# Zener diode Question

Discussion in 'General Electronics Chat' started by lordeos, Nov 2, 2015.

1. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
Hi all,

I'm looking into zener diodes and i have 2 questions . I found an article on simple zener diode circuits and there is something i don't understand (screenshot article in attachment).

1) The article states that the zener is in series with the Resistance R ? In series for me means ( + pole of a device connect with - pole of another one). But since the resistor R has no polarity how can we say it's in series then.

2) Will the zener diode block current flow ? --> so no current flow in the whole circuit until the reaches his breakdown voltage.

I'm confused if it's simultaniously or in two steps ( circuit in article). Does the Zener takes all the voltage and current (because it blocks current flow) until it reaches his breakdown voltage AND THEN current flow in the whole circuit. Or does the load follow the voltage of the zener diode ? --> Zener = 1V - Load = 1 V , Zener = 2v - Load = 2 V , etc ...

Thx Mike

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2. ### BillB3857 Senior Member

Feb 28, 2009
2,400
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First, lets clear up the meaning of series. If you are dealing with batteries, yes, Plus to Minus to Plus to Minus places batteries in series. However, in non polarized devices, such as resistors, series simply means "in a single line with".

In the circuit you show, the zener is acting as a SHUNT regulator. Imagine that you want 5.6 volts at the load and had a 10 volt source. If you know the load current, and it remained constant and the source voltage remained constant, you could calculate the value of the series resistor needed to drop the 4.4 volts, thus having 5.6 volts at the load without using the zener.

Since most loads have a variable current and an unregulated source (think battery as it discharges over time), a single fixed resistor would not work. However, if we could devise a way for the total load current to be vary by an amount that would result in a variable voltage drop across the fixed resistor, and be sensitive to voltage, a single fixed resistor would work.

Now, put a zener diode in parallel with the load, as it is shown in your drawing. The characteristics of the zener will, after it reaches its breakdown voltage, result in a variable current while maintaining, for all practical purposes, a fixed voltage across itself. That variable current will result in a variable voltage drop across the series resistor in order to maintain a constant voltage across the zener, and thus the load.

3. ### dl324 Distinguished Member

Mar 30, 2015
3,211
619
Series and parallel don't require a component to have polarity.
This would be the case for an ideal zener diode. In reality, a zener will conduct current before reaching it's nominal voltage.

BTW, that schematic you posted is drawn "backwards". Another example of some of the junk you can find on the internet...

4. ### Roderick Young Member

Feb 22, 2015
408
168
Just to clarify, it's backwards in that the convention is for supply to be on the left, and load on the right. The circuit is functional as shown.

5. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
If i want a 5V regulated output from a Zener diode from a VS of 12V , how do i calculate it ?

I assume i have to drop 7V on the series resistor to have 5V on the zener. But if the breakdown voltage of my zener is 5.6 then this will not be sufficient, correct ?

- Do i have to put a Zener diode of 4,7 breakdown voltage then ?
- Is the zener breakdown region available in any voltage range (untill it reaches it's limit) For example if i put 8 V on a Zener of 5,6 breakdown will it also have the self regulating function but then on 8 Volts ?

Thx Mike

6. ### dl324 Distinguished Member

Mar 30, 2015
3,211
619
The zener will have a nominal voltage; 4.7 and 5.1 are the closest standard values to what you want.

In your example, if zener voltage is 5V, you would set the current in the zener by using an appropriate resistor with 7V across it. For example, if you wanted 20mA in the zener, your resistor would be 7V/20mA=350Ω.
For the zener to provide voltage regulation, it needs to be biased past the "knee".
A zener diode is okay for voltage regulation at low currents because the current in the zener, in addition to biasing it beyond the knee, needs to be 10X larger than the highest current the load will draw.

For higher current requirements, you'll likely get better results using a voltage regulator.

7. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
So if i understand correctly i can't regulate for example 6V because there is no zener diode of exactly 6V ... ?

- Do i need to regulate a higher voltage (say for example 7V --> don't know if this zener exists) and put a resistor before my LOAD that drops 1V then ?

8. ### dl324 Distinguished Member

Mar 30, 2015
3,211
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No zener diode will give you it's nominal voltage under all conditions. Below is graph of IV characteristics for a group of zeners (looks like 4.7-16V).

You can see for the lowest voltage zener (probably 4.7V), the voltage varies from essentially 0V to 6V; depending on zener current.

9. ### dl324 Distinguished Member

Mar 30, 2015
3,211
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FYI, standard zener voltages around the voltages you've been referring to:

Note that the diodes need to be biased to 20mA to get the nominal voltage. If your load draws, for example, 20mA; you need to bias the zener to 40mA so zener current stays above Izt.

10. ### ian field Distinguished Member

Oct 27, 2012
4,413
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The resistor has no polarity so it doesn't matter which way round.

The zener has polarity - if you connect it the wrong way round in relation to the power supply; it'll conduct like the Vf of a regular diode instead of Vz.

Resistor R and the load are effectively in series and will act as a potential divider - the zener is in parallel with the load and see's the same voltage as is developed across the load.

As long as the voltage developed across the load is less than Vz; the zener current will only be a small leakage current. If the supply voltage increases; the load voltage will try to increase in proportion to the ratio of R and Load, as the voltage reaches Vz; the zener current increases and diverts current away from the load that would otherwise increase its volt drop.

In most cases; you'd calculate R to develop a slight over voltage across the load - and then clamp it to the required voltage by adding the zener.

You have to take the worst case load current fluctuation into account when specifying the zener power rating - for example; if the load became disconnected, the zener would have to handle the whole current fed via R.

11. ### ramancini8 Member

Jul 18, 2012
442
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You can place a diode in series with a back-biased zener to obtain specific voltages. Example: a 4.3V zener in series with a 1N400X diode yields a nominal voltage of 5V. Of course this circuit is susceptible to all zener tolerance errors and to the diode errors, choosing the correct zener and diode can result in temperature stability.

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12. ### ian field Distinguished Member

Oct 27, 2012
4,413
782
Can't remember which way round it is, but some zeners are PTC and others are NTC - I think the break even value is somewhere around 5.6V.

It has been common practice to use series combinations of above and below to get near neutral tempco.

The TL431 programmable zener can be adjusted to whatever you want up to its rated limit, and it can handle 100mA as long as you don't exceed its dissipation limit.

13. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,554
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I've always thought that zeners were terrible as voltage references due to their temperature drift sensitivity. How come a zener and an ordinary 1N400x in series tend to cancel their temp drift?

14. ### dl324 Distinguished Member

Mar 30, 2015
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Temperature drift is +/- 2mV/C, depending on whether it's a zener or avalanche diode, isn't that bad. If you use an avalanche diode in series with a silicon diode, the temperature coefficients can almost cancel each other. If you try that with a zener diode, you exacerbate the problem.

To complicate matters, avalanche diodes are usually referred to as zener diodes...

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15. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,554
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So a zener has a negative temp drift, whereas an avalanche type has a positive one?
And I'm not quite following with your last statement. Do you mean that it's bad (thermally speaking) to connect two zeners in series?

16. ### dl324 Distinguished Member

Mar 30, 2015
3,211
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Yes. The transition is around 5.6V.
Two zeners of the same voltage in series would double the temperature coefficient. A "zener" diode in series with a silicon diode would also double the tempco.

I need to check my previous post to make sure I have the tempco's correct; lest someone accuse me of being a dense troll

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17. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,554
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Hope that doesn't happen... where would that leave me!?

18. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
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I dumped my collection of Zeners, and replaced them with a bunch of TL431s. It is useful for much more than just a shunt regulator...

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19. ### #12 Expert

Nov 30, 2010
16,248
6,744
I have a book...Motorola...The Semiconductor Data Library...1973...about 1600 pages.

Anyway, page 1-22, zeners from 1N5221 to 1N5281
Coefficient of temperature drift ranges from -0.085 %/C @2.4 volts to +0.110 %/C @ 200 volts.
The 4.7V and 5.1V zeners are rated at +/- 0.03%/C
and that's as close to zero as we could get from a standard zener in 1973.

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20. ### dl324 Distinguished Member

Mar 30, 2015
3,211
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Motorola published a Zener Diode Handbook back in the 70's. I misplaced mine... Anyone have a reference to a PDF?