BJT Basics Practice Problems

Practise which values to work out first when a transistor is a switch and when it is an amplifier.

Key formulas

I_C = β·I_B

I_E = I_C + I_B

R_B = (V_drive − V_BE)/(I_B)

V_CE = V_CC − I_C·R_C − I_E·R_E

A_v ≈ −(R_C)/(R_E)

Size a switch with the minimum h_FE and give the base current 2 to 3 times margin.
Work bias in the order V_B → V_E → I_E → V_CE.
Check the region from V_CE first: below 0.3 V is saturation, equal to the supply is cut-off.

Problems 10

2 multiple-choice and 8 numeric problems. Type numbers in the unit shown next to the box (for example 2.55); SI prefixes such as 4.7k also work. Small differences from rounding along the way count as correct (±2%). When you are done, press Submit and grade at the bottom to see the answers and step-by-step solutions.

An NPN transistor with β (h_FE) = 100 runs in the active region with 20 µA of base current. What is the collector current, in mA?
mA
Answer: 2 mA
Solution
1. In the active region, I_C = β·I_B.
2. I_C = 100 × 20 µA = 2 mA
Check it with the Transistor base resistor calculator
For the same transistor (I_C = 2 mA, I_B = 20 µA), what is the emitter current, in mA?
mA
Answer: 2.02 mA
Solution
1. Both currents leave through the emitter: I_E = I_C + I_B.
2. I_E = 2 + 0.02 = 2.02 mA
3. With a large β, I_E ≈ I_C is off by only 1%.
Check it with the Transistor base resistor calculator
A transistor switch must turn on a 100 mA load. Minimum h_FE = 50, base current doubled for solid saturation, 5 V drive, V_BE = 0.7 V. What base resistor is needed, in ohms?
Ω
Answer: 1,075 Ω
Solution
1. Base current needed: I_B = (2 × 100 mA)/(50) = 4 mA
2. R_B = (V_drive − V_BE)/(I_B) = (5 − 0.7)/(0.004) = 1,075 Ω
3. Pick the standard value just below, 1 kΩ; a larger resistor gives less base current and weaker saturation.
Check it with the Transistor base resistor calculator
When a silicon switching transistor is fully saturated, V_CE is roughly

A0.1 to 0.3 VB0.7 VCHalf the supplyDEqual to the supply
Answer: A. 0.1 to 0.3 V
Solution
1. In saturation the collector-emitter path acts almost like a short, so V_CE(sat) is about 0.1 to 0.3 V.
2. 0.7 V is V_BE, and V_CE equal to the supply means the transistor is off (cut-off).
Check it with the Transistor base resistor calculator
A common-emitter stage with V_CC = 12 V, R_C = 2.2 kΩ, I_C = 2 mA and no emitter resistor. What is V_CE, in volts?
V
Answer: 7.6 V
Solution
1. Voltage across R_C: 2 mA × 2.2 kΩ = 4.4 V
2. V_CE = V_CC − I_C·R_C = 12 − 4.4 = 7.6 V
3. V_CE is well above 0.3 V, so the transistor is in the active region.
Check it with the BJT bias (Q-point) calculator
Voltage-divider bias: V_CC = 12 V, R1 = 47 kΩ, R2 = 10 kΩ, R_E = 1 kΩ, V_BE = 0.7 V. Ignoring base current, what is I_C, in mA?
mA
Answer: 1.405 mA
Solution
1. Base voltage: V_B = 12 × (10)/(47 + 10) = 2.105 V
2. Emitter voltage: V_E = V_B − 0.7 = 1.405 V
3. I_C ≈ I_E = (V_E)/(R_E) = (1.405 V)/(1 kΩ) = 1.405 mA
4. Including base current (β = 100) gives a slightly smaller 1.29 mA. The BJT bias (Q-point) calculator uses that exact form.
Check it with the BJT bias (Q-point) calculator
The same circuit has R_C = 4.7 kΩ. Taking I_C ≈ I_E, what is V_CE, in volts?
V
Answer: 3.99 V
Solution
1. V_CE = V_CC − I_C·R_C − I_E·R_E ≈ V_CC − I_C(R_C + R_E)
2. V_CE = 12 − 1.405 mA × 5.7 kΩ = 12 − 8.01 = 3.99 V
3. V_CE is about a third of the supply: active region, with room for the signal to swing.
Check it with the BJT bias (Q-point) calculator
In which region should the operating point sit when a transistor is used as a signal amplifier?

ACut-offBForward activeCSaturationDReverse active
Answer: B. Forward active
Solution
1. To amplify, I_C must follow I_B proportionally, which holds in the forward active region (B-E forward, B-C reverse biased).
2. Cut-off and saturation are the off and on states of a switch.
Check it with the BJT bias (Q-point) calculator
A transistor in the active region has V_CE = 6 V and I_C = 50 mA. How much power does it dissipate, in mW?
mW
Answer: 300 mW
Solution
1. Nearly all of the loss is in the collector-emitter path: P ≈ V_CE·I_C.
2. P = 6 V × 0.05 A = 0.3 W = 300 mW
3. That is fairly close to the 625 mW rating of a TO-92 part such as the 2N3904, so check the temperature. A saturated switch at the same current (0.2 V) dissipates only 10 mW.
Check it with the Heatsink calculator
A common-emitter amplifier has R_C = 4.7 kΩ and R_E = 1 kΩ (no bypass capacitor). What is the approximate size of its voltage gain, |A_v|?
Answer: 4.7
Solution
1. With an unbypassed emitter resistor, A_v ≈ −(R_C)/(R_E).
2. |A_v| ≈ (4.7 kΩ)/(1 kΩ) = 4.7
3. The minus sign means the output is inverted (180°). The Common-emitter gain: how good is the quick estimate? guide compares this estimate with measurements.
Check it with the BJT bias (Q-point) calculator

Not graded yet.

Scores are kept only in this browser and are never sent anywhere.