FOUNDATION · CHAPTER 2 OF 2

Reasoning with ON and OFF states

This chapter decides which diodes conduct, turns a clipper into a table of breakpoints and argues by charge about a diode with a capacitor. You need it to analyse clippers, rectifiers and peak detectors quickly.

Updated 13 Oct 2026

What you will be able to do

You will decide which diodes conduct and prove it with two checks. You will turn a clipper into a table of breakpoints. You will explain a peak detector by charge, and estimate rectifier ripple from the load current.

Before you start

Chapter 1, Nonlinear elements: operating points and linearisation
The loop and node laws for linear circuits
$Q = C V$ ; current as charge per unit time
A sinusoid's peak, period and frequency
A transformer's secondary as an AC source

Assumed states and why the checks work

Chapter 1's $V_{γ}$ battery model splits a diode's characteristic into two half-lines, with $V_{D}$ and $I_{D}$ measured from anode to cathode. ON is the half-line $V_{D} = V_{γ}$ with $I_{D} \geq 0$ . OFF is the half-line $I_{D} = 0$ with $V_{D} < V_{γ}$ . The corner $(V_{γ}, 0)$ meets ON's condition $I_{D} \geq 0$ but fails OFF's strict $V_{D} < V_{γ}$ , so each point of the characteristic lies on exactly one piece. So if each diode's solved point meets the conditions of the state you assumed, the result solves the modelled circuit.

THE TWO STATES OF A DIODE

ON: OFF: V_{D} = V_{γ}, I_{D} \geq 0 I_{D} = 0, V_{D} < V_{γ}

$V_{γ} = 0.7$ V here (the ideal model has $V_{γ} = 0$ ). $V_{D}$ in V, $I_{D}$ in A. ON is a $V_{γ}$ battery, positive at the anode. OFF is an open circuit, and its $V_{D}$ is the open-circuit voltage the circuit leaves across it. The corner $V_{D} = V_{γ}$ , $I_{D} = 0$ is ON.

Assume ON or OFF for each diode.
Replace ON diodes by $V_{γ}$ batteries and OFF diodes by open circuits, and solve the linear circuit.
Check each ON diode for $I_{D} \geq 0$ and each OFF diode for $V_{D} < V_{γ}$ .
A guess that passes every check is a solution. Otherwise flip a diode that failed and repeat; $n$ diodes allow at most $2^{n}$ guesses.

In these tables a diode is listed from its anode (From) to its cathode (To), and a source from its negative terminal (From) to its positive terminal (To). A diode's value is its $V_{γ}$ .

Element	From	To	Value
V1	ground	VCC	10 V
R1	VCC	B	1 kΩ
D1	B	ground	0.7 V
D2	B	out	0.7 V
V2	ground	out	5 V

WORKED EXAMPLE

For the circuit in the table above, find $v_{B}$ , the current in R1 and the state of each diode.

Solution

Read the netlist: D1's anode is at B and its cathode at ground; D2's anode is at B and its cathode at out. V1's positive terminal is at VCC (10 V) and V2's at out (5 V).
Guess D1 ON and D2 OFF. D1 conducts once $v_{B}$ reaches $0 + 0.7 = 0.7$ V, while D2 needs $v_{B} = 5 + 0.7 = 5.7$ V, so D1 is the likelier path for R1's current. A guess needs no proof: the checks in step 4 settle it.
D1 is a 0.7 V battery from B to ground, so $v_{B} = 0.7$ V. D2 is open, so R1 carries $(10 - 0.7) V /1 k Ω = 9.3$ mA, all of it into D1.
Check D1: $I_{D 1} = 9.3$ mA $\geq 0$ , so ON holds. Check D2: $V_{D 2} = v_{B} - v_{o u t} = 0.7 - 5 = - 4.3$ V, below 0.7 V, so OFF holds.
Rule out the other three guesses, to show this answer is the only one. Both OFF: R1 carries no current, so $v_{B} = 10$ V and $V_{D 1} = 10$ V, not below 0.7 V. D1 OFF, D2 ON: D2 is a 0.7 V battery from B to out, so $v_{B} = 5 + 0.7 = 5.7$ V and R1 carries $(10 - 5.7)$ V / 1 kΩ = 4.3 mA into D2; but then $V_{D 1} = 5.7$ V, not below 0.7 V. Both ON: D1 fixes $v_{B} = 0.7$ V while D2 fixes $v_{B} = 5.7$ V, which cannot both hold.

Answer: $v_{B} = 0.7$ V, current in R1 = 9.3 mA (all in D1), D1 ON, D2 OFF.

Clippers as a table of breakpoints

A clipper's transfer characteristic is its output $v_{o}$ against its input $v_{i}$ . At a breakpoint a diode sits at the edge of conduction ( $I_{D} = 0$ , $V_{D} = V_{γ}$ ). Between breakpoints no diode changes state, so $v_{o}$ is one straight line. The whole characteristic is then a table with one row per range of $v_{i}$ .

Element	From	To	Value
Vi	ground	S	$v_{i}$
R1	S	out	1 kΩ
D1	out	B	0.7 V
V3	ground	B	4 V
D2	E	out	0.7 V
V4	E	ground	2 V

Vi's positive terminal is at S. For each diode, $V_{R}$ is the signed potential, relative to ground, of its terminal that is not connected to node out. D1's anode is at out and its cathode at node B, which V3 (positive terminal at B) holds at +4 V, so D1's $V_{R} = + 4$ V. D2's cathode is at out and its anode at node E; V4's positive terminal is at ground, so V4 holds E at $- 2$ V and D2's $V_{R} = - 2$ V.

WHEN A CLIPPER DIODE TURNS ON

anode at out: cathode at out: ON for v_{i} \geq V_{R} + V_{γ} ON for v_{i} \leq V_{R} - V_{γ}

Conditions: no load, and the other diode stays OFF (checked below). With both diodes OFF, R1 carries no current and $v_{o} = v_{i}$ , so the diode's open-circuit voltage is $v_{i} - V_{R}$ (anode at out) or $V_{R} - v_{i}$ (cathode at out). It turns ON when that reaches $V_{γ}$ and then holds $v_{o} = V_{R} + V_{γ}$ or $V_{R} - V_{γ}$ . Voltages in V.

D1, $V_{R} = + 4$ V: ON for $v_{i} \geq 4 + 0.7 = 4.7$ V.
D2, $V_{R} = - 2$ V: ON for $v_{i} \leq - 2 - 0.7 = - 2.7$ V.
$v_{i} \geq 4.7$ V: D1 ON holds $v_{o} = 4 + 0.7 = 4.7$ V, so R1 carries $(v_{i} - 4.7)$ V / 1 kΩ $\geq 0$ into D1. D2's $V_{D} = v_{E} - v_{o} = - 2 - 4.7 = - 6.7$ V, below 0.7 V, so D2 stays OFF.
$v_{i} \leq - 2.7$ V: D2 ON holds $v_{o} = - 2 - 0.7 = - 2.7$ V, so D2 carries $(- 2.7 - v_{i})$ V / 1 kΩ $\geq 0$ through R1 to S. D1's $V_{D} = v_{o} - v_{B} = - 2.7 - 4 = - 6.7$ V, below 0.7 V, so D1 stays OFF.

$v_{i}$ (V)	States	$v_{o}$ (V)
$\leq - 2.7$	D2 ON, D1 OFF	$- 2.7$
$- 2.7$ to $4.7$	both OFF	$v_{i}$
$\geq 4.7$	D1 ON, D2 OFF	$4.7$

A diode with a capacitor

In this model a diode's current flows only from anode to cathode. A diode wired to a capacitor can therefore push charge onto it in one direction only, and something else, here a load, must take that charge away. So you can argue by charge instead of solving a differential equation. The circuit below, with its load, turns an AC input into a nearly steady voltage: it is a half-wave rectifier with a capacitor filter.

Element	From	To	Value
Vi	ground	S	$v_{i}$
D1	S	out	0.7 V
C1	out	ground	C
RL	out	ground	$R_{L}$

Vi's positive terminal is at S and $v_{i} = V_{m} sin ω t$ ( $ω$ in rad/s); when a transformer feeds the circuit, $V_{m}$ is the peak of its secondary. D1's anode is at S and its cathode at out, so D1 can only add positive charge to C1's plate at out; only RL removes it.

PEAK DETECTOR OUTPUT

v_{o} = V_{m} - V_{γ}

No load ( $R_{L}$ removed), C uncharged at first; the result holds from the first peak on. Once $v_{i}$ passes $v_{o} + V_{γ}$ , D1 is ON while $v_{i}$ rises, holds $v_{o} = v_{i} - V_{γ}$ and charges C. As $v_{i}$ falls, D1's open-circuit voltage $v_{i} - v_{o}$ drops below $V_{γ}$ : D1 is OFF and C keeps its charge. Voltages in V.

CHARGE THE LOAD TAKES BETWEEN PULSES

Δ Q = I_{L} T

Steady state, load connected. C loses charge to the load, so D1 turns ON again only when $v_{i} - V_{γ}$ climbs back to $v_{o}$ , near each peak: it conducts in short pulses. Between pulses C alone feeds the load. $I_{L}$ load current (A), taken as constant; $T$ time between pulses (s); $Δ Q$ in C.

RIPPLE FROM THAT CHARGE

Δ V = \frac{Δ Q}{C} = \frac{I _{L} T}{C}

The next pulse restores $Δ Q$ , so by $Q = C V$ the output swings by $Δ V$ peak to peak. $C$ in F, $Δ V$ in V.

WHEN THE RIPPLE FORMULA HOLDS

\frac{Δ V}{V _{p}} = \frac{I _{L} T}{C V _{p}} \approx \frac{T}{R _{L} C} ≪ 1 ⟺ R_{L} C ≫ T

Resistive load $R_{L}$ ( $Ω$ ). $V_{p}$ is the output at the top of the ripple (V): D1 is ON at the peak of $v_{i}$ and holds $v_{o} = v_{i} - V_{γ}$ , so $V_{p} = V_{m} - V_{γ}$ here, and $I_{L} \approx V_{p} / R_{L}$ . Treating $I_{L}$ as constant needs $Δ V ≪ V_{p}$ .

WATCH OUT FOR

Using the supply period for $T$ in a full-wave rectifier, which charges C on both half-cycles. Test: count the charging pulses per supply cycle. One gives $T = 1/ f$ , two give $T = 1/ (2 f)$ , $f$ the supply frequency (Hz).

WORKED EXAMPLE

In the circuit above, $V_{m} = 20.7$ V at 50 Hz, $C = 2200 μ$ F and $R_{L} = 200 Ω$ . Find the peak-to-peak ripple.

Solution

D1 is ON at the peak of $v_{i}$ and holds $v_{o} = v_{i} - V_{γ}$ , so the top of the ripple is $V_{m} - V_{γ} = 20$ V, and $I_{L} \approx 20 V /200 Ω = 0.1$ A.
One diode conducts once per supply cycle, so $T = 1/50 Hz = 20$ ms.
Between pulses C alone feeds the load and loses $Δ Q = I_{L} T = 0.1 A \times 0.02 s = 2$ mC.
Since $Q = C V$ , $Δ V = Δ Q / C = 2 mC /2200 μ F \approx 0.91$ V.
Check the condition: $R_{L} C = 200 Ω \times 2200 μ F = 0.44$ s, far above 20 ms, and $T / (R_{L} C) \approx 4.5%$ . The formula applies.

Answer: About 0.91 V peak-to-peak.

Check yourself

CHECK YOURSELF 1

A diode has $V_{γ} = 0.7$ V. Its circuit is given below. A diode is listed from anode (From) to cathode (To), and a source from its negative terminal (From) to its positive terminal (To).

Element	From	To	Value
V1	ground	VCC	0.5 V
R1	VCC	B	1 kΩ
D1	B	ground	0.7 V

Which result is correct?

(A)
D1 is ON and $v_{B} = 0.7$ V
(B)
D1 is OFF and $v_{B} = 0.5$ V
(C)
D1 is OFF and $v_{B} = 0$ V
(D)
D1 is ON and $v_{B} = 0.5$ V

Answer and solution

Answer: (B)

Assume D1 ON. Then $v_{B} = 0.7$ V and R1 carries $(0.5 - 0.7) V /1 k Ω = - 0.2$ mA.
The check $I_{D} \geq 0$ fails, so ON is not a solution.
Assume D1 OFF. D1 is open, R1 carries no current, so $v_{B} = 0.5$ V.
Check: $V_{D 1} = 0.5 - 0 = 0.5$ V, which is below $0.7$ V. OFF holds.
So D1 is OFF and $v_{B} = 0.5$ V: option B.

CHECK YOURSELF 2

A full-wave rectifier with a capacitor filter is fed from a 50 Hz supply. It charges the capacitor on both half-cycles, so there are two charging pulses per supply cycle. The capacitor is $C = 2000 μ$ F and the load current is a constant $I_{L} = 40$ mA. Use $Δ V = I_{L} T / C$ . Find the peak-to-peak ripple in volts.

Answer and solution

Answer: $0.2$ (accepted range $\pm 0.005$ )

Two pulses per supply cycle give $T = 1/ (2 \times 50 Hz) = 10$ ms.
Between pulses the capacitor alone feeds the load, so it loses $Δ Q = I_{L} T = 0.04 A \times 0.01 s = 0.4$ mC.
From $Q = C V$ : $Δ V = Δ Q / C = 0.4 mC /2000 μ F = 0.2$ V.

CHECK YOURSELF 3One or more options are correct

A half-wave rectifier with a capacitor filter has this circuit. A diode is listed from anode (From) to cathode (To), and a source from its negative terminal (From) to its positive terminal (To).

Element	From	To	Value
Vi	ground	S	$v_{i}$
D1	S	out	0.7 V
C1	out	ground	$C$
RL	out	ground	$R_{L}$

Vi's positive terminal is at S, and $v_{i} = V_{m} sin ω t$ with $V_{m} = 10.7$ V. Take $V_{γ} = 0.7$ V. Which statements are correct?

(A)
With $R_{L}$ removed and C1 uncharged at first, the output is 10 V from the first peak of $v_{i}$ on.
(B)
With $I_{L}$ and $T$ unchanged, doubling $C$ halves the ripple $Δ V$ .
(C)
With the peak output, $C$ and $T$ unchanged, doubling $R_{L}$ doubles the ripple $Δ V$ .
(D)
The ripple formula $Δ V = I_{L} T / C$ is valid when $R_{L} C$ is much less than $T$ .

Answer and solution

Answer: (A), (B)

A: with no load and C1 uncharged at first, D1 conducts while $v_{i}$ rises and C1 charges to $V_{m} - V_{γ} = 10.7 - 0.7 = 10$ V at the first peak. As $v_{i}$ falls, D1 is OFF and C1 keeps that charge. Correct.
B: $Δ V = I_{L} T / C$ , so doubling $C$ halves $Δ V$ . Correct.
C: with the peak output fixed, $I_{L} \approx V_{p} / R_{L}$ halves when $R_{L}$ doubles, so with $C$ and $T$ unchanged $Δ V$ halves. Incorrect.
D: the formula needs $Δ V ≪ V_{p}$ , which means $R_{L} C ≫ T$ , not much less. Incorrect.
So the correct statements are A and B.

CHECK YOURSELF 4

In the $V_{γ}$ model with $V_{γ} = 0.7$ V, a diode's solved point is $V_{D} = 0.7$ V and $I_{D} = 0$ . Which state does it belong to?

(A)
ON, because ON requires $V_{D} = V_{γ}$ with $I_{D} \geq 0$ , and this point meets both
(B)
OFF, because $I_{D} = 0$
(C)
Neither state
(D)
Both states, so either may be assumed and the point has no unique state

Answer and solution

Answer: (A)

ON is the half-line $V_{D} = V_{γ}$ with $I_{D} \geq 0$ . The point has $V_{D} = 0.7$ V and $I_{D} = 0$ , so it meets both conditions.
OFF needs $I_{D} = 0$ with $V_{D} < V_{γ}$ . Here $V_{D} = V_{γ}$ is not below $V_{γ}$ , so the strict condition fails.
The point therefore lies in ON only, and each point of the characteristic lies on exactly one piece: option A.

Used in these GATE topics

Diode circuits: clipping, clamping, rectifiers and regulators