Art of Electronics Exercise 2.29: Finding Small Signal Gain

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elec_system_design

Joined Jun 23, 2017
49
Hi, I'm currently stuck on part b of exercise 2.29. How do you find the small signal gain of the circuit?

The circuit I have set up is an example of bad biasing as mentioned by the book.

I have attached a picture of the circuit to the thread.

Thanks!
 

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Papabravo

Joined Feb 24, 2006
21,322
Oh sorry, I forgot to mention that the circuit has the quiescent point at .5Vcc.
With the resistors you have a Q point of Vcc/2 is not going to happen. It is more like 0.77V or less. This is because with the emitter grounded, the base will not rise above about 0.7V
This really not a very good Q point because it is right at the conduction threshold of the transistor. The implication of this is that when the AC input goes negative it will drive the transistor deeper into cutoff and the output will go to the rail at Vcc because of the collector load resistor. When the AC input goes positive the transistor will begin to conduct, the collector current will increase, and the collector voltage will decrease. In this case the circuit only exhibits "gain" on the positive peaks.

In order to bias the circuit at Vcc/2 you need an emitter resistor and a bypass capacitor.
 
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Thread Starter

elec_system_design

Joined Jun 23, 2017
49
With the resistors you have a Q point of Vcc/2 is not going to happen. It is more like 0.77V or less. This is because with the emitter grounded, the base will not rise above about 0.7V
This really not a very good Q point because it is right at the conduction threshold of the transistor. The implication of this is that when the AC input goes negative it will drive the transistor deeper into cutoff and the output will go to the rail at Vcc because of the collector load resistor. When the AC input goes positive the transistor will begin to conduct, the collector current will increase, and the collector voltage will decrease. In this case the circuit only exhibits "gain" on the positive peaks.

In order to bias the circuit at Vcc/2 you need an emitter resistor and a bypass capacitor.
The Q point there is definitely around the 0.77V range, but for the Vout, it's at Vcc/2, which was what I was referring to. With what the exercise had given me, I was pretty sure that the Q point they were referring to was that of the output rather than the base voltage. The exercise states that a student adjusted R3 until the Q point was Vcc/2. I have tried changing the R3 to 1k, but that doesn't result in a base voltage of 10V. Then I tried to get the output voltage to 10V, which worked after having a very specific R3 value.
 

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elec_system_design

Joined Jun 23, 2017
49
How can you know that ?
Also what is the gain expression given in AoE?
The exercise says that the student that adjusted the R3 until the quiescent point was 0.5Vcc.

Initially R3 was at 100 ohms, so I did the same thing that the "student" did and adjusted the quiescent point to 0.5Vcc.

I'm running the circuit on LTspice and I do get a quiescent point of 0.5Vcc on the output.

For a common emitter amplifier without emitter degeneration, the voltage gain is said to the following: Gv = -Rc/re
 

WBahn

Joined Mar 31, 2012
30,303
Hi, I'm currently stuck on part b of exercise 2.29. How do you find the small signal gain of the circuit?

The circuit I have set up is an example of bad biasing as mentioned by the book.

I have attached a picture of the circuit to the thread.

Thanks!
I don't know that small signal analysis is even going to be applicable in this circuit. The quiescent operating mode of the transistor is very ill-defined and the validity of small signal analysis relies on the total range of operation about the quiescent operating point being reasonably linear. If it is operating near cutoff or saturation, then this approximation fails badly.

The first thing that strikes me as a bit absurd is that the top resistor in the voltage divider is known to just one sig fig while the bottom resistor is given to five sig figs. Then consider that even if you had a 0.5% 1 kΩ resistor, the uncertainty in the value of that resistor if 5 Ω, or more than 10% of the value of the bottom resistor.

Assuming that the resistor values are exact, with the base current neglected, the base voltage would be about 775 mV. The current in the bias resistors is 19.23 mA.

The collector current in the transistor if it were fully saturated would be slightly under 2 mA. Assuming a saturation beta of 10 would mean that the base current would be slightly under 200 uA. That would only drop the base voltage by about 8 mV. So I suspect that this transistor is HARD in saturation and you aren't going to get signal through it at all (or only a highly attenuated one).

Oh sorry, I forgot to mention that the circuit has the quiescent point at .5Vcc.
Which point is at 0.5·Vcc? The output? We don't have any information about the transistor, so I'm having to make up hopefully reasonable numbers.

If the output is at 10 V, then the collector current is 1 mA. Assuming a beta of 100, the base current is 10 uA. That is pretty negligible compared to the nearly 20 mA current in the bias resistors, so this transistor is sitting there in the active region with a base voltage of 775 mV. I'm skeptical, but you can turn a blind eye to it and just assume that's the case. If so, then go ahead and draw your small signal circuit and analyze it.
 

Jony130

Joined Feb 17, 2009
5,539
The exercise says that the student that adjusted the R3 until the quiescent point was 0.5Vcc.

Initially R3 was at 100 ohms, so I did the same thing that the "student" did and adjusted the quiescent point to 0.5Vcc.

I'm running the circuit on LTspice and I do get a quiescent point of 0.5Vcc on the output.
I see, so now try to change the transistor type in your simulation.

For a common emitter amplifier without emitter degeneration, the voltage gain is said to the following: Gv = -Rc/re
So what is the problem then ?

Icq = 10V/10k = 1mA and re = 26mV/Ic = 26Ω
 

WBahn

Joined Mar 31, 2012
30,303
The exercise says that the student that adjusted the R3 until the quiescent point was 0.5Vcc.

Initially R3 was at 100 ohms, so I did the same thing that the "student" did and adjusted the quiescent point to 0.5Vcc.
Okay, that explains the five sig figs on the value given for that resistor.

I'm running the circuit on LTspice and I do get a quiescent point of 0.5Vcc on the output.
What value did you get for the resistor? I would expect it to be something more like 30 Ω to 35 Ω. Interested to see what you got. What transistor are you using?

For a common emitter amplifier without emitter degeneration, the voltage gain is said to the following: Gv = -Rc/re
So what is re for the transistor operating at that quiescent current?

Do you understand where that equation comes from? You will be far better served if you are comfortable with working with the small signal equivalent circuits to do your analysis and also how to derive those small signal equivalents. It's fine to use equations to do the grunt work, but only if you understand where those equations come from.
 

Thread Starter

elec_system_design

Joined Jun 23, 2017
49
Okay, that explains the five sig figs on the value given for that resistor.



What value did you get for the resistor? I would expect it to be something more like 30 Ω to 35 Ω. Interested to see what you got. What transistor are you using?



So what is re for the transistor operating at that quiescent current?

Do you understand where that equation comes from? You will be far better served if you are comfortable with working with the small signal equivalent circuits to do your analysis and also how to derive those small signal equivalents. It's fine to use equations to do the grunt work, but only if you understand where those equations come from.
The value for the resistor is 40.308 as seen on the picture of the schematic. I did try values from 30-35 ohms, but they all failed in having a output voltage close to 10V. The npn transistor I'm using is actually just the generic npn transistor from LTspice, so I'm not really sure about its specs.

The equation for re is re = 25.3mV / Ic, for a BJT in room temperature right?
So, that would mean that the re value is 25.3mV/10^-3 = 25.3 ohms.

I have an okay understanding of the equations, though I definitely need to practice on knowing how to derive the small signal equivalents.
 

Thread Starter

elec_system_design

Joined Jun 23, 2017
49
I see, so now try to change the transistor type in your simulation.



So what is the problem then ?

Icq = 10V/10k = 1mA and re = 26mV/Ic = 26Ω
So I swapped the transistor from the generic LTspice npn transistor to the commonly used 2N3904 and the Vout isn't even .5Vcc anymore. It's running at a measly 14.6mV now.

This is neat, I feel like I should test out my circuit designs with a variety of transistors now to ensure a good level of flexibility.

With those values then, the gain is around -395?
 

WBahn

Joined Mar 31, 2012
30,303
What you're seeing is one of the reasons why this is a very poorly designed circuit. A well designed circuit should be tolerant of using other transistors that are anywhere in the ballpark of being roughly the same, or of temperature variations, or of component value variations within their tolerances.
 

WBahn

Joined Mar 31, 2012
30,303
For what it's worth, here is the original problem from AoE.
View attachment 132949
This makes something a lot more palatable. I was under the impression that the 40.308 Ω value was given in the problem and I was very surprised that it just happen to result in Vcc/2 in the TS's simulation. Now it makes sense -- the problem didn't specify a value at all (other than that it is less than 100 Ω and probably somewhere in the 50 Ω range) and the 40.308 Ω came solely from the TS's particular simulation.
 
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