Diodes & transistors: practice problems with step-by-step solutions
Free Diodes & transistors practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: diode circuits, BJT bias point, BJT as a switch.
Free practice problemEasy
A green LED with a forward voltage of 2.2 V runs from a 24 V supply through a series resistor. Find the resistance R that sets the LED current to 15 mA.
A diode lets current flow in one direction only, like a one-way valve in a water pipe. The arrow in its symbol points in the direction current can flow.
When it conducts (it is forward-biased), a silicon diode keeps about 0.7 V across itself, whatever the current. This is the constant-drop model.
So a conducting diode behaves like a small 0.7 V battery that takes voltage away. The resistor gets what is left:
I = (Vs − n·0.7) / R
Here n is the number of conducting diodes in series. Reversed, the diode blocks and the current is zero.
Self-check A 9 V source drives current through a 1 kΩ resistor and two forward-biased silicon diodes in series. What is the current?
10.4 mA
7.6 mA
8.3 mA
9 mA
Answer: 7.6 mA. I = (9 − 2·0.7)/1 kΩ = 7.6 mA. Counting only one diode gives 8.3 mA, ignoring the diodes gives 9 mA, and adding their voltage gives 10.4 mA.
2The bipolar transistor
A BJT (bipolar junction transistor) has three terminals: base (B), collector (C) and emitter (E). In the NPN type, current flows into the collector and the base, and out of the emitter.
A small base current IB controls a much larger collector current IC. It works like a tap: a small effort on the handle controls a large flow.
In the normal working range, the active region, three facts hold:
VBE = 0.7 V IC = β·IB IE = (β + 1)·IB
The current gain β (beta) is typically between 50 and 300. The base-emitter junction is a diode, which is why VBE is 0.7 V.
Self-check A transistor with β = 80 in the active region has a base current of 0.05 mA. What is the collector current?
0.05 mA
0.625 µA
4 mA
4.05 mA
Answer: 4 mA. IC = β·IB = 80·0.05 = 4 mA. 4.05 mA is the emitter current (β + 1)·IB, and 0.625 µA divides by β instead of multiplying.
3Active region or saturation
The collector current flows through RC, so the voltage left across the transistor is:
VCE = VCC − IC·RC
IC cannot grow without limit. When the transistor is fully on, only a small VCE(sat) ≈ 0.2 V is left, and RC alone sets the current:
IC(sat) = (VCC − VCE(sat)) / RC
Active region: β·IB is smaller than IC(sat). Then IC = β·IB. Used in amplifiers.
Saturation: β·IB is larger than IC(sat). Then IC = IC(sat), the transistor is a closed switch.
Cut-off: no base current, no collector current. The transistor is an open switch.
To use a transistor as a switch, you give it at least IB = IC(sat)/β, often several times more to be safe.
Self-check A transistor with β = 100 has VCC = 10.2 V, RC = 2 kΩ and a base current of 0.1 mA. Take VCE(sat) = 0.2 V. What is IC?
0.1 mA
10 mA
5 mA
5.1 mA
Answer: 5 mA. IC(sat) = (10.2 − 0.2)/2 = 5 mA. β·IB = 10 mA is more than that, so the transistor is saturated and IC = 5 mA. Leaving out the 0.2 V gives 5.1 mA.
4The method step by step
For a transistor with a base resistor RB from VCC (fixed bias):
Walk the base loop: VCC drops across RB and then 0.7 V across the base-emitter junction.
Find the base current: IB = (VCC − 0.7)/RB.
Find the collector current: IC = β·IB.
Find the collector-emitter voltage: VCE = VCC − IC·RC.
Check the region: VCE must be above about 0.2 V. If not, the transistor is saturated and IC = IC(sat).
With V in volts and R in kΩ, currents come out in mA. That keeps the numbers easy.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
An NPN transistor with β = 100 has RB = 565 kΩ from VCC = 12 V to its base, RC = 2.2 kΩ and a grounded emitter. Take VBE = 0.7 V. Find IC and VCE.
1
Look at the circuit
The base current comes from VCC through RB.
2
Base current
IB = (12 − 0.7) / 565 = 0.02 mA
3
Collector current
IC = β·IB = 100·0.02 = 2 mA
4
Collector-emitter voltage
VCE = 12 − 2·2.2 = 7.6 V
5
Check the region
The saturation current would be (12 − 0.2)/2.2 = 5.36 mA. Our 2 mA is below that, and VCE = 7.6 V is well above 0.2 V.
So the transistor really is in the active region, and IC = β·IB was allowed.
Answer
IC = 2 mA and VCE = 7.6 V.
6Common mistakes
Forgetting the 0.7 V drop of each conducting diode, or of the base-emitter junction.
Using IC = β·IB when the transistor is saturated. Always check VCE.
Confusing IC and IE. The emitter carries the base current as well: IE = (β + 1)·IB.
For a switch, forgetting VCE(sat) = 0.2 V in IC(sat), or forgetting to divide by β.
Self-check A transistor with β = 100 switches a 100 Ω load from VCC = 5 V. The base is driven from 5 V through RB. Take VCE(sat) = 0.2 V. What is the largest RB that still saturates it?
10.4 kΩ
8.6 kΩ
8.96 kΩ
89.6 Ω
Answer: 8.96 kΩ. IC(sat) = 4.8/100 = 48 mA, so IB ≥ 0.48 mA and RB = (5 − 0.7)/0.48 = 8.96 kΩ. Forgetting the 0.7 V gives 10.4 kΩ, forgetting β gives 89.6 Ω, and leaving out the 0.2 V gives 8.6 kΩ.
7How exam questions look
Easy: diode circuits. A source, a resistor and one or two diodes in series, or a diode in parallel with a resistor. You find a current in mA.
Medium: the BJT bias point. Fixed bias as in the example, or voltage-divider bias with an emitter resistor. You find IC or VCE.
Hard: the BJT as a switch. You design RB for saturation, sometimes with an overdrive factor, or decide whether a given RB saturates the transistor.
A diode in parallel with a resistor clamps that node at 0.7 V. The resistor then takes 0.7/R, and the diode takes the rest of the current.
For divider bias, replace the divider by its Thévenin equivalent first:
A 3.3 V Zener diode is fed from 18 V through a 470 Ω resistor, with no load connected. Find the current IR in the resistor.
Vin = 18 VVZ = 3.3 VR = 470 Ω
Find IR(mA)
Show hints
The Zener holds the output at VZ.
VR = 18 − 3.3 = 14.7 V.
I = V/R.
Show the answer and the step-by-step solution
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
Resistor voltage
VR = 18 − 3.3 = 14.7 V
Current
IR = 14.7/470 = 31.28 mA
Answer
IR = 31.28 mA
Example 2Medium
A 5.1 V Zener regulator is fed from 12 V through 270 Ω and supplies a 3300 Ω load. Find the Zener current IZ.
Vin = 12 VVZ = 5.1 VR = 270 ΩRL = 3300 Ω
Find IZ(mA)
Show hints
IR = (12 − 5.1)/270 = 25.56 mA.
IL = 5.1/3300 = 1.545 mA.
IZ = IR − IL.
Show the answer and the step-by-step solution
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
Resistor current
IR = 6.9/270 = 25.56 mA
Load current
IL = 5.1/3300 = 1.545 mA
Zener current
IZ = 25.56 − 1.545 = 24.01 mA
Answer
IZ = 24.01 mA
Example 3Hard
An NPN transistor with β = 200 switches a 150 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 100 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.
Find the base current from the input loop: IB = (Vin − 0.7)/RB.
Compare β·IB with the largest current the load allows, IC(sat) = (VCC − 0.2)/RC.
IB = 43 µA, IC(sat) = 78.67 mA.
Show the answer and the step-by-step solution
BJT as a switch: to switch a load fully on, the transistor is driven into saturation, where VCE = VCE(sat) and the collector current is set by the load: IC(sat) = (VCC − VCE(sat))/RC. Saturation needs at least IB = IC(sat)/β. If the base current is smaller, the transistor stays active and IC = β·IB.
Base current
IB = (5 − 0.7)/100 kΩ = 43 µA
Saturated or active?
β·IB is smaller than IC(sat), so the transistor stays active.
β·IB = 8.6 mA
IC(sat) = (12 − 0.2)/150 = 78.67 mA
Collector current
IC = β·IB = 8.6 mA
Answer
IC = 8.6 mA
The transistor is active, so the load is not fully switched on.
Questions and answers
What is Diodes & transistors?
A diode lets current flow in one direction only, like a one-way valve in a water pipe. The arrow in its symbol points in the direction current can flow.
When it conducts (it is forward-biased), a silicon diode keeps about 0.7 V across itself, whatever the current. This is the constant-drop model.
How do I solve Diodes & transistors problems step by step?
Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.
What are common mistakes in Diodes & transistors?
The Zener holds 3.3 V: only Vin − VZ is across R.
The resistor voltage is Vin − VZ, not VZ.
Give the answer in mA.
What do the Diodes & transistors exercises cover?
Easy: diode circuits
Medium: BJT bias point
Hard: BJT as a switch
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.