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
+−9 VR = ?2.1 V20 mA

A yellow LED with a forward voltage of 2.1 V runs from a 9 V supply through a series resistor. Find the resistance R that sets the LED current to 20 mA.

Vs = 9 VVF = 2.1 VI = 20 mA
R = Ω
2.Easy
+−Vs​5 VR = 1.5 kΩD1​D2​I = ?

A 5 V DC source drives current through a 1.5 kΩ resistor in series with two forward-biased silicon diodes. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 5 VR = 1.5 kΩVD = 0.7 V
I = mA
3.Medium
RC​ = 1 kΩβ = 80RE​ = 2.2 kΩVCC​ = 20 VR1​ = 68 kΩR2​ = 8.2 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 20 V: R₁ = 68 kΩ from VCC to the base and R₂ = 8.2 kΩ from the base to ground. The collector resistor is RC = 1 kΩ and the emitter resistor is RE = 2.2 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 = 20 VR1 = 68 kΩR2 = 8.2 kΩRC = 1 kΩRE = 2.2 kΩβ = 80
VCE = V
4.Medium
RC​ = 1 kΩβ = 80IC​ = ?VCC​ = 15 VRB​ = 820 kΩ

An NPN transistor with β = 80 is biased by a base resistor RB = 820 kΩ from the supply VCC = 15 V. The collector resistor is RC = 1 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector current IC in mA.

VCC = 15 VRB = 820 kΩRC = 1 kΩβ = 80VBE = 0.7 V
IC = mA
5.Hard
+−≥ 18 VR = ?12 V≤ 40 mA

A 12 V Zener regulator must supply up to 40 mA to a load. The input can drop to 18 V, and the Zener needs at least 2 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 18 VVZ = 12 VIL,max = 40 mAIZ,min = 2 mA
Rmax = Ω
6.Hard
RC​ = 1 kΩβ = 100VCC​ = 24 V5 VRB​ = ?

An NPN transistor with β = 100 switches a 1 kΩ load resistor connected between VCC = 24 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 2 times the minimum needed for saturation. Find RB in kΩ.

VCC = 24 VRC = 1 kΩβ = 100Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 2
RB = kΩ

Answer key

  1. 1. R = 345 Ω
  2. 2. I = 2.4 mA
  3. 3. VCE = 17.98 V
  4. 4. IC = 1.395 mA
  5. 5. Rmax = 142.9 Ω
  6. 6. RB = 9.034 kΩ

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