# Using a zener as a means to create a voltage supply for an IC

#### mike _Jacobs

Joined Jun 9, 2021
122
So let’s say you have a 12v battery
You need 5v to turn on a small IC
You don’t have the room for a small converter so you opt for a zener as a voltage regulator to produce the 5v

here is the question.
How do you calculate the current you can pull off this voltage regulator before the zener voltages is no longer the assumed 5v regulation

circuit would would

power supply rail
A resistor then the Zener to ground
The voltage regulation point is between the resistor and Zener

#### boostbuck

Joined Oct 5, 2017
416
It's normal to allow a big multiple of the required current to flow through the resistor and the zener. Eg if the supply to the chip was 100mA then the resistor is calculated to allow 500mA or more with most going through the zener. This excess allows the zener to regulate the voltage as the demand of the IC fluctuates.

The actual excess current required through the zener depends of the fluctuation of IC current - if the IC has a completely fixed current demand (eg it's equivalent to just a resistor) then the zener is superfluous, and if the load varies a lot then there should be enough current to keep the regulation within required limits. The zener I/V curve will tell where the design limits are.

Zeners are a poor choice. These days, there are much better regulators than zeners in very small sizes.

#### mike _Jacobs

Joined Jun 9, 2021
122
It's normal to allow a big multiple of the required current to flow through the resistor and the zener. Eg if the supply to the chip was 100mA then the resistor is calculated to allow 500mA or more with most going through the zener. This excess allows the zener to regulate the voltage as the demand of the IC fluctuates.

The actual excess current required through the zener depends of the fluctuation of IC current - if the IC has a completely fixed current demand (eg it's equivalent to just a resistor) then the zener is superfluous, and if the load varies a lot then there should be enough current to keep the regulation within required limits. The zener I/V curve will tell where the design limits are.

Zeners are a poor choice. These days, there are much better regulators than zeners in very small sizes.
so how do you calculate how much “excess” current the Zener can take before its voltage is no longer its voltage

#### crutschow

Joined Mar 14, 2008
33,325
so how do you calculate how much “excess” current the Zener can take before its voltage is no longer its voltage
It's based upon it's power rating and the maximum ambient temperature, so you never want to exceed that.
The Zener power dissipated is simply its current times its Zener voltage.

Much better way to get a regulated voltage is an IC regulator such as the common LM317, or an LM7805.

#### dl324

Joined Mar 30, 2015
16,110
so how do you calculate how much “excess” current the Zener can take before its voltage is no longer its voltage
You want to make sure that the resistor in series with the zener doesn't drop enough voltage to cause the zener to stop regulating well enough.

You don't have to have 5x the load current going through the zener. You just want to keep the zener past it's knee voltage.

What is the IC and it's maximum current draw?

#### mike _Jacobs

Joined Jun 9, 2021
122
Interesting info here
There is no ic
It was just something I was thinking about as a general question

#### mike _Jacobs

Joined Jun 9, 2021
122
It's based upon it's power rating and the maximum ambient temperature, so you never want to exceed that.
The Zener power dissipated is simply its current times its Zener voltage.

Much better way to get a regulated voltage is an IC regulator such as the common LM317, or an LM7805.
so your saying if the part is rated for x watts
As long as you stay under that it will maintain its regulation point?

#### MrChips

Joined Oct 2, 2009
29,801
I think you are asking the wrong question.

The question should be, ”What is the value of the series resistor to protect the zener diode?”

For this, you need to specify the maximum current the load will ever draw. The purpose of the zener diode is to take up the spare current. For example, suppose you design for a maximum current of 500mA and the load happens to be cruising along at 300mA, then the zener diode has to conduct 200mA. But that does not matter. What matters is the design is for 500mA max when no load is connected. If the load exceeds the max current, the zener diode conducts little current and you lose zener regulation, i.e. the voltage to the load falls below the zener voltage.

Now you can calculate the resistance required.
R = (12V - 5V)/500mA = 14Ω

Now you need to calculate the safe wattage of both the resistor and the zener diode.
In this example, you will see that the resistor power rating will need to be 5W and the zener diode power rating should exceed 3W.

#### WBahn

Joined Mar 31, 2012
29,481
So let’s say you have a 12v battery
You need 5v to turn on a small IC
You don’t have the room for a small converter so you opt for a zener as a voltage regulator to produce the 5v

here is the question.
How do you calculate the current you can pull off this voltage regulator before the zener voltages is no longer the assumed 5v regulation
The answer to the specific question is that you can pull of all of the current except the minimum current needed to keep the Zener at it's Zener voltage.

The real question that decides things is what is the maximum current that the Zener needs to conduct, which occurs when the load is at its minimum current. In most designs, it is usually assumed that this is equal to the maximum load current (because it is assumed that we have to allow for the possibility that the load is removed entirely). But occasionally we can assume a minimum load, perhaps because it is not possible to actually remove the load.

So let's assume that the load has a minimum current of Iloadmin and a maximum current of Iloadmax. Let's further assume that we are using the classic simple Zener regulator, like you described, which is a high supply voltage and series voltage dropping resistor feeding the parallel connection of the Zener and the load. We need to examine the data sheet and decide on what minimum Zener current we want in order to keep adequate regulation. Call that Izreg.

We also need to account for the power supply voltage variation, so let's call that range Vsmin to Vsmax.

Now it's just a matter of seeing what the requirements that need to be met are.

The minimum current in the Zener is going to happen when the supply voltage is at it's minimum value and the load is drawing the maximum current.

Iz = (Vsmin - Vz)/R - Iloadmax > Izreg

This just says that the Zener current is current in the resistor when it has the smallest voltage across it minus the maximum current going to the load and that this must be greater than the minimum current needed to keep the Zener in regulation.

Solving for R, we get

R < (Vsmin - Vz)/(Izreg + Iloadmax) = Rmax

We also have to honor the maximum power that the Zener can handle, which will happen when the supply voltage is at it's max and the load current is at its min.

P = Iz·Vz = [(Vsmax - Vz)/R - Iloadmin]·Vz < Pzmax

Again solving for R, we get

R > (Vsmax - Vz)/[(Pzmax/Vz) + Iloadmin] = Rmin

So you can pick any resistor that lies between Rmin and Rmax.

Rmin = (Vsmax - Vz)/[(Pzmax/Vz) + Iloadmin] < R < (Vsmin - Vz)/(Izreg + Iloadmax) = Rmax

This is the general case. You can apply the assumption that the load can be completely removed by simply setting Iloadmin to zero. You can apply the assumption that we have a solid supply voltage by simply setting Vsmin and Vsmax to Vs.

Note that the Pzmax in the equation is not necessarily the power rating of the Zener. It's a bad idea to operate components at their max power. So you would pick a Pzmax that is suitably lower than the rated power. How much lower depends on the application. Lacking specific knowledge of the application, a common rule of thumb is to not dissipate more than half the rated power in a component.

• MisterBill2

#### WBahn

Joined Mar 31, 2012
29,481
I forgot to mention that, like voltage dividers, Zener regulators are good for low-power stuff, such as reference voltages. They aren't that good for use as a voltage regulator that delivers more than small amount of power -- but they are well suited to providing reference voltages for voltage regulator circuits that do.

#### MrAl

Joined Jun 17, 2014
10,881
So let’s say you have a 12v battery
You need 5v to turn on a small IC
You don’t have the room for a small converter so you opt for a zener as a voltage regulator to produce the 5v

here is the question.
How do you calculate the current you can pull off this voltage regulator before the zener voltages is no longer the assumed 5v regulation

circuit would would

power supply rail
A resistor then the Zener to ground
The voltage regulation point is between the resistor and Zener
Hi,

To add to the other great replies...

First, you can't calculate an accurate current because zeners are not very good at keeping the voltage constant under load and input variations that are significant. Most of them are not temperature compensated either. In short, it's not the best method to regulate a voltage by far. More modern ways are much better and can guarantee success. You can only calculate things if you are willing to accept not-so-good approximations.

There are so many small regulator IC's that can do this and are very small. The 78L05 or similar comes to mind. It's a series regulator that regulates to about 5 percent i think over a wide range of input and output extremes as well as some temperature extremes.
Next up is the small TL431 IC, which is a shunt regulator. With a few resistors and a capacitor or two it works almost like a zener but with much, much better regulation (BTW zeners are employed as shunt regulators usually).

The simplest way to know if a zener circuit is going to poop out on you is if the output voltage gets close to the zener voltage even without the zener in the circuit, as a calulation that is. So once you have the circuit you think will work set up and you do some calculations, remove the zener and find out what load will cause the voltage across the load to be the same as the zener voltage (again with no zener in the circuit). That's the limit of the load. You can also test the input limits this way too. With no zener in the circuit, find the input voltage with the required load where the load voltage becomes equal to the zener voltage (again no zener in the circuit for this test also).

#### boostbuck

Joined Oct 5, 2017
416
so how do you calculate how much “excess” current the Zener can take before its voltage is no longer its voltage
Its voltage is never its 'voltage'. If you look at a I/V chart on a zener diode datasheet, you'll see that the zener knee is rounded and the slope is not perpendicular - a zener's voltage is dependent on the current through it to a disappointing extent. When designing with a zener the change in zener voltage with current needs to be allowed for to determine the operating point, which is why zeners are a poor choice for a load that varies a large amount but needs close voltage regulation.

• MrAl

#### MrSoftware

Joined Oct 29, 2013
2,143
Why even consider a Zener, when there are linear regulators that are absolutely tiny. Here are 2 that are each 2mm x 2mm:

#### BobTPH

Joined Jun 5, 2013
8,068
A linear regulator in a TO92 or SMT package, with two capacitors, is not going to be significantly larger than a zener plus power resistor. And it is more efficient and regulates better. I see no reason ever to prefer a zener for voltage regulation, unless you are producing millions of boards and need to save a few cents.

• MrAl

#### MrChips

Joined Oct 2, 2009
29,801
A linear regulator is more efficient than a zener regulator.

If the zener regulation is designed for 500mA max, the total power dissipated is always 12V x 500mA = 6W.
With a linear regulator, the total power dissipated is 12V x load current. Hence if the load current is 100mA, the power dissipated is 1.2W, that is, 0.7W by the regulator and 0.5W by the load.

If the load current is 500mA, then 2.5W is dissipated by the load and 3.5W by the regulator.

If the load current is constant at 500mA, then you don't need a zener diode or a linear regulator. Just a 14Ω 5W resistor is what you need.

• BobTPH

#### Ian0

Joined Aug 7, 2020
8,937
Zeners are useful for two jobs.
1. Clamping voltages to protect other parts of the circuit (usually as TVS diodes)
2. distorting guitar signals.
Providing an accurate reference voltage isn’t one of them.

#### mike _Jacobs

Joined Jun 9, 2021
122
So, back to the zener
say you are trying to regulate a small voltage from a large source. So the drop across it is large.
say a 50V source and you want 5v

So you have 50-5=45
Now lets say 1000 ohm series resistor

45/1000 or 45mA
45mA * 5V zener = or 225mW
Which is clearly a lot.

Is there merit in stacking zeners up to spread the power load?
So hypothetically, if i took that same 50V and put (5) 5V zeners in series. I pluck my 5V supply off the very last one.
Each one dissipates part of the power? I know this is not efficient. Im just curious about the concept. In real life, i understand many other options are out there. Im just curious about the math and if im thinking about this correct in my head.

#### Ian0

Joined Aug 7, 2020
8,937
So, back to the zener
say you are trying to regulate a small voltage from a large source. So the drop across it is large.
say a 50V source and you want 5v

So you have 50-5=45
Now lets say 1000 ohm series resistor

45/1000 or 45mA
45mA * 5V zener = or 225mW
Which is clearly a lot.
. . .but not as much as the 45mA*45V which is dissipated in the 1k resistor!

#### mike _Jacobs

Joined Jun 9, 2021
122
. . .but not as much as the 45mA*45V which is dissipated in the 1k resistor!
right right i know but i just made numbers up
the question is, does stacking the Zener's up help spread the power dissipation out. or does it just create more power dissipation to deal with.

#### Ian0

Joined Aug 7, 2020
8,937
right right i know but i just made numbers up
the question is, does stacking the Zener's up help spread the power dissipation out. or does it just create more power dissipation to deal with.
makes no difference to the total dissipation. For reliability, best to dissipate power in a resistor rather than a semiconductor.