Analyze the following ideal transistor circuit. Can use general rule of thumbs for analyzing transistors b-base, c- collector, and e-emitter ports (i. e., Vbe = 0.6V, Vce = 0.2V, Gain = Ic/Ib = 10:1, etc.). Show your work. a) When Vin = 0V, what is the voltage at Vout? b) When Vin = 5V, what is the voltage at Vout? c) When Vin = 0V, what is the voltage at the base of the transistor? d) When Vin = 5V, what is the voltage at the base of the transistor?

a) Vout= 5V

b) Vout= 5V

c) Vbase= 0.6V

d) Vbase= 0.6V

Explanation:

Consider the circuit shown in attachment

a) When Vin is 0V, the base circuit is not turned, so

Ib=0 and Ic=∞ as transistor is not turned on so

Vout =5V

b) When Vin= 5 V,

Ib= (Vin-Vb)/Rb

Ib=(5-0.6)/1000= 0.0044A

Ic= 0.0044×10=0.044A

Vout= 5- 0.044×1000= not real value

Vout= Vce= 5V

c) voltage drop across Vbase= 0.6V

d) Vbase= 0.6V

In all the above cases, the transistor will not be turned on biasing base voltage and resistor values are very high compared to VCC which is 5V in the given circuit

(a) Vout = Vce = 5V (cut-off)

(b) Vout = 0V (saturation)

(c) Vb =  - 0.6V

(d) Vb =  4.4V

Explanation:

This transistor circuit depicts a common emitter configuration

Firstly we have to take note of the formulas used in a common - emitter configuration

(a) when Vin = 0V, the transistor is said to be in cut-off because it does not conduct any current , In cut-off both the base- emitter and the base- collector junctions are reverse-biased.

Vce = Vcc - Ic× RL

when Ib =0,  Ic= 0

Vce = Vcc

Vout = 5V

(b) Ib =  

; therefore  

Ib = 4.4mA

Ic = β ×Ib ; this relationship does not hold good when the transistor is in saturation

therefore we have to find the value of Ic at saturation (when the transistor is on)

Ic (sat) = Vcc / RL

= 5/ 1000

= 5mA

= 10 × 4.4× 10⁻³

Ic= 0.044A; this value is too large as Ic cannot increase more than the saturation value

Vce = Vcc - Ic× RL

= 5 - 5 ×10⁻³× 1000

= 5 - 5

Vce= 0V; Vout = 0V

(c) The diagram depicts an NPN transistor ; for an NPN

the base emitter voltage Vbe = 0.6V , whereas a PNP has a Vbe = -0.6V

Vbe = Vb - Ve

Vin = Ib ×Rb + Vbe ; Ib× Rb = Vb (voltage at the base )

therefore Vin = Vb + Vbe

making Vb as the subject of formulae

Vb = Vin - Vbe

Vb = 0- 0.6 = - 0.6V

(d) when Vin = 5V

Vb = Vin - Vbe

= 5 - 0.6

Vb = 4.4V

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explanation:

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