Answer:
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
Answer:
(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 =
[tex]\frac{V_{in}- V_{BE} }{R_{B} }[/tex]; therefore [tex]I_{B} = \frac{5- 0.6}{1000}[/tex]
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
