BJT Inverter with Feedback: Vin-Vout Characteristic & Saturation Analysis

[hogrampage]

Homework Statement

The circuit in the figure has the form of a BJT inverter, but also has a resistor connected between the emitter lead and ground. This addition provides feedback between the output loop of the circuit and its input loop.

(a) Find an exact expression for the V_in-V_out transfer characteristic over the region where Q1 operates in the constant-current region.

(b) To what does your expression reduce to for large βf?

(c) At what value of V_in does Q1 first turn on?

(d) When Q1 saturates, what will be the value of V_out?

(e) At what approximate value of V_in will Q1 first go into saturation?

(f) For V_CC = 12V, R_C = 4.7kΩ, R_E, large β_F, V_sat ≈ 0.2V, and V_f ≈ 0.7V, draw the approximate transfer characteristic of the circuit.

(g) The approximate expression obtained in part (b) will overestimate the actual gain of the circuit found in part (a). How large must β_F be for the approximation to overestimate the gain by no more than 10%?

Homework Equations

KVL

In the constant-current region:

i_C = β_Fi_B
V_out has the form V_out = mV_in + b​

The Attempt at a Solution

i_E = i_B + i_C = i_B + β_Fi_B = i_B(1 + β_F)

KVL input loop:

[1] i_B = \frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}​

KVL output loop:

[2] V_out = V_CC - i_CR_C - i_ER_E

[3] V_out = V_CC - β_Fi_BR_C - i_BR_E(1 + β_F)

[4] V_out = V_CC - i_B(β_FR_C + R_E(1 + β_F))​

Part (a)

If I combine equations [1] and [4]:

[5] V_out = V_CC - \frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}(β_FR_C + R_E(1 + β_F))​

I want to know if I'm headed in the right direction and where to go from here. Any guidance is greatly appreciated!

Thanks!

 

[rude man]

Homework Statement

The circuit in the figure has the form of a BJT inverter, but also has a resistor connected between the emitter lead and ground. This addition provides feedback between the output loop of the circuit and its input loop.

(a) Find an exact expression for the V_in-V_out transfer characteristic over the region where Q1 operates in the constant-current region.

(b) To what does your expression reduce to for large βf?

(c) At what value of V_in does Q1 first turn on?

(d) When Q1 saturates, what will be the value of V_out?

(e) At what approximate value of V_in will Q1 first go into saturation?

(f) For V_CC = 12V, R_C = 4.7kΩ, R_E, large β_F, V_sat ≈ 0.2V, and V_f ≈ 0.7V, draw the approximate transfer characteristic of the circuit.

(g) The approximate expression obtained in part (b) will overestimate the actual gain of the circuit found in part (a). How large must β_F be for the approximation to overestimate the gain by no more than 10%?

Homework Equations

KVL

In the constant-current region:

i_C = β_Fi_B
V_out has the form V_out = mV_in + b​

The Attempt at a Solution

i_E = i_B + i_C = i_B + β_Fi_B = i_B(1 + β_F)

KVL input loop:

[1] i_B = \frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}​

KVL output loop:

[2] V_out = V_CC - i_CR_C - i_ER_E
​

No. Vout is collector voltage with respect to ground. Vout is not Vc - Ve.​

 

[hogrampage]

Oh, oops. V_out = V_CC - i_CR_C, so:

V_out = V_CC - \frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}R_Cβ_F

I still don't see what it would reduce to with a large beta :|.

 

[berkeman]

Oh, oops. V_out = V_CC - i_CR_C, so:

V_out = V_CC - \frac{V_{in} - V_{f}}{R_{B} + R_{E}(1 + β_{F})}R_Cβ_F

I still don't see what it would reduce to with a large beta :|.

Just put in a large beta. What does the equation simplify to?

 

[hogrampage]

Maybe I'm blind, as I don't see what it would simplify to. Would it cause R_B + R_E to be ignored in the denominator? Then, the two beta values would cancel from the top/bottom? So:

\frac{(V_{in} - V_{f})R_{C}}{R_{E}}

 

[berkeman]

Maybe I'm blind, as I don't see what it would simplify to. Would it cause R_B + R_E to be ignored in the denominator? Then, the two beta values would cancel from the top/bottom? So:

\frac{(V_{in} - V_{f})R_{C}}{R_{E}}

Yes, Rc/Re would be the gain.