# Is using Zener diode with a voltage divider sensible?

I need to convert a floating 12 V voltage to more or less stable 3.3V. So I bought the appropriate Zener diode. Now, since there's such a difference in voltage, I want to make Zener's life simpler and reduce the chance of it being blown up by using a 1:2.5 voltage divider. So I calculate the required Zener resistor value using 11 V - the lowest possible source voltage: (V/2.5 - 3.3)/Iz = (4.4 V-3.3 V)/0.005 A = 220 Ohm. And then I use this value to calculate the value of the other resistor required for the 2.5 ratio (147 Ohm).

simulate this circuit – Schematic created using CircuitLab

The question is: does this make any sense? Will this work, or will the R2 interfere with Zener's operation? Is 4.4 V high enough source for 3.3 V Zener?

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To answer you question I need 2 bits of information. What accuracy do you need from your 3.3V supply? How much current do you need to draw from the 3.3V supply? – user1582568 Feb 10 at 19:57
What is the minimum and maximum current you need at 3.3 volts? You have to take that range of load current into account when calculating the required series resistor (R1) and choosing the Zener diode power rating. R2 is just wasting power. – Peter Bennett Feb 10 at 19:59
@user1582568: Anything above 3.0 V and up to 5.0 will be OK; the higher the voltage - the better. I need to draw ~10 mA. – Violet Giraffe Feb 10 at 20:00
R2 just degrades the already terrible regulation. – Spehro Pefhany Feb 10 at 20:01
I don't think spikes are your main problem. That part can take 100usec at 6A non repetitive. With a 510 ohm resistor (same zener current) that's over 3kV. Not a big deal. But 3.3V zeners don't regulate very well and your output voltage might end up being >4V at that current, you might well run into the abs max voltage of whatever you're connect it to on the spikes. Use a TVS, filter and a regulator like everyone else, unless your requirements are in some way constrained. – Spehro Pefhany Feb 10 at 20:27

Given that your current requirement is low and you do not need great accuracy then a zener is a cheap and viable solution. I would suggest filtering the spikes and using a circuit like this.

simulate this circuit – Schematic created using CircuitLab

The values can be calculated when we have all the information.

R1 = 150ohms, R2 = 150Ohms C1 = 40V, 10uF.

The worst case automotive transient you will see will be load dump, most likely 40V for 100s of milliseconds (this is an unlikely event and modern vehicles have central load dump protection, so very unlikely to see more than 40V for any length of time). Coupled transients can be higher voltage but last for 10s of microseconds and will be removed by the R and C. If the zener is 500mW or greater is will withstand load dump condition for any length of time). Use 1W resistors to give good transient capability. The circuit will work below 9V in and still allow 5mA to bias the zener. A 3.9V zener will meet your maximum 5V requirement and not be in danger of failing to give 3V (a 3.3V zener may be a bit close at low input volts).

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That looks interesting! You've split the zener resistor into 2 in order to manipulate the capacitor's time constant? – Violet Giraffe Feb 10 at 20:19
R1 allows the source to increase or even drop to zero momentarily while the voltage on the capacitor stays repetitively constant. The time constant will be the parallel combination of R1 and R2 times C = 1.5ms. R2 will keep the zener current almost constant during short transients so the output will remain steady. You should of course include supply decoupling capacitors in you load circuit to decouple the 3.8V output. – user1582568 Feb 10 at 20:34
The time constant is R1*C1, nothing to do with R2. I would add another shunt C after R2. – EJP Feb 10 at 21:47
I didn't get the "decoupling capacitors in my load" part. What is there left to decouple on this circuit's output that C1 doesn't already take care of? – Violet Giraffe Feb 10 at 22:08
Does it make sense to limit the input voltage by means of a TVS diode parallel with D1 (or maybe with C1)? – Violet Giraffe Feb 10 at 22:08

R2 doesn't do a lot of good, except to increase the load current and decrease the overall efficiency. Consider, your circuit could be redrawn like this:

simulate this circuit – Schematic created using CircuitLab

However what you can do is calculate the maximum current your load will require, then select R1 so that at maximum current, it drops most of the excess voltage from V1. Of course if the load current decreases, then less voltage is dropped across R1, but that's why you have the zener.

Or another way to think of it: if you omit R2, then you already have a voltage divider, formed by R1 and your load. So you just need to select R1 such that R1 and your load make the voltage divider you want, accounting for the range of possible currents required by the load, and the operating range of the zener.

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No.

Consider that R2 is in parallel with your load. Removing it will be like reducing your load, which in turn means you don't need as much current through R1.

A larger R1 will mean that any transients on V1 result in a lower current spike at the zener, than would have been the case with R1 and R2.

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I can't freely manipulate R1 since I need to keep the zener in its working current range, no? – Violet Giraffe Feb 10 at 20:17
Removing R2 will allow a higher value for R1 while keeping the Zener in its working range. – Peter Bennett Feb 10 at 20:20
@PeterBennett: yes, because the voltage will be 2.5x higher, and current will be the same - which also means higher power dissipation without the divider. Am I at fault? – Violet Giraffe Feb 10 at 22:05
Let's assume a proper design with just R1 passing enough current for the load. Now add R2. R1 must be reduced to make up for the current R2 draws, which means it will get hotter, more current at the same voltage drop. For the same load and zener current range, the zener dissipation will be the same. – Neil_UK Feb 10 at 22:21
You will have higher power dissipation with the divider, as you will have current through R2, in addition to that through the Zener, and through the load. The addition of R2 will increase the current, and therefore power dissipation, in R1, as well as adding some power dissipation in R2. – Peter Bennett Feb 10 at 22:45

This page seems like much ado about nothing. It shouldn't be a big problem to create a zener shunt regulator for this particular situation.

Your R2 serves no practical purpose. I'd replace it with an electrolytic capacitor to lower the output impedance and provide a more constant output voltage that is more immune from spikes and brown-outs on the input. For a 10mA load, a value of 47uF will be just fine.

You didn't say how much your load current varies (if at all) from the 10mA nominal. Let's say it could potentially double to 20mA. And let's say the input voltage could potentially fall to as low as 10V. And let's say we want the zener current to never fall below 5mA otherwise voltage regulation will just get worse. Finally, let's use a 3.9V zener, as this will suit your requirement better.

R1's value must be calculated for minimum input voltage at maximum output current. So R1 = (10V - 3.9V) / (20mA + 5mA) = 244 Ohms. (Use a standard value of 220R)

The maximum current that will flow through R1 will be with the highest input voltage while the output is short-circuited. Assuming an automotive application, Vin could be as much as 14.5V. Maximum power dissipation in R1 will then be 14.5V * 14.5V / 220R = 0.96W, so use a 1W resistor.

Under normal running conditions the dissipation in the resistor will be: (12V - 3.9V)^2 / 220R = 298mW, which means the 1W size resistor will stay fairly cool.

The dissipation in the zener under nominal conditions will be: (((12V - 3.9V) / 220R) - 10mA) * 3.9V = 105mW, which is well within its capabilities.

Maximum dissipation in the zener will be at maximum input voltage and with the load disconnnected: ((14.5V - 3.9V) / 220R) * 3.9V = 187mW, which is still below its maximum rating. (I believe it's about 400mW.)

So, make R1 a 220R 1W resistor, D1 a 3.9V zener diode, and replace R2 with a 47uF electrolytic capacitor, and you're all good to go.

Or... you could just switch to using something like an LM317 regulator with a couple of resistors to set the output voltage, and a capacitor for stability, but that's another story.

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I'm wary of using LM317, I suspect the transients might break it. Need an automotive-grade voltage regulator, I've only found one so far from Infineon. Quite possibly, I will go with it after all, just to be on the safe side... – Violet Giraffe Feb 11 at 8:20