Diodes & transistors worksheet with answers

Free printable Diodes & transistors worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
+−Vs​24 VR = 1.5 kΩD​I = ?

A 24 V DC source drives current through a 1.5 kΩ resistor in series with a forward-biased silicon diode. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 24 VR = 1.5 kΩVD = 0.7 V
I = mA
2.Easy
+−15 V330 Ω5.6 VIR​ = ?

A 5.6 V Zener diode is fed from 15 V through a 330 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 15 VVZ = 5.6 VR = 330 Ω
IR = mA
3.Medium
+−12 VR = ?3 × 2.1 V25 mA

3 yellow LEDs (forward voltage 2.1 V each) are connected in series with a resistor to a 12 V supply. Find the resistance R that sets the current to 25 mA.

Vs = 12 VVF = 2.1 Vn = 3I = 25 mA
R = Ω
4.Medium
RC​ = 2.7 kΩβ = 80RE​ = 1 kΩVCC​ = 12 VR1​ = 68 kΩR2​ = 22 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 12 V: R₁ = 68 kΩ from VCC to the base and R₂ = 22 kΩ from the base to ground. The collector resistor is RC = 2.7 kΩ and the emitter resistor is RE = 1 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector–emitter voltage VCE.

VCC = 12 VR1 = 68 kΩR2 = 22 kΩRC = 2.7 kΩRE = 1 kΩβ = 80
VCE = V
5.Hard
RC​ = 470 Ωβ = 200VCC​ = 5 V3.3 VRB​ = ?

An NPN transistor with β = 200 switches a 470 Ω load resistor connected between VCC = 5 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the largest base resistor RB, in kΩ, that still drives the transistor just into saturation.

VCC = 5 VRC = 470 Ωβ = 200Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 V
RB = kΩ
6.Hard
V (V)I (mA)0.6533Q = ?

A diode is modelled as 0.65 V in series with 10 Ω (it conducts above 0.65 V). It is in series with a 1000 Ω resistor across a 3 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 3 VR = 1000 ΩV0 = 0.65 Vrd = 10 Ω
ID = mA

Answer key

  1. 1. I = 15.53 mA
  2. 2. IR = 28.48 mA
  3. 3. R = 228 Ω
  4. 4. VCE = 5.205 V
  5. 5. RB = 50.92 kΩ
  6. 6. ID = 2.327 mA

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