Is using Zener diode with a voltage divider sensible?

Question

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?

Answer 1

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 hundreds 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 tenths 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).

Answer 2

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.

Answer 3

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.

Answer 4

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.