Question

With the circuit in the figure below we obtained the waveforms that can be visualized in the graph (Vsec peak: 17V / 60Hz, C = 653 uF, R = 60Ω, RL = 60Ω, Vz = 5V and rz = 5Ω). Vd on = 0.7V

D1, 110 Vac Vsec 这个字: Det

a) Analyzing the circuit and waveforms, which components are not working Justify.
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Then the components that were not working are replaced, answer :
b) What is the peak voltage (Vp) in the capacitor and in the zener? Justify.
c) What is the amplitude of the ripple voltage (Vrp) in the capacitor and in the zener? Justify

Answer 1

first of all the circuit given is of a 5V regulated power supply

a) from analyzing the circuit and waveform the issue is with the zener diode it's reverse breakdown voltage is 5V and so obviously the waveform across it should be  5V rather than 7.6V as shown in graph.

b)When the zener diode is replaced with working one , the Vp across capacitor will not have any change as we can see from the graph itself the voltage across it is correct , will explain how

secondary peak voltage Vsec peak is =17V

and as two diodes are conducting on both half cycles so the drop would be=Vd on$\times$2= 0.7$\times$2=1.4V

and peak voltage(Vp) in capacitor =Vsec voltage-Vd on$\times$2= 17V-1.4V=15.6V

and this correct in waveform.

and the peak voltage(Vp) in  the Zener diode will be it's breakdown voltage, (used here as regulator -

that is reverse biased)=5V

C)The amplitude of ripple voltage(Vrp) in the capacitor will be same as that of shown in figure= 15.6V-12.6

=3V.

and the zener diode regulator will not have any ripples and so that amplitude of ripple voltage (Vrp) =0V

the same can be observed by simulation, for reference i attached here the same circuit done in LT spice with secondary voltage 17V,60HZ the circuit and please note that here 6.2V zener is used as 5V precise zener is not available, but we can observe and understand how it works .

LTspice XVII - Draft18.asc Eile Edit Hierarchy View Simulate Tools Window Help 92 % 82t@} + 3 YDY CEME An op Draft18.asc Draft18.raw Draft18.raw X V(n003) 6.6V- 6.0V- 5.4V- 4.87- 4.2V- 3.6V- 3.0v- 2.4V- 1.87- 1.2v- 0.6V- 0.00+ Oms 3ms 6ms 9ms 12 ms 15ms 18ms 21ms 24ms 27ms 30ms Draft18.asc x D1 V1 D4 iNgip3 C60 D5 PRL 60 D SINE(O 1P6) D2 653UF EDZV6_2B .tran 32ms x = 18.86ms y = 1.253V

we can observe that  as zener is 6.2 we obtain 6.2V wave across it as shown above ,in our case it will be 5V that is the only difference, and ripple voltage is  0 .