Transistor Biasing: Calculating Input & Output Resistance

[likephysics]

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.

 

[likephysics]

Anyone?

 

[CEL]

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.

 

[likephysics]

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?

 

[CEL]

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.

 

[vk6kro]

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?

 

[likephysics]

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.

 

[vk6kro]

Are you sure?

The emitter voltage depends on the base current you are trying to calculate, but multiplied by 100.

 

[likephysics]

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?

 

[vk6kro]

Yes, but you don't know Ib, so how can you know Ic or Ve?

 

[likephysics]

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

 

[CEL]

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.

 

[likephysics]

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.

 

[vk6kro]

Yes. That is much better.