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I have a question regarding \$C_1\$ capacitor - do we really need it in this circuit if we know, that the input signal will have no offset (i.e. \$\overline{V_{IN}}=0V\$)?

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do we really need it in this circuit if we know, that the input signal will have no offset

You mean, can we remove C1 (replace it with a short) if the DC voltage of Eg is 0 V ?

Well in this case: no.

Look at the circuit and think what would happen if you would remove C1. Let's look only at the situation for DC. Since the DC voltage of Eg = 0 V, we can remove Eg as well. Now Rg connects the base of the NPN to ground.

Under normal operation, in the circuit as you have drawn it (with nothing removed), what is the DC voltage at the base of the NPN?

Is it 0 V or something different?

Now what would happen if we connect Rg between that base of the NPN and ground?

What will happen to the base current Ib? Will it still all flow into the base of the NPN? I think not. I think part of the current will flow through Rg.

So what does C1 do? It blocks Ib from flowing into Rg and Eg.

Only if you would make sure that there is no DC voltage (VDC = 0 V) present at the base of the NPN then C1 could be removed. But that would require a negative supply voltage. Another solution is to give Eg a DC voltage (offset) that is exactly the same as the DC voltage at the base of the NPN.

But that DC voltage depends on Vbe, Ic and Re. Especially Vbe is very temperature dependent, making it difficult to always have the right offset. It is much easier to just use a DC blocking capacitor: C1.

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