Transcript Diode Approximations

Chapter 3:
Diode Approximations
There are three (3) Diodes Approximations:
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Ideal (1st-Approximation)
Second approximation
Third Approximation
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For the 1st-approx. assume the diode drop voltage is
zero (Perfect closed switch)
For the 2nd-approx. assume the diode drop voltage of
0.7 volts
For the 3rd –approx. assume the diode drop voltage of
0.7 volt and consider the forward bulk resistance of the
diode:
Vd = 0.7 V + Id x Rb (3-5)
Ignoire bulk resistance of the diode if Rb < 0.01 Rth
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Forward Bulk Resistance
Calculate the Bulk Resistance (Rb) from the formula given below:
Rb = (V2 – V1)/(I2 – I1)
Forward bulk resistance (Rb) is very small less than 1 Ohms or 10
Ohms. The bulk resistance is the dynamic resistance while current
pass through the diode, but static
resistance that usually measured by meter is relatively small
compare to the bulk resistance.
The reverse resistance of the diode is very large greater than
1 M-ohms.
Example of 2nd-approximation
Thevenin’s Equvalect Circuit
Second Approximation Example:
In this Example the diode is forward-biased to act as a closed switch to
turn ON the 5 V @ 5W Light Bulb.
Calculat the Lamp current and compare that to the
measured lamp current by MultiSim.
The load current can be calculated by using the 2nd-approximation and
Ohms' Law:
Since this is a complex circuit (more than one loop) we must apply
Thevenin's Theorem first to convert that to a single loop:
Vth = [6/ (10+6)] (18V) = 6.75 V
Rth = 6|| 10 = (6 x 10)/(6+10) = 3.75 Ohms
Lamp Resistance (RL) can be calculated from: IL = P/V = 1W / 5V =
0.2 A
IL = 0.2 A = 200 mA
===> Lamp resistance (RL) = VL/RL = 5V/0.2 A = 25 Ohms
From the Thevenin's Equivalent circuit diode current (load current)
can be calculated by Ohm's Law:
Id = IL = (Vth - Vd)/(Rth + RL) = (6.75V - 0.7 V) / ( 3.75 +25)
Id = IL= 6.05 / 28.75 = 0.21 A = 210 mA (calculated Lamp current)
IL(Measured) = 0.207 A = 207 mA ====> They are very close !