Diodes, LEDs and Rectification

• The diode is a one-way valve with a priceForward biased it conducts with a roughly constant drop; reverse biased it blocks until breakdown. The forward drop is not a fixed 0.7 V — it is logarithmic in current and drops with temperature, which matters for precision and for thermal runaway.Silicon signal diode: ~0.6-0.7 V. Schottky: ~0.2-0.4 V. LED: 1.8 V (red) to 3.4 V (blue/white).
• Reverse recoveryA silicon PN diode does not stop conducting instantly when reversed; stored charge must be swept out. At high switching frequencies this becomes a real loss and a noise source. Schottky diodes have essentially no reverse recovery, which is why they dominate switching supplies.Schottky trr ~ 0 ns; 1N4007 trr ~ 30 us. That is not a typo.
• Zener and shunt regulationA Zener is designed to operate in reverse breakdown at a specified voltage. With a series resistor it makes a crude shunt regulator: cheap, always burning current, and with poor load regulation. Good for clamping, poor for powering anything real.
• Rectification and rippleHalf-wave uses one diode and wastes half the cycle; full-wave bridge uses four and doubles the ripple frequency, halving the reservoir capacitor needed. Ripple voltage on the reservoir cap is set by load current, capacitance and the time between peaks.V_ripple ~ I_load x t / C, with t = 1/(2 x f_mains) for a bridge.
• Protection diodes: reverse polarity, TVS and clampingA series diode gives reverse-polarity protection at the cost of a forward drop; a P-channel MOSFET does it better. A TVS diode clamps transient overvoltage on inputs and supplies; it is specified by standoff voltage, clamping voltage and peak pulse power, not by 'volts'.
• Diode OR-ing and steeringTwo supplies feeding a load through diodes means the higher one wins automatically — the basis of battery backup. The cost is the forward drop and the leakage; ideal-diode controller ICs exist precisely because that drop is often unacceptable.

ProblemA 12 V RMS transformer secondary feeds a bridge rectifier and a reservoir capacitor, supplying 500 mA to a 9 V linear regulator that needs at least 11 V at its input. Mains is 50 Hz. Choose the capacitor.

1. 1
   Peak secondary voltage: 12 V RMS x sqrt(2) = 17.0 V.
2. 2
   A bridge puts two diode drops in the path: 17.0 - 2 x 0.8 = 15.4 V peak at the capacitor (using 0.8 V for a bridge diode at 500 mA, not 0.7 V).
3. 3
   The regulator needs 11 V minimum, so the allowed ripple is 15.4 - 11.0 = 4.4 V peak-to-peak. Use 3 V to leave margin for low mains.
4. 4
   A full-wave bridge recharges the cap twice per mains cycle, so the discharge interval t = 1/(2 x 50) = 10 ms.
5. 5
   C = I x t / V_ripple = 0.5 x 0.010 / 3 = 1667 uF.
6. 6
   Round up to a standard 2200 uF. Actual ripple = 0.5 x 0.010 / 2200e-6 = 2.3 V.
7. 7
   Voltage rating: peak is 17 V with no load and can be higher at +10% mains, so 17 x 1.1 = 18.7 V. Choose a 35 V electrolytic, not a 25 V one.
8. 8
   Now check ripple current, which is what actually kills electrolytics: the cap only conducts in short charging pulses, so RMS ripple current is roughly 1.5 to 2x the DC load, i.e. about 1 A RMS. The chosen part must have an Irms rating above that at 100 Hz.

Answer2200 uF, 35 V, ripple-current rating >= 1 A at 100 Hz. Resulting ripple 2.3 Vpp, minimum input to the regulator 13.1 V.

Tool: ngspice - rectifier transient

• Create bridge.cir:

  Bridge rectifier
  Vac a b SIN(0 17 50)
  D1 a p DMOD
  D2 b p DMOD
  D3 n a DMOD
  D4 n b DMOD
  C1 p n 2200u IC=0
  Rload p n 24
  Vn n 0 DC 0
  .model DMOD D(Is=1e-9 N=1.8 Rs=0.02)
  .tran 100u 200m UIC
  .end

• Rload = 24 ohm draws roughly 500 mA at ~12 V, matching the worked example.
• Run and: plot v(p,n) — watch the capacitor charge over the first few cycles and settle into a ripple pattern.
• Measure the ripple: meas tran vmax MAX v(p,n) FROM=150m TO=200m, then meas tran vmin MIN v(p,n) FROM=150m TO=200m. Subtract. Compare to your 2.3 V prediction.
• Plot the diode current: plot i(D1) — note it flows in narrow high-amplitude pulses, not smoothly. This is why ripple current rating matters.
• Change C1 to 470u and re-run. Confirm the ripple scales inversely with capacitance.
• Change the model to a Schottky-like device (N=1.05, Is=1e-6) and observe the higher output voltage from the lower forward drop.

• Semiconductor Devices - Theory and Application (Fiore)open licencereading~150 minOpen textbook, CC BY-NC-SA. Read the PN junction and diode application chapters (rectifiers, clippers, clampers, Zeners).
• All About Circuits - Diodes and Rectifiersfreereading~60 minStrong on rectifier topologies and practical diode ratings.
• 1N4148 and 1N5819 datasheets (any manufacturer)freedatasheet~30 minPull both and compare side by side: Vf at 100 mA, reverse leakage, trr, and maximum reverse voltage. Build the habit of comparing real parts.
• Littelfuse TVS diode selection guidefreereference~30 minLearn the vocabulary: reverse standoff voltage, breakdown voltage, clamping voltage, peak pulse current. Needed for any board with an external connector.

Design the front end of a small 12 V DC-input board: reverse-polarity protection, transient protection, a power-path OR between a 12 V jack and a 9 V battery, and a status LED. Every part is specified from a real datasheet.

Deliverables

• A schematic sketch (hand-drawn or in KiCad) of the full front end, with every part labelled with a real manufacturer part number.
• Reverse-polarity protection: implement it two ways — a series Schottky and a P-channel MOSFET — and compare them numerically on forward drop and power dissipation at 1 A. State which you would ship and why.
• TVS selection: choose a part for a 12 V rail, showing that its reverse standoff voltage is above the maximum normal rail voltage and its clamping voltage is below the absolute maximum of the downstream regulator.
• Diode OR: show the switchover behaviour with numbers — at what jack voltage does the battery take over, and how much does the drop cost you?
• An ngspice simulation of the OR-ing circuit sweeping the jack voltage from 0 to 12 V, plotting which source supplies the load.
• A one-page 'what I would change for a production board' note.

Acceptance criteria — how you know it is good enough

• Every diode is a real part with datasheet numbers quoted, not a generic symbol.
• The Schottky vs P-FET comparison includes power dissipation at 1 A for both.
• The TVS standoff and clamping voltages are checked against both the normal rail and the downstream absolute maximum.
• The OR-ing simulation clearly shows the crossover point.
• The LED calculation from the first skill is reused correctly for the 12 V rail.

If you want to push further

• Add inrush current limiting with an NTC or a soft-start P-FET and calculate the peak current into the input capacitance without it.
• Model the TVS clamping action against a simulated 100 V, 1 us transient and check the clamp voltage.

• I can explain the diode I-V curve and why Vf is not a fixed 0.7 V.
• I can choose between silicon and Schottky using reverse recovery and forward drop as the criteria.
• I can size a reservoir capacitor for a given ripple and check its ripple-current rating.
• I can specify a TVS from standoff, clamping and peak pulse power.
• I can design reverse-polarity protection two ways and defend the trade-off.
• I never leave an inductive load or an external connector unprotected.