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Basic transistor amplification circuit


I'm posting a theoretic question.

Given the basic transistor amplification circuit - I attached it as a screenshot.

The circuit works fine unless I remove the capacitor at the input voltage source.

As soon as it's removed, obviously, the circuit stops working, as the source voltage is not high enough to open the basis circuit.

But what exactly does the capacitor do?

In the books and articles I found they only say that it removes the direct current components from the input voltage, but this time, the input is a nice sinus voltage, without any direct current components, the capacitor is still an essential part of the circuit!

Could you explain me the real role of the capacitor?


The capacitor at the output doesn't stop the circuit working, but removing it produces also a different behaviour, although most books I found mark it's role similarly to the other capacitor that it has to remove direct current components.


So what are the two capacitors really for?

Thanks in advance!

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Message 1 of 3

The input coupling capacitor isolates the absolute level (DC level) between both side and keep AC signals passing.


When you remove the cap and connect both sides directly, the level of the signal work on the circuit directly and bias changes.

So the amplifier will not work as expected.


Your can put a probe before and after the cap to see their difference.

Message 2 of 3


Hi Makovsky2000 and tonysun1,



Coupling capacitors remove the DC component of a signal. Both input and output capacitors block the DC component on one side from entering the other. When both input and output are another stage of amplifier, the input and output capacitors play a similar role of isolating the bias levels of each side. At the ends of amplifier chain the coupling capacitors will have some variations in the role that the play.


For output capacitor you pointed out that removing it produces a different behavior but the circuit does not stop working. By different behavior, you may be referring to the different DC levels of the output signal. Removing the output coupling capacitor can prevent an amplifier circuit from working properly. Your circuit does not have a load so the bias level of the collector remains the same while the output signal is simply shifted upward. If there is a load of significantly lower resistance, the collector bias level will be affected and this can result to improper operation. Another factor to consider about direct output coupling is the nature of the load. For example, dynamic speakers can be damaged/destroyed by direct current.


For input capacitor, where the input comes from the signal source and not from an amplifier stage, its purpose is to let the signal ride on the DC bias level of the amplifier input. In your circuit the input bias level is established by R1, R2, and R4. With the input capacitor, the voltage applied to the base of transistor is the sum of DC bias and the AC input signal. Without the input coupling capacitor, the input signal will be applied directly to the transistor's base. Assuming an ideal source, the DC bias level will be disregarded or overridden. It can be easily seen that at the negative alternation of the source (Vin) the base-emitter junction will be reverse-biased and this is not the desired mode of operation.



Best regards,

G. Goodwin


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Message 3 of 3