Transistor Biasing: Calculating Input & Output Resistance

  • Thread starter Thread starter likephysics
  • Start date Start date
  • Tags Tags
    Biasing Transistor
Click For Summary

Discussion Overview

The discussion revolves around calculating input and output resistance in a transistor biasing circuit, focusing on the application of Thevenin's theorem and the relationships between base current, collector current, and voltage drops across various components. The context includes homework-related problem-solving in electronics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that input resistance is R1||R2 but is unsure how to calculate output resistance.
  • Another participant suggests that output resistance is Rc in parallel with hoe and describes the Thevenin equivalent circuit for the base current calculation.
  • A participant proposes a formula for calculating base current (IB) based on the voltage at the base and the Thevenin equivalent resistance, including the base-emitter diode drop.
  • Concerns are raised about the voltage across the emitter resistor being influenced by the relationship between base current and collector current, with a participant questioning whether this has been accounted for.
  • There is a discussion about the calculation of base current (Ib) and its dependence on the voltage divider and the voltage across the emitter resistor, leading to a realization about the voltage drop across Re affecting the calculations.
  • Participants engage in clarifying and correcting each other's understanding of the relationships between the currents and voltages in the circuit.

Areas of Agreement / Disagreement

Participants express differing views on the calculations of base current and the effects of the emitter resistor voltage. While some calculations are agreed upon, there remains uncertainty and debate regarding the correct approach to determining these values.

Contextual Notes

Participants highlight dependencies on various assumptions, such as the values of resistances and the relationship between base and collector currents. The discussion reveals unresolved mathematical steps and the complexity of the circuit analysis.

likephysics
Messages
638
Reaction score
4

Homework Statement


See attachment.



Homework Equations





The Attempt at a Solution



Input resistance is R1||R2
I don't know how to calculate output resistance.

To calculate the base current, I calculated the voltage at the resistor divider (5V), thevinized the base circuit and applied KVL. But Ie is not known.
I am stuck now.
 

Attachments

  • bjt.jpg
    bjt.jpg
    38.4 KB · Views: 412
Physics news on Phys.org
Anyone?
 
Output resistance is Rc in parallel with hoe.
In the input bias mesh you have the Thévénin equivalent of Vcc, Rb1 and Rb2 in series with the base-emitter diode. You can calculate IB. From IB and beta you can calculate Ic.
 
CEL, so its basically (Vb-0.7)/Rth
where, Vb is the voltage at base; Rth is the thevinin equivalent and 0.7 is base-emitter diode drop.
Correct?
 
likephysics said:
CEL, so its basically (Vb-0.7)/Rth
where, Vb is the voltage at base; Rth is the thevinin equivalent and 0.7 is base-emitter diode drop.
Correct?

Correct.
 
There is a voltage across the emitter resistor caused by a base current and a collector current 100 times as great as the base current.

Have you allowed for this?
 
vk6kro said:
There is a voltage across the emitter resistor caused by a base current and a collector current 100 times as great as the base current.

Have you allowed for this?

yup.
 
Are you sure?

The emitter voltage depends on the base current you are trying to calculate, but multiplied by 100.
 
The voltage drop across the emitter resistor is Ie*Re.
Ie is almost equal to Ic
Ic = beta*Ib

Is this what you are talking about?
 
  • #10
Yes, but you don't know Ib, so how can you know Ic or Ve?
 
  • #11
Well, Ib is Voltage divider voltage(5v) -Vbe drop(0.7v)/Thevinin equivalent of the 2 resistors - Rb1 and Rb2.

Ib= (5v-0.7v)/Rth

I then calculate Ic, which is beta*Ib
 
  • #12
likephysics said:
Well, Ib is Voltage divider voltage(5v) -Vbe drop(0.7v)/Thevinin equivalent of the 2 resistors - Rb1 and Rb2.

Ib= (5v-0.7v)/Rth

I then calculate Ic, which is beta*Ib

No, Ib is Voltage divider voltage(5v) -Vbe drop(0.7v)/Thevinin equivalent of the 2 resistors - Rb1 and Rb2 - voltage across Re (beta+1)Ib.
You have an equation with one unknoen: Ib.
 
  • #13
CEL said:
No, Ib is Voltage divider voltage(5v) -Vbe drop(0.7v)/Thevinin equivalent of the 2 resistors - Rb1 and Rb2 - voltage across Re (beta+1)Ib.
You have an equation with one unknoen: Ib.

That's what I initially thought, but to find the current thru a resistor
I = V/R

[STRIKE]The voltage across Rth is (5-0.7)v
That divided by Rth should give I flowing thru Rth, which is nothing but base current.
Correct?[/STRIKE]

(just realized)
Dang! I get it. The voltage at right side of Rth is not 0.7v, it is 0.7+drop across Re.
Thanks.
 
  • #14
Yes. That is much better.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K