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This is a completely made up imaginary question to ask the question easier. Let's say I have a 0.1W 5V Zener diode. And I want to use this to regulate from 9V to a 10k load. I draw this as in the below diagram:

enter image description here

So above if I want to set Rs. I do the following:

First calculate zener current as 01W/5V = 20mA. The load current Iload = 5V/10K = 0.5mA.

So the current passing through Rs = 20+0.5 = 20.5mA.

The voltage drop across Rs = 9-5 = 4V.

Rs = 4V / 20.5mA = 195k.

Is my way of sizing Rs correct?

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There is a 1000-factor error in the last step. You wouldn't have much voltage left across Rl with a 195k resistor in series (e.g. ignore the zener, and think about the Rs and Rl arrangement as a voltage divider).

In any case, it doesn't make much sense to start the design process of a zener regulator by arbitrarily choosing the power specs of the zener upfront, and using this value as your initial input for your calculations. The initial inputs should only be the load and supply charateristics, and the zener power requirements should be deduced from that. It's the other way around.

So if we apply what I just said to your specific case (a fixed 10k load and a fixed 9V supply), what will it lead to? Well, I can straight away tell you: it will lead to a zener power requirements of exactly zero watts. Because if the input voltage is fixed, and the load has a fixed resistance, you simply need to size Rs appropriately to get your fixed voltage across the load, without any zener required. Just make the design considering Rs and Rl as a simple voltage divider.

A zener is required only if the resistance of the load varies, and/or if the supply voltage varies. The zener will then be used to "absorb" these variations so you can have a constant output voltage.

Here is the overall design process:

  • You need first to size Rs. The limit for Rs needs to be calculated considering the highest current the load may require (or minimum resistance it is equivalent to), the minimum supply voltage you can have, and ignoring the zener. Because these are the conditions where the voltage drop across Rs is maximal and the zener current is minimal. So if you consider that, in these conditions, the zener current is zero and you solve Rs to ensure the correct output voltage, that will guarantee that for all other conditions within your operating range, the target output voltage is met.

    $$Rs_{max} = \frac{Vs_{min} - Vout}{Iload_{max}}$$

    This will give you the maximum Rs to ensure proper operation, so choose Rs lower than this. The lower the value, the stronger the requirements on the zener and Rs power will be (and the supply current), but the steadier the output voltage will be (the operating conditions will be further away from the zener knee voltage).

  • Then you can deduce the zener power. Take the maximum input supply you may have, the minimum load current you may have, the chosen Rs value, and solve. That will give you the worst case maximum current you may have in the zener.

    $$Iz_{max} = \frac{Vs_{max} - Vout}{Rs} - Iload_{min}$$

    Deduce the zener power requirement:

    $$P_{Z} = Iz_{max} \times Vout$$

  • As the final step, calculate the Rs power requirements too (same input conditions as the zener power requirements).

    $$P_{Rs} = \frac{(Vs_{max} - Vout)^2}{Rs}$$

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  • \$\begingroup\$ You wrote: Rs = (Vsupplymin - Vout) / Iloadmax. But isn't the current passes through Rs is (Iloadmax+Izener)? Are you neglecting current sinked by the Zener? \$\endgroup\$ – atmnt May 20 at 12:24
  • \$\begingroup\$ @atmnt I'm neglecting the current through the zener, because in these conditions, the zener current can be zero (and you'll still have the output voltage you want). The zener starts to be useful when Vsupply > Vsupplymin and/or Iload < Iloadmax. So to calculate Rs, I choose the conditions that will make the zener current zero, so I can totally ignore it. \$\endgroup\$ – dim May 20 at 12:30
  • \$\begingroup\$ Im trying to understand this. But if Iz is zero how come the Zener can function as a regulator? \$\endgroup\$ – atmnt May 20 at 12:44
  • \$\begingroup\$ Ah okay that time the voltage divider will set ti to 5V. Takes time to digest this. It is more steps than I thought. The books I read never mentions such tricks methods, \$\endgroup\$ – atmnt May 20 at 12:46
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    \$\begingroup\$ @atmnt I appreciate the compliment... But I'm not nearly knowledgeable enough to write books. Maybe I could teach electronics to middle school children? Nah, I'd smash them after explaining the same thing for the fourth time... Nevermind, I'll just stick to stackexchange, then. \$\endgroup\$ – dim May 20 at 13:03
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Specify:

  • The MINIMUM Vin = Vinl that this will work for.

  • The MAXIMUM Vin = Vinu this will work for.

  • Decide max load current = Iu and

  • Minimum load current Il

At Vinl the resistor must drop Vinl-5.
It must do this at Iu.
So Rs = V/I = (Vinl-5)/Iu.
ie the resistor will drop just the minimum voltage allowed at max I expected.

NOW Max I = Iinu = V/R = (Vinu-5)/Rs.
This sets the max current the resistor will ever see.

Now determine the max current the Zener will ever handle
= Izm = Iinu - Iu.
ie this is max current in R - max current in load.

Zener rating min wattage = V x I = 5 x Izm.

Resistor power rating = I^2R = Iinu^2 x Rs.

QED.

E&OE.

Notes:

To fully generalise this replace V=5 with Vload.

Make resistor and zener wattages at least double the calculated wattages.

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Your method is basically correct, but you have a math error at the very last step. The resistor value you calculated is much too large.

Keep in mind that if the load is disconnected then the full 20.5mA will flow through the zener diode. The diode power will then exceed 0.1W. It would be much more reliable if the normal zener power dissipation is less than 100% of the rated power. In this case you could choose 15mA for the zener current and be on the safe side.

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  • \$\begingroup\$ I see what you mean. But is there a way to determine the reasonable current? In your case 15mA. I thought 0.1W 5V zener means if the zener passes 20mA it will regulate to 5V best. 15mA makes sense as you mention but 3mA is not okay. So is it from the datasheet we can set that current. In short where does in real 15mA estimate come from? Is it normally estimated from a plot? \$\endgroup\$ – atmnt May 20 at 11:25
  • \$\begingroup\$ The power rating of the diode is its maximum power. Look for the section in the datasheet where the zener voltage is specified, and it will also tell you the current level that was used to make the voltage measurement. That current value is probably your best choice since that is how the diode is specified. But, if it's possible that the load can be disconnected then you must consider the maximum current through the diode and stay below the maximum power and current ratings of the diode. \$\endgroup\$ – Elliot Alderson May 20 at 11:33
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5V across 10k load gives 0.5mA load current.

Allow for, say, 5mA Zener current to allow the Zener to regulate properly.

Total current drawn from supply via Rs = 5.5mA.

Value of Rs = (9-5)/5.5mA Ohms.

Power dissipated by Zener = 5 * 5mA = 25mW.

Power dissipated by Zener with load removed = 5 * 5.5mA = 27.5mW.

Both these power dissipations are well within the zener's 100mW power specification.

Power dissipated by Rs = 4V * 5.5mA = 22mW.

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