Common Emitter Amplifier Design Help

Thread Starter

Lime4

Joined Oct 11, 2018
7
Problem 4.79. I am having trouble finding values for my capacitors and for my input voltage and frequency. I have all of my resistor values ready but I am stuck on this portion of the design process.
 

MrChips

Joined Oct 2, 2009
19,416
Welcome to AAC!

There are three capacitors shown in the circuit schematic.
Label the capacitors C1, C2, C3.
What is the function served by each capacitor?
How does the behaviour of the circuit depend on the value of each capacitor and the signal frequency?

Show your circuit and all your work for determining the values of all resistors and capacitors.
 

Jony130

Joined Feb 17, 2009
5,022
This problem does not give any information about the amplifier bandwidth. So you do not need to calculate the capacitor values.
C = ∞ mens the Xc of a capacitor is a much less than the resistance seen by the cpacitor.
For exampel for C1 ---> you whant Xc1 to be much less than the input impedance of the stage at the lowest frequency you care about.
 

Thread Starter

Lime4

Joined Oct 11, 2018
7
The capacitor at the input (C1) couples the signal source to the base.
The capacitor at the output (C2) couples the amplified signal at the collector to the load.
The capacitor parallel to Re (C3) is a bypass capacitor that provides low impedance for the ac emitter current.

Large capacitors provide low impedances at signal frequencies but with low frequencies the capacitors will reduce the gain of the amp.

I understand that for the most part but I just can't figure out how to find the capacitor values. I've tried using the input impedance and output impedance to find values for C1 and C2 but when I put those into my spice program the circuit fails.
 

Thread Starter

Lime4

Joined Oct 11, 2018
7
For instance I chose Cap values as 1 pF, Vin is 100mV and f is 1kHz. My output voltage is 0 at those settings.
 

Thread Starter

Lime4

Joined Oct 11, 2018
7
Okay, I just realized I was reading what you wrote incorrectly and I figured it out. I was using the equation for impedance, frequency and capacitance incorrectly so I needed to increase the capacitance.
 

Jony130

Joined Feb 17, 2009
5,022
I needed to increase the capacitance.
Exactly
C1 = 1/ (2 * pi * F * (Rin + Rsource) ) ≈ 0.16/(F * Rin) so for Rin = 2kΩ and Fc = 20Hz we get C = 0.16/(20Hz * 2kΩ) = 4.7μF.
C3 ≈ 0.16/(F * 1/gm) ≈ 0.16/( 20Hz * 25mV/2mA) = 680μF
 
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