In the circuit shown, the silicon BJT has β = 50. Assume V BE = 0
In the circuit shown, the silicon BJT has β = 50. Assume VBE = 0.7 V and VCE(sat) = 0.2 V. Which one of the following statements is correct?
A. For R<sub style="">C</sub> = 1 kΩ, the BJT operates in the saturation region
B. For R<sub style="">C</sub> = 3 kΩ, the BJT operates in the saturation region
C. For R<sub style="">C</sub> = 20 kΩ, the BJT operates in the cut-off region
D. For R<sub style="">C</sub> = 20 kΩ, the BJT operates in the linear region
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
In the linear region:
IC = β IB
VBE = 0.7 V
Applying KVL from the base to the emitter ground, we get:
\({I_B} = \frac{{{V_{i\;}} - {V_{BE}}}}{{{R_B}}}\)
Also:
\({V_{CE}} = {V_{CC}} - {I_C}{R_C}\)
Where IC is collector current, IB is base current, Vi is the supply for the base, VCC is the supply for the collector, RB is a base resistor, RC is collector resistor & β is transistor gain.
In the saturation region, the above relation doesn’t hold.
Analysis:
Assuming that the transistor is operating in the linear region, we get the base current as:
\({I_B} = \frac{{5 - 0.7}}{{50\; \times\; {{10}^3}}} = 86\;\mu A\)
IC = 50 × 86 = 4.3 mA
If RC = 3 KΩ then,
\({V_{CE}} = 10 - 4.3 \times {10^{ - 3}} \times 3*{10^3}\)
\({V_{CE}} = - 2.9\;V\)
The negative value of VCE indicates that the transistor is in saturation.
Important Points/Short Tricks:
- Always assume initially that the transistor is in the linear region and then proceed for calculations.
- Alternatively, you can calculate the collector current IC assuming the transistor to be in the saturation and then calculate base current as done above.
- If IC < βIB, then the transistor is in saturation.