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Questions about Transistor Amplifier Circuit
- Thread starter skinner89
- Start date
First, it is a valid circuit. It would work if you built it and asked it to amplify.
R1 and R2 set the bias voltage on the base at 12V/11
C1 keeps that voltage from going out the input terminal.
The bias voltage minus .7 is the voltage across R4. It's about 831 microamps.
That current through R3 causes about 3.9 volts across R3.
C2 takes R4 out of the gain equation. The Low frequency limit is 34 Hz.
C3 stops the DC voltage on the collector from getting out the output terminal.
R1 and R2 set the bias voltage on the base at 12V/11
C1 keeps that voltage from going out the input terminal.
The bias voltage minus .7 is the voltage across R4. It's about 831 microamps.
That current through R3 causes about 3.9 volts across R3.
C2 takes R4 out of the gain equation. The Low frequency limit is 34 Hz.
C3 stops the DC voltage on the collector from getting out the output terminal.
Most of your answers are close.
1) Yes. With an AC signal at the input, current will alternate between flowing in the I1 node (and out the I2 node) and vice-versa.
2) Yes. R1 and R2 establish the DC bias, or the operating point of the amplifier when there is no input. The intent is to keep the transistor in the active region as long as the input signal stays within a compatible region. However, it is not the intent to keep it in the active region regardless of the input signal -- you can apply signals that will drive it out of the active region (and do so without damaging anything).
3) C1 not only prevents the DC levels generated in the bias circuit from making it back to the signal source, but they permit the signal source to have a DC component that will be blocked from affecting the amplifier's bias. It decouples in both directions.
4) Again, it decouples both ways -- it keeps the bias signal at the transistor collector from affecting the follow-on circuit and it prevents the DC level of the follow-on circuit from affecting the bias of the amplifier. I would recommend not saying "bias we added at R1/R2" and, instead, saying something like "the bias voltage at the collector, as established by all four of the resistors." While R1/R2 are only there to play a role in establishing a bias level, the same is true for R3. Contrary, R4 plays a role both is establishing the bias and in the amplifier AC gain.
5) Remember these are called "circuits" for a reason. For a DC signal to flow, it must have a continuous path for current to flow in. You only have to break the path in one place to block the DC signal. You could, in fact, put capacitors in the I2 (and O2) paths and thereby completely decouple the input from the amplifier from the output. There are situations in which you want to do this. Apparently none of those apply in this case, but the analysis is identical in either case.
6) The gain of the amplifier is, roughly, the ratio of the collector resistance to the emitter path resistance. For the DC bias signal, which is very large, you need a small gain and so a relatively large Re. But for the small AC signals, you want a high gain and, hence, a low value of Re. The capacitor effectively removes it, leaving you with only the small effective emitter resistance of the transistor itself.
As for why you can't just use a diode for C1, remember that a diode only permits flow in one direction. The capacitor permits flow in both directions, which is needed by an AC signal, but blocks the DC flow in either direction. If you had a diode, not only would you corrupt the AC signal, but it would permit the DC signal to get through in one direction.
1) Yes. With an AC signal at the input, current will alternate between flowing in the I1 node (and out the I2 node) and vice-versa.
2) Yes. R1 and R2 establish the DC bias, or the operating point of the amplifier when there is no input. The intent is to keep the transistor in the active region as long as the input signal stays within a compatible region. However, it is not the intent to keep it in the active region regardless of the input signal -- you can apply signals that will drive it out of the active region (and do so without damaging anything).
3) C1 not only prevents the DC levels generated in the bias circuit from making it back to the signal source, but they permit the signal source to have a DC component that will be blocked from affecting the amplifier's bias. It decouples in both directions.
4) Again, it decouples both ways -- it keeps the bias signal at the transistor collector from affecting the follow-on circuit and it prevents the DC level of the follow-on circuit from affecting the bias of the amplifier. I would recommend not saying "bias we added at R1/R2" and, instead, saying something like "the bias voltage at the collector, as established by all four of the resistors." While R1/R2 are only there to play a role in establishing a bias level, the same is true for R3. Contrary, R4 plays a role both is establishing the bias and in the amplifier AC gain.
5) Remember these are called "circuits" for a reason. For a DC signal to flow, it must have a continuous path for current to flow in. You only have to break the path in one place to block the DC signal. You could, in fact, put capacitors in the I2 (and O2) paths and thereby completely decouple the input from the amplifier from the output. There are situations in which you want to do this. Apparently none of those apply in this case, but the analysis is identical in either case.
6) The gain of the amplifier is, roughly, the ratio of the collector resistance to the emitter path resistance. For the DC bias signal, which is very large, you need a small gain and so a relatively large Re. But for the small AC signals, you want a high gain and, hence, a low value of Re. The capacitor effectively removes it, leaving you with only the small effective emitter resistance of the transistor itself.
As for why you can't just use a diode for C1, remember that a diode only permits flow in one direction. The capacitor permits flow in both directions, which is needed by an AC signal, but blocks the DC flow in either direction. If you had a diode, not only would you corrupt the AC signal, but it would permit the DC signal to get through in one direction.
Unless you are using polarized capacitors, this is not an issue. Let's assume our input voltage looks like this:
Vin = Vdc + Vac(sin(wt))
where w is high enough to be in the AC range of this amplifier.
It doesn't matter what Vdc is, it could be +100V or -100V (as long as the resulting voltage across the capacitor doesn't exceed the limits of the capacitor used). But Vac, which is the signal that will get through the coupling capacitor, does have to be small enough to avoid moving the voltage of the base of the transistor around too much. In fact, you generally want to keep the voltage excursions pretty small in order to keep the properties of the amplifier more-or-less constant (i.e., not change the operating point of the transistor too much).
Now, consider if Vdc were chosen to match Vb, the bias voltage produced by R1 and R2. Under those conditions, Vac will result in the polarity of the voltage on the input coupling capacitor to change twice each cycle. But that's completely fine.
Vin = Vdc + Vac(sin(wt))
where w is high enough to be in the AC range of this amplifier.
It doesn't matter what Vdc is, it could be +100V or -100V (as long as the resulting voltage across the capacitor doesn't exceed the limits of the capacitor used). But Vac, which is the signal that will get through the coupling capacitor, does have to be small enough to avoid moving the voltage of the base of the transistor around too much. In fact, you generally want to keep the voltage excursions pretty small in order to keep the properties of the amplifier more-or-less constant (i.e., not change the operating point of the transistor too much).
Now, consider if Vdc were chosen to match Vb, the bias voltage produced by R1 and R2. Under those conditions, Vac will result in the polarity of the voltage on the input coupling capacitor to change twice each cycle. But that's completely fine.
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