Suggestions for selecting a zener diode

Hi all,

I have a circuit which consists of some quad op-amps and other low power ICs.
As for the input supply, I am going to use the supply from microprocessor development boards (either 3.3V or 5V).
As an extra protection for the circuit, I was planning to include a 5.1V zener diode so it won't be damaged in case of accidental use of voltages above 5V (up to 12V).
As for the zener diode, I was planning to use the TZM5231B-GS08, together with a 400ohms resistor.
Could anyone please suggest if this is the right circuit to implement and if the zener and resistor combination is correct? Also, this zener is listed as "small signal diode". Will it work for regulating the supply (few mA)?

Thanks in advance.

Yes, it should work fine.

Ideally you want a small current to flow through the Zener all the time. You have not said whether the load current to your other components is steady or varies but the basic calculation is R = V/I where R is 'R0' in your schematic, V is the difference between input voltage and 5.1V and I is the current drawn by you load plus say 2mA through D0.

If the load current is variable, use the highest current you will draw to calculate R0 and check that D0 doesn't pass excessive current at minimum load current.

Brian.

PS: I did not simulate the circuit as this zener diode is not available in the library, but I did a simulation using the 1N5231B and the results were kind of promising (3.3Vout when input was 3.3V, 4.95Vout when input was 5V and 5.21Vout when input was 12V)

You obviously simulated the circuit with zero load current. You need to define a minimum and maximum load current and check if the regulation is still O.K. with low input voltage.

Thanks for your reply @betwixt.
My main concerns are the following:

1) Is the max current of the zener 200mA? It is not listed in the datasheet, but the description on Mouser states that Zener Current is 200mA.
2) You mentioned that I have to use R = V/I where R is 'R0' in your schematic, V is the difference between input voltage and 5.1V. In my case Vin may very from 3.3V up to 12V. Should I calculate using 12V?
3) I will have a switch after the zener circuit (i.e. load current can be 0mA). Unfortunately I cannot place the switch before the zener. How can this affect the circuit/my calculations?

Hi,

Also, this zener is listed as "small signal diode". Will it work for regulating the supply (few mA)?

Click to expand...

--> read the datasheet.

You talk about two different things:
* regulating the supply
* overvoltage protection.

In the first case the zener needs contionous current, in the other case there is just current on error condition.
Both cases need a different calculation of part values.

I wonder: Your schematic looks like you use a simulation tool.
May I ask why you don't simulate your operating cases to get the answers on your own.
I assume you expect us to do the simulation or at least the calculations...

It shouldn't be a problem to simulate for different input voltages and different load currents.
Btw: "a few mA" is no information we can calculate with (nor can your simulation tool). Please give values.

Other informations you need to decide:
* What load voltage (output) range do you want for your circuit?
* is the output current considered to be DC, or are there times with higher or lower current?
* What power dissipation / temperature rise can you accept?

Mind: This type of voltage regulation is called "shunt regulator". It continously needs to draw more current than the highest load current you expect. This means: in times of low load current a lot of power is wasted ... and produces heat.

Overvoltage protection:
In most cases one needs a two stage overvoltage protection, beacuse the V-I curve is not steep enough.

Klaus

Hi KlausST.

Thanks for your reply. As already stated, no I did not simulate the circuit as most of the parts being used (including the zener diode) are not included in the simulation library. The name of the zener was edited by myself and not selected from the part list.
No I am not expecting anyone to simulate nor calculate anything for me. I just opened this thread as I got confused by the information I found on the internet.

Sorry for the confusion, I need the circuit for over-voltage protection. The circuit is intended to be powered from 3.3V or 5V of the development board. I need to include the zener just in case a supply up to 12V is accidentally connected.

Other informations you need to decide:
* What load voltage (output) range do you want for your circuit?
* is the output current considered to be DC, or are there times with higher or lower current?
* What power dissipation / temperature rise can you accept?

Click to expand...

1) The circuit can operate at 3.3V or 5V, hence why I am using a 5.1V zener to protect the circuit from any voltage above this level.
2) The output current should not vary. I will try to calculate the average current from the datasheets because as already mentioned, most of the ICs were not found in the simulation part list.
3) Power dissipation and temperature rise should be stable

The output current should not vary.

Click to expand...

Unlikely. Besides temperature dependent variations, the current consumptions of most linear circuits is to some extent modulated by the signals level and operation mode transitions. In any case, even a constant load current causes a voltage drop across the series resistor. Question is how much you can tolerate?

The problem is simple Ohms Law. If your circuit draws ANY current, it will drop a voltage across the resistor and you will get less at the load. Even if you apply a lower voltage than the Zener, you will still get less out than you put in. True, applying higher than the Zener voltage will clamp (hence the term 'shunt regulate') the output as long as voltage drop in the resistor is taken into account. It doesn't mean that if you feed say 6V in, you will still get 5.1V out, that would only happen if your switch was open and no other current was drawn through the resistor. If the other load current dropped more than 0.9V in the resistor, the Zener wouldn't do anything at all.

The maximum current through the Zener is easy to calculate. It is a 500mW rated device so as W=VxI, and you know V is 5.1V the current must not exceed 98mA. That is it's rated maximum though, in practice, aim for something considerably lower.

Brian.

Hi,

1) The circuit can operate at 3.3V or 5V, hence why I am using a 5.1V zener to protect the circuit from any voltage above this level.

Click to expand...

Review your requirements.
If the circuit can operate at a maximum voltage of 5V, then a 5.1V zener will not keep the voltage below this level.
Either your circuit is able to operate at higher voltage, or choose a lower voltage zener.

As the others said: there is voltage drop in the resistor. At least: I_load x R. This means every mA will cause 400mV of drop.

Zener current:
Dissipated power generates heat. Thus it makes sense to specify the max power dissipation for a zener.
This specification is independent of zener voltage.
You simply can calculate the current by: I = P / V.
But this is only true for the given thermal parameters.

Klaus

You don’t need exact part numbers to simulate the circuit. Zeners are simple. Pick any 5.1V in the library and you can simulate your circuit.

@KlausST, Let me explain my circuit better as it looks like my previous explantation was not so clean.
I have a simple signal conditioning circuit (filtering, rectifying and enveloping) which can operate at 3.3V to 5V.
As for the power of this circuit, I intend to have three options (switch selectable) as shown below:

**broken link removed**

Option 1 and 2) Use an unregulated power supply (7V up to 12V) as Vin and select either the 3.3V or the 5V regulated supply as the output.

Option 3) Use a regulated 3.3V or 5V supply from a development board (or any other regulated supply source) to power the circuit. The voltage regulators for this option are not required and the input is shorted directly to the output.

Now I need to design a simple protection circuit for the third option. Lets assume by plug a 12V unregulated supply with the intend to use option 1 or 2, but by mistake the switch is set to option 3 (direct short from the input to the output). This will damage all the active components, hence why I was planning to include a zener diode to clamp any voltages higher than 5.1V (all active components can tolerate 5.5V max, so they will not be damaged with a 5.1V supply)

Reading the comments and doing some simulations using a different part number for the diode, it looks like this zener diode configuration will not work for my application.

We can see what you are trying to achieve but a resistor and Zener will not work as you intend. Adding any resistor will cause a voltage drop across it and so will upset the voltage you want. If you remove the resistor (so it is just a link), the circuit WILL work but excess voltage will be dumped as power into the Zener alone. How much current will depend upon the power source feeding it but there's a good chance if it is rated more than 100mA or so it will cause damage.

There is a method of protecting against over-voltage without dropping any through the circuit, it's called a 'crowbar' (seriously!) and it consists of a Zener diode feeding the gate of an SCR which has it's anode and cathode directly across the supply. For safety a resistor (~10K) should also be wired from the gate to ground to leak away residual current in the Zener. The idea is the circuit does nothing until the Zener voltage is reached, when it starts to conduct, it lifts the gate voltage of the SCR and 'fires' it. The effect is to almost short out the supply so downstream equipment is protected. It does more or less what you intended with the Zener alone but has the advantage that the SCR can handle high current but only needs a small current through the Zener so it's size can be small.

Brian.

Hi,

I can't open the attachment of post#12. Just use the "insert image" button to upload pictures.

I'd use an LDO. It safely limits the output to a precise voltage. It does nit waste that much current and you may choose one with low droput with low input voltages.

Klaus

Thanks for the replies.

@KlausST, Coincidentally I was considering the LDO option.
Following some search, I am now considering the LD2981 (3.3V and 5V versions). These are ultra-low dropout regulators, so they should work fine for my appliction.

A quick question, since Iwill be using two regulators, can I spare the input and output capacitors as shown below:



Also, would I be good practice to include a 100nF capacitor at the input and output?

PS: I did use the Insert Image button (image in .JPG format)

Absolutely NO!

You can have the regulators share common input capacitors as long as you fit them as close as possible to their inputs and ground. The outputs must also have capacitors connected all the time so you should fit them to the outputs of both regulators before the switch instead of after it. It would be a good idea to also add a resistor across the output of each regulator, something like 22K should be adequate. The reason is that it is possible that when unloaded, leakage through the regulator may make the output pin voltage rise higher than expected and switching to that regulator can spike that voltage into the load.

A word of caution regarding output resistors, some (not all) regulators have a minimum load requirement so if you choose one that does, fit a resistor that passes the smallest specified current.

Brian.

Hi,

I will be using two regulators

Click to expand...

Why two regulators?

If you want a 3.3V supply, then use a single 3.3V regulator,
If you want a 3.3V ... 5.0V supply, then use a single 5V regulator.

Klaus

@betwixt, Thanks for your suggestions. I did read the datasheet and it is not mentioned that a minimum output current is required. What is this parameter usually refered to please? Also, are the 100nF caps necessary?

@KlausST, I need the circuit to operate at either 3.3V or 5V (sorry, in my previous post I stated 3.3V to 5V). For this reason, two regulators are required. I did find regulators with dual outputs of 3.3V and 5V, but the drop-out voltage was in the 150mV range for every 10 to 20mA of output current.

Hi,

I need the circuit to operate at either 3.3V or 5V

Click to expand...

Please explain: "The circuit" means:
* the input supply voltage --> then just one regulator is sufficient
Or
* the supply for the Opamps --> pleae tell us why they need to be operated on different voltages

Klaus

For the LD2981 series there is no minimum current so it isn't specified in the data sheet. What I meant was some regulators do have a minimum so be aware of that if you use a different type. If there is a minimum, it is usually only one or two mA so calculate the resistance needed to pass it, you know the voltage so use Ohms law to find the best value.

The 100nF capacitors are not essential but highly recommended. A ceramic 100nF is better at filtering high frequencies while electrolytics at better at low frequencies, between them they cover the whole range. What is very important is their placement, you should keep them as close as possible to the body of the regulator.

Brian.