# Regulating voltage with a zener diode

I need to supply a comparator in a bipolar way, which admits bipolar voltages of up to +-18V, like the LM139 (open collector). In its datasheet it says that the High Level Output Current is 1uA and the Low Level Output Current is 16mA. The idea at first was to do it with Arduino. However, the Arduino can't give negative voltage values, so I have to find another way to power the comparator. I use one 9V battery to give -9V. To regulate the voltage of the battery I have thought of using a zener diode acting as a voltage regulator, as shown in the following image. I choose the 1N750, which has a breakdown voltage of 4.7 V and a minimum operating current of 20 mA. Therefore, the necessary resistance to polarize it and drop a voltage of 4.7V on it is: R=(9V-4.7V)/20mA=215 ohms. Would this circuit be enough to get a stable power supply for the comparator or are there better alternatives?

UPDATE

The circuit with the zener would feed the next circuit. The BJT is at the output of the comparator while the two resistors and the diode serve to raise the voltage and prevent it from taking negative values.

• Your second schematic is drawn upside down. Positive voltages point up. Battery lifetime at 20 mA will be short. How much current does your load take? Apr 10, 2022 at 21:59
• The second schematic is necessary to draw it like this to obtain the 4.7 V, right? The cells with the zener diode would power a comparator, such as the LM139 (open collector). In its datasheet it says that the High Level Output Current is 1uA and the Low Level Output Current is 16mA. Apr 10, 2022 at 22:11
• 2nd schematic is correct; just drawn in an unconventional way. Apr 10, 2022 at 22:17
• It does not affect how the circuit functions, it’s just a convention for human readable schematics. Similar to writing from left to right. Apr 10, 2022 at 22:17
• please use save and insert in the schematic editor instead of taking a screenshot ... the inserted schematic will be editable and will have no grid lines Apr 10, 2022 at 22:31

Supply current for the LM339 (LM139 is the military temperature range version) is 2mA maximum (800uA typical), what output current you use will depend on the pullup resistor (which is your choice).

Using batteries with a zener regulator is very inefficient, you'll drain them quickly. You have about 20mA draw from each battery when fresh. If you assume battery life is down to 10mA (meaning you cannot draw more supply current @4.7V than a bit less than 10mA) the battery voltage will fall to ~6.8V, probably around 25-35 hours of life with an expensive name-brand alkaline 9V battery set.

Probably a better way would be to use a 5V supply and a 5V:-5V charge pump converter such as a 7660. Even if you have to use batteries, alkaline AA cells are much less expensive in terms of energy per unit of fiat.

• So I guess there is no point in using the zener diode for this. I could power the 7660 with Arduino without any problem right? Apr 10, 2022 at 22:56
• Yes, the 7660 itself draws very little current (just the - supply load current). If you want to combine a comparator with +/- supplies with an Arduino you would be better served by a comparator with a separate GND pin from the - supply. Otherwise you will have an open collector to the - supply which is inconvenient and requires a few more parts to safely interface. Apr 10, 2022 at 23:00
• But wouldn't the GND pin be the - supply if the comparator supply is dual? Apr 10, 2022 at 23:07
• The idea would be to connect to the supply - the 7660 to have the -5V. So the emitter of the BJT would be connected. And then, on the other hand, to the collector we would connect a pair of resistors with a diode, necessary to raise the signal and prevent it from taking negative values. Wouldn't this be okay? Apr 10, 2022 at 23:16
• Sounds like a plausible approach. Apr 10, 2022 at 23:47

Your circuit will work, but may not be necessary. The LM139 can operate from a supply as low as 2 V; also it is not necessary that the output (open collector) be returned to the same positive supply as the comparator.

Steps:

1. determine what load and V swing you need on the comparator output. Likely you have a R pull-up to some positive voltage. Try to make that R as high as possible to reduce current consumption.

2. Calculate the max. current the output can sink == (V+_Supply-V-_Supply)/R

bias your -4.7 V zener with a current ~ 50 % greater than this. 4) Your +4.7V zener won't have much load in this circuit -- you can use 1 mA to bias it.

1. Why do you need the 4.7 V Is it for some other circuit ? The LM139 can run directly from +/- 9 V.
• The circuit that provides the 4.7V does not need to be used because the Arduino can positively feed the comparator, but the circuit that provides the -4.7V is necessary. Would that be it? As a load (the resistors connected to the BJT), there is a pair of resistors and a diode to pull the signal up and prevent it from going negative. Another thing I wanted to comment on is that I need the comparator supply to be bipolar, even though it can be single, because the voltages at the input can exceed the -0.3V that the comparator admits at least. Apr 10, 2022 at 22:47

Why is there a need to regulate the supply? With enough AC bypassing, those comparators will work quite well without any supply voltage regulation. In fact, voltage regulation wasn’t as ubiquitous in analog designs as one might think. Battery-powered devices sometimes used local voltage stabilization, but were mostly designed to just run through the full battery discharge voltage range, without additional regulation.

And you don’t need to feed the comparator the voltages directly. You can use a resistor network to divide the voltages down, perhaps a trimmer potentiometer to equalize the division ratios, and just use a 5V-powered rail-to-rail input comparator. As long as you scale the voltages down to 0-5V range you’ll be fine. You’ll need a source of 2.5V as a virtual ground – an op-amp buffer will deal with it fine.

For use with an LM139, battery supply works directly: just use two 9V batteries or a single 5V supply and a small 5V-to-+/-9V DC-DC converter brick.

• Thanks for the tips. I would like to understand it better. Could you give me more details on how to design the 5V to +/-9V converter you are talking about? Apr 11, 2022 at 22:14