Diodes

Clipper

A clipper (limiter) is a diode circuit that removes (clips) the part of the input signal above or below a chosen level, without distorting the rest of the waveform. It uses a diode, a resistor and sometimes a DC battery to set the clipping level.

• Series clipper: diode in series with the load.
• Shunt (parallel) clipper: diode across the output.
• Types: positive, negative, biased and combination (two-level) clippers.

Example: shunt positive clipper (ideal Si diode, 0.7 V)

 vin o--[ R ]--+------o vo
               |
              _|_  D (anode up)
              \ /  cathode to ground
               |
 o-------------+------o

For vin = 10 sin ωt: in the positive half the diode conducts once vin > 0.7 V, so vo is held at 0.7 V; in the negative half the diode is off and vo = vin. Output swings from +0.7 V to −10 V. Adding a battery in series with the diode moves the clipping level (e.g. 3 V battery → clip at 3.7 V).

Uses: wave shaping, protecting circuits from over-voltage, noise limiting.

Clamper

A clamper (DC restorer) shifts the whole waveform up or down by adding a DC level, without changing its shape or peak-to-peak value. It needs a capacitor, a diode and a resistor; the time constant RC must be much larger than the signal period so the capacitor does not discharge noticeably.

Example: positive clamper

 vin o--| C |--+------o vo
               |
              _|_  D (cathode up)
              /_\  anode to ground
               |    R across output
 o-------------+------o

For vin = ±10 V square wave: during the negative half the diode conducts and C charges to 10 V; then vo = vin + 10 V, so the output swings from 0 to +20 V (ideal diode). Peak-to-peak stays 20 V. Reversing the diode gives a negative clamper (0 to −20 V).

Uses: TV receivers (DC restoration), voltage multipliers.

A half-wave rectifier converts AC into pulsating DC by passing only one half-cycle of the input and blocking the other, using a single diode.

Circuit

        Tr          D
 AC  o-)||(--+---->|----+------+
 mains )||(  |          |      |
       )||( vs         RL     vo
       )||(  |          |      |
     o-)||(--+----------+------+

Operation

• Positive half-cycle: the top of the secondary is positive, the diode is forward biased and conducts. Current flows through RL, and vo ≈ vs (minus 0.7 V for a Si diode).
• Negative half-cycle: the diode is reverse biased and acts as an open switch. No current flows, so vo = 0. The diode must withstand the peak inverse voltage PIV = Vm.

Waveforms

 vs  |  _        _
     | / \      / \
   0 +/---\----/---\----> t
     |     \__/     \__/

 vo  |  _        _
     | / \      / \
   0 +/---\____/---\____> t

The output is unidirectional but pulsating: one pulse per input cycle, so the ripple frequency equals the supply frequency (50 Hz).

Important values (ideal diode)

It is simple and cheap but has high ripple and low efficiency, so it is used only in low-power circuits; a filter capacitor across RL is added to smooth the output.

A Zener regulator uses a series resistor Rs and a Zener diode in parallel with the load. The Zener keeps the load voltage at Vz; Rs drops the extra input voltage.

 Vi o--[ Rs ]--+-------+
     30-50 V   |       |
             Zener    RL = 1 kΩ
             Vz=20V    |
 o-------------+-------+

Given: Vz = VL = 20 V, RL = 1 kΩ, Vi = 30 V to 50 V. Assume an ideal Zener (Iz(min) ≈ 0, rz = 0), as in the usual textbook design (Boylestad).

Load current (constant)

IL = VL/RL = 20/1000 = 20 mA

Choosing Rs (worst case: minimum input)

At Vi(min) = 30 V the Zener must still be in breakdown, so the current through Rs must at least supply the load (Iz ≥ 0):

Rs = (Vi(min) − Vz)/IL
   = (30 − 20)/20 mA
   = 500 Ω

Maximum Zener current (maximum input)

I_Rs(max) = (Vi(max) − Vz)/Rs
          = (50 − 20)/500 = 60 mA
IZM = I_Rs(max) − IL = 60 − 20 = 40 mA

Answer: Rs = 500 Ω, IZM = 40 mA.

Zener power rating needed: PZM = Vz × IZM = 20 × 40 mA = 0.8 W, so select a 20 V Zener of at least 1 W.

Rs dissipates (50 − 20)²/500 = 1.8 W at maximum input, so use a 2 W (or higher) resistor. In practice Rs is chosen a little below 500 Ω (e.g. 470 Ω standard value) so a small minimum Zener current Iz(min) flows at 30 V input and regulation is kept.

Clamper

A clamper (DC restorer) is a circuit of a capacitor, a diode and a resistor that shifts the whole input waveform up or down by a DC level without changing its shape or peak-to-peak value. A battery in series with the diode sets the level at which the waveform is clamped. The time constant RC must be much larger than the half-period of the input.

Output waveform

Assumption: a clamper needs a series capacitor, so the series element is taken as a capacitor C (large RC). Si diode: V_D = 0.7 V. The diode with the 3 V battery conducts when vo ≥ 3 + 0.7 = 3.7 V.

Positive half (vin = +6 V): the diode is forward biased. The output is held at

vo = 3 + 0.7 = 3.7 V
C charges to: Vc = vin − vo = 6 − 3.7 = 2.3 V
(left plate +)

Negative half (vin = −6 V): the diode is reverse biased (open). The capacitor keeps its 2.3 V, so

vo = vin − Vc = −6 − 2.3 = −8.3 V

Check: peak-to-peak output = 3.7 − (−8.3) = 12 V = input peak-to-peak ✓.

 vin                     vo
 +6 |‾‾‾|   |‾‾‾|      +3.7 |‾‾‾|   |‾‾‾|
  0 +---+---+---+--      0 +---+---+---+--
 -6 |   |___|   |___        |   |   |   |
                       -8.3 |   |___|   |___

Answer: the output is a square wave switching between +3.7 V and −8.3 V (shifted down by 2.3 V), same shape and 12 V peak-to-peak.

Note: if the series element is really a resistor (as drawn), the circuit acts as a biased clipper, not a clamper: vo = +3.7 V during the positive half (diode on) and vo = −6 V during the negative half (diode off).

The large signal model of a pn junction diode is an approximate equivalent circuit used when the voltage and current swings are large (as in rectifiers, clippers and switches), so the diode is treated as switching between ON and OFF states rather than working at a small region around a Q-point.

The real diode follows the exponential law I = Is (e^(V/ηV_T) − 1). This is replaced by straight-line (piecewise-linear) pieces. Three common levels are used:

1. Ideal diode model

• Forward bias: short circuit (V_D = 0, any current).
• Reverse bias: open circuit (I = 0).
• Used when supply voltages are much larger than 0.7 V.

2. Constant voltage drop (simplified) model

• ON when V_D ≥ V_γ: a battery V_γ (0.7 V for Si, 0.3 V for Ge) in series with an ideal diode.
• OFF when V_D < V_γ: open circuit.
• Most often used in hand analysis.

3. Piecewise-linear model

• ON: battery V_γ in series with the forward resistance r_f (slope of the straight line, typically a few ohms to tens of ohms) and an ideal diode.
• OFF: open circuit (or a very large reverse resistance).
• I_D = (V_D − V_γ)/r_f for V_D > V_γ.

Model         Equivalent (ON)        I-V shape
Ideal         --o--                  vertical at 0
Const. drop   --|+ -|--  (0.7 V)     vertical at 0.7 V
Piecewise     --|+ -|--/\/\--        slope 1/r_f
                 V_γ    r_f          from 0.7 V

  I |      /   piecewise linear
    |     /
    |    /  slope = 1/r_f
    |___/____________ V
    0  V_γ

The diode is first assumed ON or OFF, the circuit is solved, and the assumption is checked (I_D > 0 for ON, V_D < V_γ for OFF). The large signal model is different from the small signal model, which uses only the dynamic resistance r_d = ηV_T/I_D around a Q-point.

The Zener diode holds the load voltage at Vz = 10 V (it is in breakdown because the open-circuit load voltage 18 × 1000/1250 = 14.4 V is greater than 10 V).

 Vs=18V o--[ Rs=250Ω ]--+--------+
                        |        |
                      Zener    RL=1kΩ
                      Vz=10V     |
        o---------------+--------+

Current through Rs

Is = (Vs − Vz)/Rs = (18 − 10)/250 = 0.032 A = 32 mA

Load current

IL = Vz/RL = 10/1000 = 0.010 A = 10 mA

Zener current (KCL)

Iz = Is − IL = 32 − 10 = 22 mA

Power dissipated in the Zener

Pz = Vz × Iz = 10 × 0.022 = 0.22 W = 220 mW

Answer: Iz = 22 mA, Pz = 220 mW.

(Power in Rs = 8 × 0.032 = 0.256 W; power in the load = 0.1 W.) The Zener should be rated above 220 mW, e.g. a 10 V, 0.5 W Zener.