# Bias Stable Circuit Design with Capacitors

For the circuit, design a bias stable circuit such that Icq=.8mA and Vceq=5V. Let Bf=100. Using the results, determine the percentage change in Icq if Bf is in the range of 75<=Bf<=150. Repeat this if Re=1kOhms.

Rth=.1(1+b)re=.1*101*.5Kohms=5050 Ohms

I can do KVL around the b-e loop to find Vth, but how do I split the two resisters out of Rth? Anything could be r1*r2/(r1+r2)

The capacitors are confusing me. If they act as opens, then RL is irrelevant? By the way, what is the function of the emitter capacitor? (just curious)

Is the part B, do you think the author means the difference between the upper value of Icb and the lower value of Icb? Or what percent of icblower is icb upper? It's a little confusing.

Thank you.

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Please find your solution below.

part1.

Neglect the capacitors. Once you find the Rth, you should use the (r2/r1 +r2) to find Vth. becasue as ...

#### Solution Summary

For the circuit, design a bias stable circuit such that Icq=.8mA and Vceq=5V. Let Bf=100. Using the results, determine the percentage change in Icq if Bf is in the range of 75<=Bf<=150. Repeat this if Re=1kOhms.