Common emitter amplifier Electronics

In summary, this circuit has problems and the standard procedure for finding the operational DC point does not work.
  • #1
Alexa431
5
0
media%2F6c0%2F6c06edb6-6a1d-4b15-9fb8-489f29e37250%2FphpzK0HEa.png

Rb=40 ohms
RL=100 ohms
Vcc=15V

I need help in figuring this problem out, I understand if anyone here doesn't want to give me the answer but I would appreciate if someone would give me a formula in which I could solve the problem myself.
 
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  • #2
Alexa - are you sure about the value of Rb=40 Ohms? This is equivalent to a base current in the order of 100mA.
Perhaps Rb=40kohms?
 
  • #3
Yeah Its 40 ohms I checked again to make sure.
 
  • #4
This looks like a description of a lab exercise. Is this an experiment that you've carried out? What sort of help are you looking for?

Determining the gain theoretically will involve modelling the transistor to some extent.
 
  • #5
I agree with LvW -- that circuit looks to have problems. Is this for a lab? Or are you supposed to simulate it? Have you calculated its quiescent operating point?EDIT -- Dang it! Beaten out by gneill again! :smile:
 
  • #6
Well I was in charge of doing the theory part (formulas) and my partner the actual experiment if his numbers are wrong then, I would repeat the experiment myself but I just need to figure out how to solve the v0 on paper.
 
  • #7
You'll need to know the frequency range that was used and the capacitor value (its impedance will be of the same order of magnitude as the circuit's resistors unless the frequency is pretty high). You'll also need to know something about the transistor used, either its specific part number or know its β or have its characteristic curves on hand.
 
  • #8
The capacitor was 0.1uF, the B was 100
 
  • #9
All in all I just want a standard formula which could be used to solve this circuit and circuits like these, all I need is the theory part of it
 
  • #10
Alexa431 said:
The capacitor was 0.1uF, the B was 100
Well, with only the β you can start by finding the approximate DC operating point. You'll have to assume a typical value for the base-emitter voltage. Since the base current is going to be pretty substantial for this circuit (I'd guess on the order of a couple of milliamps thanks to the collector current pulling down the voltage at the collector and limiting the voltage at the "top" of RB), that base-emitter voltage will probably turn out to be somewhat higher than "typical".
 
  • #11
Alexa431 said:
All in all I just want a standard formula which could be used to solve this circuit and circuits like these, all I need is the theory part of it
Unless you can find this circuit already analyzed somewhere you won't find a "standard formula". There are many different configurations for transistors, each requiring their own analysis and producing different formulas. The approach here would be to replace the transistor with a simple equivalent model and then apply circuit analysis to the resulting circuit.
 
  • #12
Alexa431 said:
media%2F6c0%2F6c06edb6-6a1d-4b15-9fb8-489f29e37250%2FphpzK0HEa.png

Rb=40 ohms
RL=100 ohms
Vcc=15V
The 'standard way' to analyze a circuit like this is the following:
1. assume infinite beta so base current = 0
2. therefore, the input current is Vin/Zc = current thru the feedback resistor Rb. Zc = 1/ωC.
3. this gives you Vc = collector voltage.
4. sum currents to zero at the collector. This gives you Ic = collector current.
You can use this approach for either ac or dc analysis.
Problem is, your resistor values are way too low to make this approach work in the lab. Who came up with 40 and 100 ohms? What is the range of ω? What is Vcc? I predict a puff of smoke for this circuit ...
 
  • #13
Alexa431 said:
All in all I just want a standard formula which could be used to solve this circuit and circuits like these, all I need is the theory part of it

Yes - there is something like a "standard formula"; better: Standard procedure for finding the ioperational DC point:
You have to solve a system of two equations with two unknowns (assuming a base-emitter dc voltage Vbe=0.8V):

* Base current Ib=Ic/B=(Vc-0.8)/RB
*
Ic+Ib=Ic(1+B)=(Vcc-Vc)/RL

The two unknowns are the collector voltage Vc and the current Ic.
However, as mentioned already - this is a pure academical exercise without much practical relevance.
 
  • #14
Alexa, I have simulated the circuit using a power transistor - and it works!
However, because the voltage Vce is in the order of 1V only, the input signal must not exceed a value of app. 10mV.
This applies for a very large coupling capacitor (100uF) and a frequency which is sufficiently high (some kHz).
 
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Related to Common emitter amplifier Electronics

1. What is a common emitter amplifier?

A common emitter amplifier is a type of electronic circuit that amplifies a small input signal by using a transistor in common emitter configuration. It is commonly used in audio circuits, and can also be used for power amplification in radio frequency circuits.

2. How does a common emitter amplifier work?

A common emitter amplifier works by using a transistor to amplify a small signal. The input signal is applied to the base of the transistor, and the amplified output signal is taken from the collector. The emitter is connected to ground, and a biasing circuit is used to set the operating point of the transistor.

3. What are the advantages of a common emitter amplifier?

Some advantages of a common emitter amplifier include: high voltage gain, low output impedance, and relatively simple circuit design. It also has a high input impedance, meaning it can take in a wide range of input signals.

4. What are the limitations of a common emitter amplifier?

Some limitations of a common emitter amplifier include: high distortion levels, poor frequency response, and temperature sensitivity. It also requires careful design and selection of components to ensure stability and proper operation.

5. How can I improve the performance of a common emitter amplifier?

You can improve the performance of a common emitter amplifier by using negative feedback, which can reduce distortion and improve frequency response. You can also use higher quality components and adjust the biasing circuit to optimize the operating point of the transistor.

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