DIODE CIRCUIT MODELS

9.7 Find voltage vL in the circuit of Figure P9.7, where D is an ideal diode, for positive and negative values of vS. Sketch a plot of vL versus vS.

1. The diode is given as ‘ideal’ i.e. we can replace it with a switch such that

At vD < 0 (negative) => the switch is open and,

At vD ≥ 0 (positive) => the switch is closed

2. For computing vL: find equivalent resistance replacing the diode with a switch as per the above conditions and then computing the voltage across RL

3. Show, graphically, how vL will change with varying values of vs.

The voltage at V1 will decide if the diode is OFF or ON

SS

VRR

RV .1

11 +=

At v1 < 0 ⇒ the diode switch is OPEN

Hence no current flows in RL . Thus, VL =0V

At v1 >= 0 ⇒ the diode switch is CLOSED

KCL at node 1 ⇒ iS - i1 - iL = 0

LS

L

S

L

RRRRR

VV

||||

1

1

+=

DIODE CIRCUIT MODELS

9.8 Repeat Problem 9.7, using the offset diode model.

1. This question is similar to Q 9.7, just that a different diode model is used. The diode model to be used is offset diode model (described in pg 471), i.e. we can replace it with a switch such that

At vD < vγ (negative) => the switch is open and,

At vD ≥ vγ (positive) => the switch is closed

Where vγ is the offset voltage.

2. For computing vL: replace the diode with a switch with a voltage source (of offset voltage value) as per the above conditions and then compute the voltage across RL

3. Show, graphically, how vL will change with varying values of vs.

4. Another example of offset diode model can be found in example 9.3, pg 471 and 9.4, pg 473

The offset diode could be depicted as an ideal diode in series with barrier potential Vγ. The voltage at V1 will decide if the diode is OFF or ON

SS

VRR

RV .1

11 +=

At ⇒ the diode switch is OPEN

Hence negligibly small current flows in RL . Thus, VL ~ 0V

At ⇒ the diode switch is CLOSED

KCL at node 1 ⇒ iS - i1 - iL =0

γVVVL −= 1

)1(.1R

RVV S

S +< γ

)1(.1R

RVV S

S +≥ γ

LS

LSS

RRRRVRV

V/1/1/1

//

11 ++

+= γ

DIODE CIRCUIT MODELS

9.13 In the circuit of Figure P9.13, find the range of Vin for which D1 is forward-biased. Assume ideal diodes.

The diodes are given as ‘ideal’. For forward biasing, we canreplace D1 with a switch such that

At vD ≥ 0 (positive) => the switch is closed

Additional: Can you see that voltage across 500Ω resistor cannever exceed 1.5V? (hint: when both diode startsconducting, meaning to say the “switches” are closed, the1.5V source is parallel to the 500Ω resistor.

DIODE RECTIFIER CIRCUITS

9.27 In the rectifier circuit shown in Figure P9.26, v(t) = A sin(2π 100t) V. Assume aforward voltage drop of 0.7V across the diode when it is conducting. If conductionmust begin during each positive half-cycle at an angle no greater than 50, what is theminimum peak value A that the AC source must produce?

Note that the diode is NOT an ideal type. Similar to problem 9.8.

For the diode to conduct:

Condition 1: we are told that it needs minimum 0.7 V (forward voltage drop)

Condition 2: we are told that it needs minimum (no greater than) 50 in the positive cycle.

From 1 and 2 ⇒

Thus, Amin could be computed as 8.03V

Ans: Amin = 8.03 V

DIODE RECTIFIER CIRCUITS

9.29 A half-wave rectifier, similar to that of Figure 9.25 in the text, is used to provide aDC supply to a 100-Ω load. If the AC source voltage is 30 V (rms), find the peak andaverage current in the load. Assume an ideal diode.

Ans: Ipeak = 0.424 A and Iavg=0.135 A

RVIVV

RVIVV

avgavgpeakavg

peakpeakrmspeak

//

/.2

=⇒=

=⇒=

π

For computing peak and average currents; use the AC relationships:

See example 9.8 pg 486 for another similar example on half-wave rectifier

9.37 In the full-wave power supply shown in Figure P9.37 the diodes are 1N4001 with a rated peak reverse voltage (also called peak inverse voltage) of 50 V. They are fabricated

from silicon. VLine = 170 cos(377t) V, n = 0.2941, C = 700 μF, RL = 2.5 KΩDetermine the actual peak reverse voltage across each diode.

1. This is a standard bridge rectifier circuit. Note that the diodes are not ‘ideal’ but they are identical (same IC number); so they would have similar characteristics (voltage- drops/current etc). We also know that at any instant of time (positive or negative cycle), two of them are conducting and two are open.

2. Note, n<1, so you have a step-down transformer. Find VS = n.Vline

= ~ 50.cos (377t) V

For peak value cos(0)=1

3. Since diodes are identical, consider any pair, let’s say D1 and D3.

0)0cos(.)0( 31 =−++ SOL VVVV γγ

Write KVL equation for conduction (ωt = 0) for this pair as shown on R.H.S

Similarly, Write KVL equation for non-conduction (ωt =π) for the same pair:

0)cos(.)()()( 31 =−−− ππππ SODDL VVVV

Evaluate VD1=VD3=?

The peak reverse voltage is the voltage across the diodes when they are reverse biased (open circuit while no current flows). It is reverse polarity - negative.

Ans: - 49.3 V

VL(0) = 50 - 0.7 - 0.7 = 48.6V

VL(0)=VL(π)= 48.6V

Assuming the forward voltage drop/barrier potential to beVγ =0.7 V

)6.4850(31 +−=+ DD VV