A 5 V supply, a forward-biased silicon diode (0.7 V drop) and a 1 kΩ resistor are in series. What current flows, in mA?
mA
Answer: 4.3 mA
Solution
Use the constant-drop model: the diode takes 0.7 V. I = ( 5 − 0.7 )/( 1 kΩ ) = ( 4.3 V )/( 1 kΩ ) = 4.3 mAThe real forward voltage shifts a little with current, roughly 0.6 to 0.75 V.
In the same circuit, the diode is turned around (reverse-biased). What current flows?
A 5 mA B 4.3 mA C Almost none (nA to µA leakage) D 0.7 mA
Answer: C. Almost none (nA to µA leakage)
Solution
A reverse-biased diode passes almost no current until it reaches breakdown. Nearly all 5 V appears across the diode and almost none across the resistor; only a tiny leakage current flows.
A 12 V peak sine wave is half-wave rectified by one silicon diode (0.7 V). What is the peak output voltage, in volts?
V
Answer: 11.3 V
Solution
In a half-wave rectifier the current passes through one diode. Vp,out = 12 − 0.7 = 11.3 V On the negative half-cycle the output is 0 V and the diode blocks 12 V in reverse.
A 12 V peak sine wave goes through a bridge rectifier (four diodes, 0.7 V each). What is the peak output voltage, in volts?
V
Answer: 10.6 V
Solution
In a bridge, current always passes through two diodes in series, on either half-cycle. Vp,out = 12 − 2 × 0.7 = 10.6 V Subtracting 2.8 V for four diodes is a common slip: only two conduct at a time.
A full-wave rectifier gives 10.6 V peak from 60 Hz mains into a 1 kΩ load with a 470 µF smoothing capacitor. Using the usual estimate, what is the peak-to-peak ripple, in mV?
mV
Answer: 187.9 mV
Solution
The full-wave ripple estimate is Vr ≈ ( Vp )/( 2fRC ) Vr ≈ ( 10.6 )/( 2 × 60 × 1,000 × 470 × 10^( −6 ) ) = ( 10.6 )/( 56.4 ) = 0.188 VThat is about 187.9 mV . A half-wave rectifier drops the 2 and doubles the ripple.
A 5.1 V zener regulator runs from 12 V and feeds a 10 mA load. To keep at least 5 mA in the zener, what series resistor is needed, in ohms?
Ω
Answer: 460 Ω
Solution
The series resistor carries both the load and zener currents: 10 + 5 = 15 mA. Rs = ( Vs − Vz )/( IL + Iz ) = ( 12 − 5.1 )/( 0.015 ) = ( 6.9 )/( 0.015 ) = 460 ΩChoose the standard value just below, 430 Ω; a larger one would leave less than 5 mA for the zener.
20 mA flows through a 5.1 V zener diode. How much power does the zener dissipate, in mW?
mW
Answer: 102 mW
Solution
P = Vz ·Iz = 5.1 V × 0.020 A = 0.102 W = 102 mW If the load is removed, its current also goes through the zener, so size the rating for the no-load case.
At the same current, what happens to a silicon diode's forward voltage as it gets hotter?
A It falls by about 2 mV per °C B It rises by about 2 mV per °C C It stays at 0.7 V regardless of temperature D It falls by about 0.1 V per °C
Answer: A. It falls by about 2 mV per °C
Solution
A silicon junction's forward voltage has a temperature coefficient of about −2 mV/°C. So a diode at 0.65 V at 25 °C reads about 0.55 V at 75 °C. The effect is used in temperature sensors and is why BJT bias drifts with heat.
A 3.3 V microcontroller pin should drive 10 mA through a red LED (2.0 V). What series resistor is needed, in ohms? (Ignore the pin's own drop.)
Ω
Answer: 130 Ω
Solution
R = ( 3.3 − 2.0 )/( 0.010 ) = ( 1.3 V )/( 0.010 A ) = 130 ΩOnly 1.3 V is left across the resistor, so small changes in the LED voltage move the current a lot. A real pin's output voltage also sags under load.
For an ideal diode (n = 1, VT ≈ 25.9 mV at room temperature), how much does the forward voltage rise when the current increases tenfold?
A About 6 mV B About 60 mV C About 0.6 V D It doubles
Not graded yet.