Question

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?

Answer 1

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

(b) Vout = 0V (saturation)

(c) Vb = - 0.6V

(d) Vb = 4.4V

Step-by-step explanation:

This transistor circuit depicts a common emitter configuration

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

• Vce = Vcc - Ic× RL
• when Ib =0, Ic= 0
• Vce = Vcc
• Taking the emitter -base circuit , we have that Ib =
• 
  `(V_(in)- V_(BE) )/(R_(B) )`
• 
  `(V_(in)- V_(BE) )/(R_(B) )`

(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 =

`(V_(in)- V_(BE) )/(R_(B) )`; therefore
`I_(B) = (5- 0.6)/(1000)`

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

Answer 2

a) Vout= 5V

b) Vout= 5V

c) Vbase= 0.6V

d) Vbase= 0.6V

Step-by-step 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