Voltage amplification done by a transistor amplifier

In summary, the conversation discusses the equivalent circuit for a p-n junction in forward bias and reverse bias. It is determined that the forward bias junction is equivalent to an input resistance in series while the reverse bias junction is equivalent to an output resistance in parallel. The output circuit acts as a current source with current ##I_C##. The simplified transistor model is also mentioned, with the input resistance being a result of combining various factors into one amplifier parameter. The issue of the connection at B is brought up, with it being determined that ##I_B >>I_C##.
  • #1
Pushoam
962
52

Homework Statement


upload_2017-9-17_16-12-33.png
upload_2017-9-17_16-41-8.png


Homework Equations

The Attempt at a Solution


The equivalent circuit is drawn below.
It is assumed that the potential drop across the pn junction in forward bias is 0 V(I can't take it 0.7 V as the input voltage is 0.5 V). In this case the p-n junction is equivalent to a connecting wire-piece.
Since ##\alpha ## is nearly equal to 1, ##I_B# is taken to be 0.

upload_2017-9-17_16-59-14.png


So, the current flowing through ##R_{in}## is ##I_E = \frac{500~mV}{20~\Omega} = 25 ~mA##
In fact, the above figure is wrong.Right?
It is the p-n junction in forward bias which should get replaced by Rin and np junction in reverse bias which should get replaced by Rout.
Then, the potential drop across the diode in forward bias is equal to the potential drop across the Rin i.e. 500 mV and the potential drop across the reverse bias is the potential drop across the Rout.
Potential drop across Rout = ## I_C R_{out} = 2500 V ##........(1)
Using Kirchoff's Voltage loop law, this voltage drop should be equal to the voltage drop across## R_C##.
But, due to Ohm's law, voltage drop across ##R_C = I_C R_C = 25V##......(2)
So, there is a contradiction.
In (1), I have considered the n-p junction as a linear element. I think this is wrong. The n-p junction could not be replaced by Rout. Then , in this case, there is no need to know Rout. Right?
So, does it mean that , in general, in transistor, the forward bias p-n junction could be taken as a linear element, but the reverse bias p-n junction has to be taken as a non-linear element?

The voltage amplification is : ## \frac{25V}{0.5V}=50##
 

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  • #2
I'm a bit rusty but...

Potential drop across Rout = ICRout=2500VICRout=2500V I_C R_{out} = 2500 V ........(1)

No that's not correct. The output looks like a dependant current source with the 100K output impedance in parallel. So the 1k load resistor is in parallel with the 100K output impedance.

300px-Common-base_small_signal.svg.png



I agree that IE = 25mA
IB is small so IC = IE = 25mA
The output impedance is high compared to the load so the output impedance can be ignored.
So the voltage across RC is 25mA * 1K = 25V
So the voltage gain is 25/0.5 = 50.
 
Last edited:
  • #3
CWatters said:
No that's not correct. The output looks like a dependant current source with the 100K output impedance in parallel.
O.K.
So, in the equivalent circuit, the forward bias p-n junction gets replaced by an input resistance in series while the reverse bias junction gets replaced by an output resistance in parallel.
The output circuit acts as a current source with current ##I_C##.

upload_2017-9-19_10-54-50.png


Is this right?
 
  • #4
No, I would simplify the transistor model to an amplifier model and draw it like this...

Common Base Amp.png


When the problem statement says the input resistance is 20R you shouldn't think of the circuit as a transistor and a separate 20R resistor. The 20R results from combining the effects of the transistor, it's bias circuit, feedback, base current (eg the whole thing) into one amplifier parameter for the input.
 
  • #5
I got it.
The problem is with the connection at B in my circuit. The connection at B implies that ##I_B >>I_C##.
Thank you.
 

Related to Voltage amplification done by a transistor amplifier

1. What is voltage amplification?

Voltage amplification is the process of increasing the amplitude or strength of an electrical signal. This is typically done using electronic components, such as transistors, to boost the voltage level of a signal.

2. How does a transistor amplifier work?

A transistor amplifier uses a small input signal to control a larger output signal. The input signal is applied to the base of the transistor, which then amplifies the signal and outputs it through the collector terminal. The amount of amplification is determined by the transistor's characteristics and the circuit design.

3. What is the role of a transistor in voltage amplification?

A transistor is a semiconductor device that is commonly used in electronic circuits. In voltage amplification, the transistor acts as an amplifier by controlling the flow of current between its collector and emitter terminals. By varying the input signal, the transistor can amplify the output signal to a desired level.

4. What are some advantages of using a transistor amplifier for voltage amplification?

Transistor amplifiers offer several advantages over other methods of voltage amplification. They are small in size, have low power consumption, and can operate at high frequencies. They also offer high gain and low distortion, making them ideal for use in audio and radio frequency applications.

5. What are some common types of transistor amplifiers used for voltage amplification?

There are several types of transistor amplifiers used for voltage amplification, including common emitter, common collector, and common base configurations. Each type has its own advantages and limitations, and the best choice will depend on the specific application and requirements.

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