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calculation of change in voltage in zener diode

bhuvanesh

Aug 29, 2013
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i dont know how to do the calculation in the attached image.10 for dynamic resistance i dont know 0.1 and 0.01 are used plzz help.thank you in advance
 

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KrisBlueNZ

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This section is showing you how to calculate the change in voltage across a zener diode when the current into the zener diode is varied, using the zener diode's dynamic resistance defined at a particular current.

This shows you how stable the zener voltage will be when the current varies. Generally, the zener current will vary because the input voltage (to the zener regulator) varies, and this causes a varying voltage drop across the series resistor, which causes the zener current to vary.

Ideally, the zener voltage would be absolutely steady regardless of variations in the zener current, but zeners are not perfect, and this variation in current will produce a small variation in voltage.

The quantities in the formula are:
  • delta V is the change in voltage across the zener as a result of the change in zener current, in volts;
  • Rdyn is the dynamic resistance of the zener diode at a specified current of 10 mA, in ohms;
  • delta I is the change in current that the zener will experience, in amps ("I" in formulas is always in amps, not milliamps or microamps).
In the calculation section of the formula:
  • 10 is Rdyn, expressed in ohms;
  • 0.1 is the change in current (10%) expressed as a decimal;
  • 0.01 is the actual zener current, 10 mA, expressed in amps;
The "0.1 x 0.01" section of the formula is calculating the change in zener current. The change is defined as a 10% change in a 10 mA current, so the actual amount of the change is 10% x 10 mA which is 0.1 x 0.01 amps.

The result of the formula is 0.01V which is shown at the end as 10 mV.

This means that if the zener is operating at 10 mA current and the current is increased (or decreased) by 10%, the voltage will increase (or decrease) by about 10 mV.

If the nominal zener voltage is 5V, can you calculate the error that a change of ±10 mV represents, as a percentage?
 
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bhuvanesh

Aug 29, 2013
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0.10 is error.5/100=0.05 then =>0.10/0.050=0.2 percent error am i right ?
 

KrisBlueNZ

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I'm not sure about your calculations, but that's the right answer.

So that's telling you that if your zener current is nominally 10 mA but varies by ±10% (i.e. ±1 mA), the zener voltage will vary by ±0.2%. This is intended to show you that variations in zener current DO cause variations in voltage. The effect is not large, and there are other sources of error that may be more of a problem when you use a zener diode for regulation, but you do need to be aware of it.
 
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