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Lecture #07, Electronic Devices

The document discusses Zener diode voltage regulators, detailing their application as DC voltage regulators with no load and under load conditions. It includes problem-solving examples for designing circuits with specific voltage and current requirements, as well as performance parameters such as line regulation, load regulation, and ripple rejection ratio. The calculations for component selection and performance metrics are provided to illustrate the concepts.

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21 views7 pages

Lecture #07, Electronic Devices

The document discusses Zener diode voltage regulators, detailing their application as DC voltage regulators with no load and under load conditions. It includes problem-solving examples for designing circuits with specific voltage and current requirements, as well as performance parameters such as line regulation, load regulation, and ripple rejection ratio. The calculations for component selection and performance metrics are provided to illustrate the concepts.

Uploaded by

hasanmehedi26696
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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EEE-2103: Electronic Devices and

Circuits
Dept. of Computer Science and Engineering
University of Dhaka

Prof. Sazzad M.S. Imran, PhD


Dept. of Electrical and Electronic Engineering
sazzadmsi.webnode.com
Zener Diode Voltage Regulators
Regulator circuit with no load:
Zener diode application = dc voltage regulator

Reference voltage source 


supplies very low current to output
RS limits Zener diode current
𝐼
IZ >≈ IZK and IL << IZ

Loaded regulator:
Zener diode regulator supplies load current IL
IS = IL + IZ
IZ >≈ IZ(min) to keep diode in reverse breakdown
IS < IZ(max)

Circuit current equation 


𝐼 𝐼
Zener Diode Voltage Regulators
Problem-12:
A 9 V reference source is to use a series-connected Zener diode and resistor connected to a
30 V supply. Select suitable components, and calculate the circuit current when the supply
voltage drops to 27 V. Assume that a 9.1 V Zener diode has knee current of 20 mA.

Assume, IZ ≈ IZK = 20 mA
When ES = 30 V
R1 = (ES – VZ)/IZ = (30–9.1)/20×10-3 = 1.05 kΩ ≈ 1 kΩ
PR1 = I12R1 = (20×10-3)2×1×103 = 0.4 W

When ES = 27 V
IZ = (ES – VZ)/R1 = (27–9.1)/1×103 = 17.9 mA
Zener Diode Voltage Regulators
Problem-13:
Design a 6 V dc reference source to operate a 16 V supply as shown in Fig. 13. The circuit is
to use a low-power Zener diode and is to produce the maximum possible load current.
Calculate the maximum load current that can be taken from the circuit. Assume that for the
Zener diode VZ = 6.2 V, IZ(min) = 5 mA and PD = 400 mW.

IZ(max) = PD/VZ = 400×10-3/6.2 = 64.5 mA


IL(max) + IZ(min) = 64.5 mA

R1 = (ES – VZ)/IZ(max) = (16 – 6.2)/64.5×10-3


= 152 Ω ( use 150 Ω standard value)
PR1 = I12R1 = (64.5×10-3)2×150 = 0.62 W

Select IZ(min) = 5 mA
IL(max) = IZ(max) – IZ(min) = 64.5×10-3 – 5×10-3 = 59.5 mA
Zener Diode Voltage Regulators
Regulator performance:
Performance parameters 
1) source and load effects
2) line and load regulations
3) ripple rejection ratio = ratio of output to input ripple amplitude

ac equivalent circuit 
replace Zener diode with its dynamic impedance ZZ
simple voltage divider circuit

1) Source effect
When ES changes by ΔES 
∆
without load, ∆𝑉
∆ ∥
with load, ∆𝑉
∥
Zener Diode Voltage Regulators
Regulator performance:
2) Load effect
Assuming RS = 0, from regulator Thevenin equivalent circuit
Ro = R1 || ZZ
ΔVo = ΔIL(R1 || ZZ)

3) Ripple rejection ration


without load,
∥
with load,
∥
Vro = output ripple amplitude
Vri = input ripple amplitude
Zener Diode Voltage Regulators
Problem-14:
Calculate line regulation, load regulation, and ripple rejection ratio for the voltage regulator
in Fig. 14. Assume that dynamic resistance of the diode is ZZ = 7 Ω and there is 10% change
in the voltage source.

Source effect:
ΔES = 10% of ES = 16×10/100 = 1.6 V
RL = Vo/IL = 6.2/59.5×10-3 = 104 Ω
∆ ∥ . ∥
∆𝑉 = 67 mV
∥ ∥
Line regulation = (ΔVo for 10% change in ES)×100%/Vo
= 67×10-3×100/6.2 = 1.08%
Load effect:
ΔVo = ΔIL(ZZ||R1) = 59.5×10-3×(7||150) = 398 mV
Load regulation = (ΔVo for ΔIL(max))×100%/Vo = 398×10-3×100/6.2 = 6.4%

∥ ∥
Ripple rejection, = 4.19×10-2
∥ ∥

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