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# Help me to analysis that regular Circuit

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#### hbaocr

##### Full Member level 4
Hello
Plz help me to solve my problem. What Is the uasage Of C1 cap on Your circuit be low. If there is not C1 , How The quality of Circuit changes!

the capacitor C1 charges to the DC voltage set by the voltage divider and the capacitor prevents any change in voltage given by voltage divider due to the impedance of the base of the darlington pair.....

hi,A.Anand Srinivasan
thank for your helping,but Your idea I , Can I think C1 charge DC voltage set by Votage Divider, And discharge to R4 (load of that circuit). C1 due to R4 as the resistor which value is R4*gain of Q1*gain of Q2.So Time of RC is t=R.C =R4*gain of Q1*gain of Q2*C1.
And If I remove C1 from that Circuit, How will the quality of that circuit change? Because, arcording to my calculation, We can Control voltage output Const depends on range of correlative LOAD without depending on C1 value.

changing the capacitor wont have much effect on the circuit because the base drive of a darlington pair is very low and hence the capacitor is not gonna have appreciable effect.... if you try with a single transistor you'll surely know the difference...

hi, thank for you helping! But Arcoding to you the larger The Cap C1 we choose , The better the quality of It is.So what is suitable Value we can choose for C1

the larger capacitance value increases the smoothening of the output and hence it is good but with larger capacitance the leakage increases and hence it involves a trade off.... the best value can be found only by simulating or trial and error....

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