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Above is a zener diode circuit used for voltage regulation.

In Art or Electronics textbook theres a following exercise about choosing the right zener: enter image description here

Always Vz = Vout and 10mA is the zener current wanted;

so R = (Vin(max) - Vz) / 10mA = (25 - 10) / 10mA = 1.5k Ohm.

and for Pz = ( Vin(max) - Vout) / R - Iout(min) ) * Vz

Pz = ( (25 - 10) / 1.5k - 0mA ) * 10 = 100mW

But actually if we can use Pz = Vz * Iz

Pz = 10V * 10mA = 100mW

I found different answers to this question in different forums. Am I doing something wrong here?


marked as duplicate by Dmitry Grigoryev, Voltage Spike, PeterJ, Daniel Grillo, Vladimir Cravero Sep 21 '16 at 20:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ The worst case for the zener is when there is no load so all the current goes through the zener, i.e. 100 + 10mA. So P = 10V * 110 mA = 1.1W \$\endgroup\$ – JIm Dearden Sep 20 '16 at 15:00
  • \$\begingroup\$ but doesnt it say 10mA is the worst case zener current? Where did 100 come from? \$\endgroup\$ – user16307 Sep 20 '16 at 15:02
  • \$\begingroup\$ When load current is 0A (load disconnected) the Zener current is now equal to Izmax = (25V - 10V)/1.5k = 10mA so Pmax = 10V*10mA = 0.1W. But now your regulator will never be able to provide 100mA in to the load. You choose resistor wrongly. \$\endgroup\$ – G36 Sep 20 '16 at 15:04
  • \$\begingroup\$ Haven't done this in years, but: 25V/1500Ohm = 16.7mA, so you can't get 100mA load current with 1500Ohm there. \$\endgroup\$ – Arsenal Sep 20 '16 at 15:05
  • \$\begingroup\$ For loads from 0 to 100mA with a minimum of 10mA through the zener - that means the zener has to conduct 110mA if no load current is taken. The question asks about the power rating of the zener. Now worst case for the resistor is if supplyis at 20V so it needs to supply 110mA with a 10V drop. R = 91 ohms \$\endgroup\$ – JIm Dearden Sep 20 '16 at 15:16

Might as well summarize Jim's observations:


simulate this circuit – Schematic created using CircuitLab

At the lowest input voltage, which is \$+20\;\textrm{V}\$, the output will be assumed at \$+10\;\textrm{V}\$ (the zener is doing its job) and if the worst case load of \$100\;m\textrm{A}\$ is present, then \$R_1\$ must allow for all of the load's \$100\;m\textrm{A}\$ plus the required zener's \$10\;m\textrm{A}\$, for a total of \$110\;m\textrm{A}\$ through \$R_1\$. Since \$R_1\$ cannot be allowed to drop more voltage than \$20\;\textrm{V}-10\;\textrm{V}=10\;\textrm{V}\$, it must be the case that its value is \$R_1=\frac{10\;\textrm{V}}{110\;m\textrm{A}}\approx 90.91\Omega\$.

From there, you can now take the opposite approach and say that there is no load present and ask what happens to the zener in this case, now that the value for \$R_1\$ is known. (We aren't using a standard value, but an exact value now, since the problem doesn't require standard value resistors.) To get the worst case current into the zener, we also now have to assume that the input voltage is no longer \$+20\;\textrm{V}\$, but now \$+25\;\textrm{V}\$, too. This leads to a current in \$R_1\$ of \$I_{R_1}=\frac{25\;\textrm{V}-10\;\textrm{V}}{90.91\Omega}\approx 165\;m\textrm{A}\$. So, with no load, all of that current must go through the zener. So the power dissipated by the zener in this case is \$P_{D_1}=10\;\textrm{V}\cdot 165\;m\textrm{A}=1.65\;\textrm{W}\$. As a bonus, the resistor, \$R_1\$, will be dissipating \$P_{R_1}=15\;\textrm{V}\cdot 165\;m\textrm{A}=2.475\;\textrm{W}\$.

The only reason I'm posting this is because Jim hasn't.

EDIT: The above design probably calls for a \$5\;\textrm{W}\$ resistor. You can also get \$2.5\;\textrm{W}\$ zeners. (A 1N5347B is a \$10\;\textrm{V}\$ zener specified for \$5\;\textrm{W}\$.) Also, a standard resistor value of \$91\;\Omega\$ would be so close as to not make much difference. So, the design is achievable with real parts.

However, most folks would find another way. It is a lot of unnecessarily wasted power and it is not hard nor expensive to add an emitter follower NPN BJT and track down a slightly different zener voltage (and/or to make other slight adjustments.)

I don't think The Art of Electronics authors are recommending the design. They just wanted you to figure it out for yourself.

  • \$\begingroup\$ Thanks. Now I understood the question. First we need to be sure at worst case scenario enough current can be supplied to the load. So this happens when the lowest input voltage 20V try to output 110mA in total. For that to happen R becomes maximum 90.9 ohm. But know if we set it up and one increases Vin to 25V and open the load, the zener would pass (25 - 10)/90.9 = 165mA. It means Pz = 10V * 165mA... The only problem here the resistor is not quarter watt resistor. Isnt that a problem for design? \$\endgroup\$ – user16307 Sep 20 '16 at 19:14
  • \$\begingroup\$ @user16307 You can get a \$5\;\textrm{W}\$ resistor. You can also get \$2.5\;\textrm{W}\$ zeners. So, it's not a problem unless you don't want to dissipate all that power when there's no load. Most folks would find another way, though. It is a lot of unnecessarily wasted power. I don't think the art of electronics authors are recommending it. They just wanted you to figure it out for yourself. \$\endgroup\$ – jonk Sep 20 '16 at 19:18

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