# Diode Circuit Analysis & Losses

## Diodes in DC Circuits

To analyze diode circuits, the state of the diode (on or off) must first be found. The diode can then be replaced by the switch equivalent circuit. However, in some circuits it may be difficult to figure out which switch equivalent to use (for example, in circuits with more than one source or with more than one diode in series). In these circuits it is helpful to replace the diodes manually with a resistive element and note the resulting current direction due to the applied voltage. If the resulting current is in the same direction as the arrow on the diode symbol, the diode is on.

Example 1

For the circuit shown in Figure 1, find the diode current (ID), the diode voltage (VD), and the voltage across resistor (VR).

Solution:

Since the current established by the source flows in the direction of the diode’s arrow, the diode is on and can be replaced by a closed switch.

Voltage across the diode               VD = 0 V

Voltage across the resistor           VR = VS – VD = 20 - 0 = 20 V

Example 2

Reverse the diode in Figure 1 and repeat Example 1.

Solution:

The direction of current is now opposite to the arrow. The diode is off and can be replaced by an open switch.

Current through the diode           ID = 0 A

Voltage across the resistor           VR = ID x R = 0 V

Voltage across the diode               VD = ES - VR = 20 – 0 = 20 V

Example 3

For the circuit shown in Figure 2, find the current (I) and voltages V0, V1 and V2.

Solution:

The two sources are aiding each other in the closed loop; the diode is on and can be replaced by a closed switch.

Applying Kirchhoff’s voltage law (KVL)

E1 – V1 – V2 + =E2 = 0

E1 – I(R1) – I(R2) + E2 = 0

Solving for I,

I = (E1 + E2) / (R1 + R2) = 25 / 7 = 3.5 mA

V1 = I x R1 = 17.5 V

V2 = I x R2 = 7.0 V

## Diodes in AC Circuits

AC circuits have a voltage that varies with the time. Therefore, there may be times when the AC voltage forward biases a diode and times when it reverse biases the same diode. Circuit analysis can be done separately for positive and negative half cycles. It must be noted when the voltage polarity across the diode forward biases it and when it reverse biases it. The diode can then be replaced by its switch equivalent circuit

Example 4

Find the switch equivalent circuit of a diode with an AC source voltage VS, as shown in Figure 3.

Solution:

During the positive half cycle, the anode is more positive than its cathode, and therefore the diode is forward biased. We can replace the diode with a closed switch.

During the negative half cycle, the anode is more negative than its cathode, and therefore, the diode is reverse biased. We can replace the diode with an open switch.

Example 5

For the circuit shown In Figure 4, draw the waveforms of the voltage across the resistance (VR) and the voltage across the diode (VD).

Solution:

During the positive half cycle, the diode is forward biased and can therefore be replaced by a closed switch. The voltage across the diode is zero, and the voltage across the resistor is the same as the source voltage. During the negative half cycle, the diode is reverse biased and can therefore be replaced by an open switch. The voltage across the resistor is zero, and the voltage across the diode is the same as the source voltage.

## Diode Losses

The total power loss that occurs in a diode is the sum of the on state, off-state, and switching losses.

PT = PON + POFF + PSW

Where

PON = VF x IF x (tON / T)

POFF = VR x IR x (tON / T)

PSW = PSW(ON) x PSW(OFF)

PSW(ON) = 1/6 VF(MAX) x IF(MAX) x tF x f

PSW(OFF) = 1/6 VF(MAX) x IF(MAX) x tR x f

In these equations

VF = Forward voltage

IF = Forward current

VR = Reverse voltage

IR = Reverse leakage current

tON = time of diode conduction

tOFF = time during which diode is reverse biased

IF = Switching time in forward direction

IR = Switching time in reverse direction

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