Input Resistance of this BJT Amplifier

  • Thread starter IronaSona
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  • #1
IronaSona
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Homework Statement:
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Relevant Equations:
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Hi i got a question ,how do i calculate the input resistance ,of the RS resistor ,so i would get half of the voltage of the voltage source (500mA) .I've created the circuit in ltspice and found out that its around 95K ohms to get 250mA,but how would i calculate it .

Ive also attached my calculations which i used to get 1.5vpp
 

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Answers and Replies

  • #2
tech99
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Homework Statement:: .
Relevant Equations:: .

Hi i got a question ,how do i calculate the input resistance ,of the RS resistor ,so i would get half of the voltage of the voltage source (500mA) .I've created the circuit in ltspice and found out that its around 95K ohms to get 250mA,but how would i calculate it .

Ive also attached my calculations which i used to get 1.5vpp
Do you mean microamps not milliamps?
 
  • #3
tech99
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Do you mean microamps not milliamps?
The input resistance of a common emitter transistor tends to be around 1 k, so your figure of 95k does not sound correct. A useful website here: https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/amplifier-impedances/
 
  • #4
jasonRF
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The input resistance of a common emitter transistor tends to be around 1 k, so your figure of 95k does not sound correct. A useful website here: https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/amplifier-impedances/
Nope. That site has a common emitter with no emitter resistor, while the OP has a large emitter resistor. I get that the input impedance looking into the base of Q1 is much larger than 1k Ohm.

IronaSona, based on your spice model your signal is 500 mV, not 500 mA.

jason
 
  • #5
tech99
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For DC signaals yes, but for AC signals the emitter resistor is bypassed by a capacitor, so it seems to me that it does not have an effect.
 
  • #6
jasonRF
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For DC signaals yes, but for AC signals the emitter resistor is bypassed by a capacitor, so it seems to me that it does not have an effect.
Perhaps I did the arithmetic wrong, but at 1 kHz a 50 nF capacitor has an impedance of about -i3185 Ohms, so at 1kHz you cannot treat it as a short.
 
  • #7
IronaSona
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Do you mean microamps not milliamps?
milivolts
 
  • #8
IronaSona
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Nope. That site has a common emitter with no emitter resistor, while the OP has a large emitter resistor. I get that the input impedance looking into the base of Q1 is much larger than 1k Ohm.

IronaSona, based on your spice model your signal is 500 mV, not 500 mA.

jason
yes sorry milivolts
 
  • #9
IronaSona
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The input resistance of a common emitter transistor tends to be around 1 k, so your figure of 95k does not sound correct. A useful website here: https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/amplifier-impedances/
But if i change the resistor to anything else besides 90k to 96k , ill get smaller or bigger output voltage than i want right.?
 
  • #10
jasonRF
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But if i change the resistor to anything else besides 90k to 96k , ill get smaller or bigger output voltage than i want right.?
When I did a back of the envelope calculation I recall getting something between 80k and 90k Ohms. It depends on the assumed gain of the transistor. The spec sheet says between 200 and 450 (that is what the ‘B’ designation means). What does your spice model use? It will be listed as ‘Bf=some number’ in the .model statement.
Jason
 
  • #11
jasonRF
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In any case, the heart of the question is about computing the input impedance of a common emitter amplifier. Have you covered this in your class, or does your textbook explain it?
 
  • #12
IronaSona
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In any case, the heart of the question is about computing the input impedance of a common emitter amplifier. Have you covered this in your class, or does your textbook explain it?
Ive tried looking again but couldn't find anything ,but i did find somewhere in my notes which says that RS=RC1=8625 so that means that RS has to be 8625 too right ?
 
  • #13
berkeman
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Ive tried looking again but couldn't find anything ,but i did find somewhere in my notes which says that RS=RC1=8625 so that means that RS has to be 8625 too right ?
No, that makes no sense to me.

What do you find for Zin for the BJT amplifier circuit? You are simulating the circuit after all...
 
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