Determine the input impedances of a circuit

• Engineering
• Granger
In summary: I went ahead and made my own transistor model and I found that the input impedance is actually closer to 3.3k x Q2-β.The answer depends on the β of Q2 which can vary widely from one transistor of the same type to another. So Ri2 could be anywhere from about 80 x 3.3K||1K to about 300 x 3.3||1K.
Granger

Homework Statement

I'm studying a circuit with BJT's and I'm asked to determine the input resistances of the two amplification steps of the circuit. The circuit I'm analyzing is the following one:

Homework Equations

3. The Attempt at a Solution [/B]

To determine the input resistances I performed an AC analysis and obtained the following results:

I know that the values are not perceptible but they are Rin1=23.272231 kohm and Rin2=220.77106 kohm.
I decided to choose a somewhat random value in the zone of the graphic where the curves are constant (I'm not sure if this is correct...).

Well this value are very different from what I obtained when I analyze the theoretical values of the input resistances.

Using:

$$R_{i2}=(β+1) R_{E3}//R_L+r_{π2}$$

where r_{π2} is 9.375 kohm and beta is 300.
I obtained R_i2=240.375 kΩ

And for Ri1

$$R_{i1}=R_1//R_2//(r_π1+R_{E1} (1+β))$$
I obtained R_i1=20.183kohm.

I used the small-signal analysis models to determine this...

Now this is odd. I'm getting an error of about 15% on Ri1 and 7% on Ri2. Am I doing something wrong or are this deviations perfectly normal and dependent on the methods used on the theoretical analysis and on LTSpice. Can someone help me clarify this?

Attachments

• 8MBk7.png
8.5 KB · Views: 833
• LiCf2.png
16.1 KB · Views: 566
220K is obviously way off; the input impedance will br less than the parallel combination of R1 and R2 which is 24.8K ohms.
If you assume a high enough frequency where C1 and CE are essentially infinite then the 24.8K ohms should be paralleled with β times RE1.

rude man said:
220K is obviously way off; the input impedance will br less than the parallel combination of R1 and R2 which is 24.8K ohms.
If you assume a high enough frequency where C1 and CE are essentially infinite then the 24.8K ohms should be paralleled with β times RE1.

Hi! I did a new measure considering the peak of the output voltage as the frequency to measure my resistances and obtained 19.6 k for Ri1 (which is pretty satisfactory)... But 200 kohm for Ri2! What am I doing wrong?

Granger said:
Hi! I did a new measure considering the peak of the output voltage as the frequency to measure my resistances and obtained 19.6 k for Ri1 (which is pretty satisfactory)... But 200 kohm for Ri2! What am I doing wrong?
What is Ri2? If the output impedance you also need to specify if RL is incuded. I would think not.

Look at the circuit: if you change RL how does that change the output voltage? Assume C2 is large.

rude man said:
What is Ri2? If the output impedance you also need to specify if RL is incuded. I would think not.

Look at the circuit: if you change RL how does that change the output voltage? Assume C2 is large.

Ri2 is the INPUT impedance of the second step of amplification. Yes, RL charges the output of the circuit

What makes you think Ri2=200K is wrong? It's in the right ball park.

rude man said:
What makes you think Ri2=200K is wrong? It's in the right ball park.

The fact that I obtained 240K in my theoretical analysis... It's an error of 17%, isn't this too much? I'm afraid 240k is the wrong value, but I revised my calculations and I don't see the mistake.

Do you have any idea what's going wrong?

17% is actually a pretty good correspondence between simulation and hand analysis. Keep in mind if you built a bunch of these circuits and measured them you'd get a pretty big range of input impedances due to resistor tolerance and transistor parameter spread.

As for what is wrong, well for one thing the equations for r_pi assume the transistor is unilateral and it isn't in reality.

Also, did you make the transistor model yourself? If you didn't it contains subtleties not included in your hand analysis (e.g. beta is a function of Ic but you are most likely taking it as a constant).

Granger said:
The fact that I obtained 240K in my theoretical analysis... It's an error of 17%, isn't this too much? I'm afraid 240k is the wrong value, but I revised my calculations and I don't see the mistake.
The answer depends on the β of Q2 which can vary widely from one transistor of the same type to another. So Ri2 could be anywhere from about 80 x 3.3K||1K to about 300 x 3.3||1K.

The idea is that for a ΔV change at Q2-e a change in Q-b current of ( ΔV/3.3K) /Q2-β is required. Since Q2-b and Q2-e follow each other pretty closely, the input impedance is close to 3.3k x Q2-β.

analogdesign said:
As for what is wrong, well for one thing the equations for r_pi assume the transistor is unilateral and it isn't in reality.

Also, did you make the transistor model yourself? If you didn't it contains subtleties not included in your hand analysis (e.g. beta is a function of Ic but you are most likely taking it as a constant).

Hi! Yes I considered the theoretical model given in class, that takes beta as a constant. I also noticed that the LTSPICE model of the transistor includes a finite value for the Early voltage, while in the theoretical analysis we are instructed to assume that this voltage is infinite (therefore, we don't include the the resistor r_o).
I don't really understand what you said about the transistor being unilateral, can you give me more details about it?
I'm relieved to hear that my values are not that absurd. If I have to justify it what do you think are the main reasons for this to happen. I'm thinking about the Early voltage aspect I mentioned earlier and also the fact that I'm consider a current that is slightly above the real quiescent point current when I'm calculating r_pi. Beta is also slightly inferior in LTSPICE. However that is common to both input impedances. I'm starting to think that the main reason of this discrepancy is that when I'm doing the theoretical calculus I'm considering the second stage of amplification separated from the circuit, and basically determining it using associations of resistor. On the other hand LTSPICE computed this impedance using the plots of currents and voltages of the global circuit, and that will affect the calculations since there nodes and meshes created by the association influences both circuits. Is this correct? Do you think there's more that shall be mentioned. Thank you for your help!

Unilateral means that the signal transmission only goes one way. Bilateral means it can also go from input to output. In fact, the capacitance between the collector and base can couple signal back and modify the impedances, that's why I meant. You can try setting that capacitance to zero in your spice model and see if you get closer to what you expect.

I think that yes, the finite early voltage would change things somewhat. This is primarily because it will couple the two gain stages in ways that are difficult to calculate, so yes, your thinking that treating the two stages as independent (i.e. treating them as unilateral amplifiers) is part of the difference between calculated and simulated.

What is the definition of input impedance?

Input impedance is a measure of the resistance that an electrical circuit presents to an input signal. It is the ratio of the input voltage to the input current, and is typically measured in ohms.

Why is it important to determine the input impedance of a circuit?

Determining the input impedance of a circuit is important because it allows us to understand how the circuit will respond to an input signal. It also helps us design and optimize circuits for specific applications.

How is the input impedance of a circuit calculated?

The input impedance of a circuit can be calculated by dividing the input voltage by the input current. However, in more complex circuits, it may be necessary to use mathematical equations or simulation software to determine the input impedance.

What factors affect the input impedance of a circuit?

The input impedance of a circuit can be affected by several factors, such as the type and number of components in the circuit, the frequency of the input signal, and the configuration of the circuit.

How can the input impedance of a circuit be measured experimentally?

The input impedance of a circuit can be measured experimentally by using an impedance analyzer or by using a multimeter to measure the voltage and current at the input of the circuit. These measurements can then be used to calculate the input impedance.

• Engineering and Comp Sci Homework Help
Replies
0
Views
606
• Engineering and Comp Sci Homework Help
Replies
6
Views
2K
• Engineering and Comp Sci Homework Help
Replies
8
Views
1K
• Engineering and Comp Sci Homework Help
Replies
4
Views
2K
• Engineering and Comp Sci Homework Help
Replies
2
Views
986
• Engineering and Comp Sci Homework Help
Replies
9
Views
1K
• Engineering and Comp Sci Homework Help
Replies
29
Views
4K
• Engineering and Comp Sci Homework Help
Replies
2
Views
2K
• Electrical Engineering
Replies
20
Views
791
• Engineering and Comp Sci Homework Help
Replies
2
Views
1K