 100
RNF = (RE * ((RC * RL) - (Av * RC * re') - (Av * RL * re'))) / ((Av * RC * re') + (Av * RC * RE) + (Av * RL * re') + (Av * RL * RE) - (RC * RL)) = 5 ohms
I hope that is correct.
(3) How do I compute the ac current flowing in RNF as I want to calculate the wattage
for this resistor.
I think that I am now in the home stretch!
David
(1) My Heathkit course manual calculates C1 And C2 using the following:
C1 = 3.18 / (f1 * Rin) = 3.18 / (100 Hz * 3.94K) = 8.07 uF
C2 = 3.18 / (f1 * RL) = 3.18 / (100 Hz * 5.6K) = 5.68 uF
Why wouldn't they use 0.159 rather than 3.18 for C1 and C2?
C3 = 0.159 / (f1 * RNF) = 0.159 / (100 Hz * 5 Ohms) = 318.47 uF
(2) I came up with a formula for computing the value of RNF:
RE = 500 ohms
RL = 5.6K
RC = 2.5K
re' = 37mV / 3mA = 12.3 ohms
Av = 100
RNF = (RE * ((RC * RL) - (Av * RC * re') - (Av * RL * re'))) / ((Av * RC * re') + (Av * RC * RE) + (Av * RL * re') + (Av * RL * RE) - (RC * RL)) = 5 ohms
I hope that is correct.
(3) How do I compute the ac current flowing in RNF as I want to calculate the wattage
for this resistor.
I think that I am now in the home stretch!
David
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