AC Circuit Analysis
AC analysis is pretty tricky to get a grasp on an first but with a good understanding of the principals and a bit of practice it's a piece of cake.
I prefer to use the Hybrid pi model for AC analysis but there are other models equally as valid. The Hybrid p model for the transistor is shown below:
rp is used to model the input resistance of the transistor. It's actually modelling the AC resistance of the forward biased Base-Emitter PN junction. vp is just the voltage across rp. Also note that the AC base current, ib, flows through rp.
Lower case letters are always used for AC signals. For example, ib is the AC signal base current whereas Ib is the DC base bias current. The exception to this is Vp, which I always seem to write using a captial V, but that's just my personal choice.
The diamond shaped box with the arrow is a 'Voltage Controlled Current Source'. It's important that you understand what this is. It is a current source, whose current is controlled by a voltage which is somewhere else in the circuit. In this case, the current is controlled by the voltage across rp, which is Vp. gm is just a constant which determines how much of a change in current is caused by a certain change of the controlling voltage. The units of gm is amps per volt or, A/V.
An example,
for the voltage controlled current source shown above, a Vp of 5V peak to peak with a gm of 0.5A/V would result in a current flow of:
0.5A/V x 5V = 2.5A peak to peak.
It's also worth mentioning that from previous subjects, you learned that the resistance of an ideal current source is infinity. Hence whilst the input resistance to the circuit is rp, the output resistance of the circuit is infinity. In practice this is not true and the output resistance is just very large. This effect is modeled by another resistor ro connected between collector and emitter at the output but is not shown on the above diagram. It can be omitted as a simplification to the model.
So if gm is a constant for a given amplifier circuit, how do you work out it's value. There is a relationship relating gm to the collector current, it is:
gm=Ic/Vt
VT is another constant which is given by:
Lower case letters are always used for AC signals. For example, ib is the AC signal base current whereas Ib is the DC base bias current. The exception to this is Vp, which I always seem to write using a captial V, but that's just my personal choice.
The diamond shaped box with the arrow is a 'Voltage Controlled Current Source'. It's important that you understand what this is. It is a current source, whose current is controlled by a voltage which is somewhere else in the circuit. In this case, the current is controlled by the voltage across rp, which is Vp. gm is just a constant which determines how much of a change in current is caused by a certain change of the controlling voltage. The units of gm is amps per volt or, A/V.
An example,
for the voltage controlled current source shown above, a Vp of 5V peak to peak with a gm of 0.5A/V would result in a current flow of:
0.5A/V x 5V = 2.5A peak to peak.
It's also worth mentioning that from previous subjects, you learned that the resistance of an ideal current source is infinity. Hence whilst the input resistance to the circuit is rp, the output resistance of the circuit is infinity. In practice this is not true and the output resistance is just very large. This effect is modeled by another resistor ro connected between collector and emitter at the output but is not shown on the above diagram. It can be omitted as a simplification to the model.
So if gm is a constant for a given amplifier circuit, how do you work out it's value. There is a relationship relating gm to the collector current, it is:
gm=Ic/Vt
VT is another constant which is given by:
Vt=kT/q=25mV Aprox
This equation should be familiar from diode study. Hence:
gm=Ic/Vt=40,Ic(10)
So all you have to do to find gm is to calculate the DC current Ic and multiply by 40, remembering the units should be in A/V.
rp is easy to calculate as well, it is given by:
rp=b/gm(11)
Another useful thing to mention is that the Voltage controlled current source, "gm.Vp" is sometimes expressed in another form:
gm.Vp=b/rp.Vp=b.Ib(12) , where equation 11 has been rearranged and substituted in for gm and then Vp / rp is of course just equal to ib --> You can see this from ohms law looking at figure 9. The b.ib model for the voltage controlled current source comes in useful for the 'common collector' and 'common emitter with emitter resistor' amplifiers.
So that's the Hybrid p model, all you have to do is replace the transistor in your circuit with this model, and do some analysis.
This equation should be familiar from diode study. Hence:
gm=Ic/Vt=40,Ic(10)
So all you have to do to find gm is to calculate the DC current Ic and multiply by 40, remembering the units should be in A/V.
rp is easy to calculate as well, it is given by:
rp=b/gm(11)
Another useful thing to mention is that the Voltage controlled current source, "gm.Vp" is sometimes expressed in another form:
gm.Vp=b/rp.Vp=b.Ib(12) , where equation 11 has been rearranged and substituted in for gm and then Vp / rp is of course just equal to ib --> You can see this from ohms law looking at figure 9. The b.ib model for the voltage controlled current source comes in useful for the 'common collector' and 'common emitter with emitter resistor' amplifiers.
So that's the Hybrid p model, all you have to do is replace the transistor in your circuit with this model, and do some analysis.
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