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If supply is producing max 20mA at 9V what will be the value of Rz? To improve the current capability, how can we use NPN transistor instead of diode here?

Ans: According to my understanding, it is step down transformer and the output of the circuit should be 9V DC. Zener diode is performing the function of the voltage regulator providing constant 9V at the output side. Power Dissipated at zener, P=VI=9*20mA=0.18W To find the value of RZ, at no load condition all 20mA current will flow through 9V zener diode 12.5V - 9V/Rz = 20mA therefore Rz=175ohm

Stuck with the second part where ciruit could perform same operation with NPN transistor, need help. Thank you

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    \$\begingroup\$ I don't know any practical way to use an NPN transistor instead of a zener, but you could use both together. You could use an NPN transistor as a ~9V zener but it would not improve the current capability. \$\endgroup\$ – Spehro Pefhany May 15 '20 at 23:50
  • \$\begingroup\$ okay. Thanks. and for the first part, does my understanding of the circuit and corresponding calculation seems right to you? I am not sure about the same \$\endgroup\$ – Prajakta Bhutkar May 15 '20 at 23:54
  • \$\begingroup\$ Yes, it looks correct, for the simplified conditions as described. Be careful how you write (12.5V-9V)/Rz. \$\endgroup\$ – Spehro Pefhany May 15 '20 at 23:58
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The assumption of 12.5Vdc after the diode is wrong.

If you had a 12V transformer that is the rated Vac rms at its rated load @ tbd VA, you would not get 12.5V. It is slightly < 10Vac.rms input but that's not my point.

It is unlikely your transformer is rated at 20mA @ 12V or 0.24VA so the output voltage Vac and VA rating must be given. Normally the transformer losses are typ. 10%, so the no load voltage will always be this amount higher.

The 3rd source of error comes from full rated load with high reactive loads with poor power factor. This results in higher peak currents and square law losses thus the temperature will rise so the VA rating at 85'C must be reduced by about 30%.

Therefore your assumption of 12.5V is wrong, therefore your conclusion will have errors. (not huge , but misleading when you compute the wrong secondary Vrms)

Finally, Zeners have high voltage ripple with AC load current due to input ripple voltage voltage Vacpp into Zener threshold resistance Zzt and the knee resistance Zzk. This must be specified in your design for Vmin and Vmax in addition to Pd and Trise.

yes there are better ways to regulate and reduce ripple or reduce temp rise but linear half wave is not the best way to power a 20mA load.

Final Words of Advice.

Try to specify the source better and then the load specs for ripple, Vmin, Vmax, Vdc as the ripple is not symmetric. Also component temperature, T_rise is best to choose 50% of max rated power, as Pmax which is often rated at 125'C case temp or 100'c C rise.

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  • \$\begingroup\$ I don't think the question states that it uses a 12 volt secondary. \$\endgroup\$ – Andy aka May 16 '20 at 8:20
  • \$\begingroup\$ @Andy Perhaps, the more obvious is the issue is it doesn't state ANY secondary Vac and the rectifier increases the RMS voltage by 30 to 40% depending on load factor adding minor errors to the calculation of optimum series R. It is also an error to compute based on constant current in the Zener as the current is not sinusoidal so RMS heat loss is not linear. I should omit that unnecessary point. The desired method is secondary Vac @ Ic and Vout Max/min to get both mean and ripple. Then state the load and any variation in load. \$\endgroup\$ – Tony Stewart EE75 May 16 '20 at 11:01
  • \$\begingroup\$ But then , that's probably not his fault, but the author of the question. \$\endgroup\$ – Tony Stewart EE75 May 16 '20 at 11:11

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