# Zener Diode fundamental question

Thread Starter

#### mike _Jacobs

Joined Jun 9, 2021
7
So i am a bit confused.
When i look at the knee of a zener diode while its reversed bias, i can see that as i push more current through it its zener voltage should be closer and closer to its rating. aka 5V zener with a test current of 500uA say as i approach 500uA i get closer to my 5V.

If i am barley turning on the zener aka current limiting so heavily that im putting say 1uA through it, its zener voltage will be something much less aka closer to the origin of the plot.
All that makes sense

Here is what i dont get.

If i dont current limit a zener and run it in reverse bias lets just say i put a 10V source on the cathode of a 5V Zener, i would get a ton of current flow.
However, what i see for the zener voltage is something far greater then the zener voltage. I see like 10V for example. Why is this the case given that as stated above, the more current i run into a zener to CLOSER it is to its actual zener voltage.
What am i missing?

Thanks

#### Audioguru again

Joined Oct 21, 2019
3,173
You are not supposed to fry a zener diode (it gets too hot if its current is too high). You should operate it at its rated test current by limiting its current with a series reaiator.

Thread Starter

#### mike _Jacobs

Joined Jun 9, 2021
7
Here is an example of what i mean
This is a 6V zener and just a regular diode.

It says there is 6A flowing thorugh it and the voltage is 8.2V
Why is that.

If im putting more current into the zener, shouldnt its voltage be closer to its zener voltage not further away

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Thread Starter

#### mike _Jacobs

Joined Jun 9, 2021
7
You are not supposed to fry a zener diode (it gets too hot if its current is too high). You should operate it at its rated test current by limiting its current with a series reaiator.
I am well aware
I am trying to understand why its reverse voltage drop is not its zener voltage when driving more current.

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,745
You can think of the zener diode as a regular diode who's breakdown voltage is controlled. As long as the current is limited, it's non-destructive to the diode.

#### crutschow

Joined Mar 14, 2008
27,184
I am trying to understand why its reverse voltage drop is not its zener voltage when driving more current.
It's simple.
The diode is not ideal, it has a small intrinsic resistance, so voltage past the knee goes up with the current.

If you want a more ideal Zener, you can use an IC shunt reference, such as a TL431, which uses an internal bandgap reference with feedback to make the "Zener" voltage very accurate and relatively independent of temperature and current (until its current limit is reached).
It can also be programmed with two resistors to any Zener voltage you want, up to its voltage limit.

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#### Papabravo

Joined Feb 24, 2006
16,103
Zener diodes are far from ideal components. With small currents any reverse bias voltage is possible from zero up to the knee. It is at the knee where the "constant" voltage part of the characteristic comes into play. The reason I put "constant" in quotes is that as more current goes through the diode the voltage will continue to increase albeit at a very slow but greater than zero rate. As you can see from the following, the characteristic is far from vertical.

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#### MrChips

Joined Oct 2, 2009
23,515
Look at the I-V characteristic curve of any P-N junction device such as the zener diode below.

There is no clearly defined zener voltage VZ.
VZ is only defined for a given IZ.
IZ goes up with VZ.

Apply 10V to a 5V zener diode and you get 10V and a blown zener diode if you exceed the maximum current.

Thread Starter

#### mike _Jacobs

Joined Jun 9, 2021
7
but this is what i dont get....
As the current ramps up the zener voltage does not change by this curve. It hits a wall. The current continues to grow and grow and the voltage stays at the zenver voltage. It does not go beyond it as seen by the red line i made

#### MrChips

Joined Oct 2, 2009
23,515
Depends on the internal resistance of the power supply.

If the internal resistance is 0Ω then the voltage on the diode will be the supply voltage.
When the internal resistance is significant then there is some give and there will be a voltage loss owing to the internal resistance.

#### Papabravo

Joined Feb 24, 2006
16,103
but this is what i dont get....
As the current ramps up the zener voltage does not change by this curve. It hits a wall. The current continues to grow and grow and the voltage stays at the zenver voltage. It does not go beyond it as seen by the red line i made

View attachment 240860
This drawing is "highly idealized". Look at any real device and look at the simulations. You won't see anything like that near vertical slope. That is, was, and always has be an outrageous fabrication. The forward characteristic is a much better approximation to "near vertical". In the simulation I did. The slope looks pretty linear at about 8 mA/Volt after the knee

#### BobTPH

Joined Jun 5, 2013
3,308
I love it when someone says exactly what I was going to say. Saves a lot of typing.

Bob

#### Papabravo

Joined Feb 24, 2006
16,103
I love it when someone says exactly what I was going to say. Saves a lot of typing.

Bob
All I can say Bob is that every once in a while great minds have the identical thoughts. I've come to expect this on this forum. Maybe I'm just spoiled.

#### MrChips

Joined Oct 2, 2009
23,515
The bottom line is: Design a series resistor into the circuit to limit the current.

#### Papabravo

Joined Feb 24, 2006
16,103
The bottom line is: Design a series resistor into the circuit to limit the current.
If you really want to amaze your little friends, you use a current source for the Zener Test current and you measure the "Zener" voltage.

#### Audioguru again

Joined Oct 21, 2019
3,173
The datasheet for the tiny 6.8V zener diode says it is 6.8V when its current is 5mA. Its absolute maximum allowed heating is 0.25W. You said the simulation tried to destroy it with 6A. Do the math, that is 6A x 8.2V= 49.2W which is 196.8 times more than the maximum allowed 0.25W.
Simulators do not care about overloading and frying things.

#### sparky 1

Joined Nov 3, 2018
539
Using a 1/2 Watt like 1N829A yes ! you would definitely want current control and more precisely 7.5mA
it is not supposed to put out alot of current just enough to transfer to a circuit that can handle the heat.
So more and more parts are required finally you have a discrete version of a voltage regulator.

You can build an amplified zener making sure that the temperature handling is done away from the zener as much as is practical.
Concider the limits that heat sinking the zener can actually do. Keeping in mind the temperature of the diode should be a few degrees over average annual ambient temperature inside the enclosure. Stable.

Thread Starter

#### mike _Jacobs

Joined Jun 9, 2021
7
This drawing is "highly idealized". Look at any real device and look at the simulations. You won't see anything like that near vertical slope. That is, was, and always has be an outrageous fabrication. The forward characteristic is a much better approximation to "near vertical". In the simulation I did. The slope looks pretty linear at about 8 mA/Volt after the knee
ok this makes a lot of sense thank you

Also the other thing i was not understanding is that the reverse biased zener really has a very low impedance.
But the thing i really missed i guess was thinking the slope was near vertical

Thanks all!

#### Audioguru again

Joined Oct 21, 2019
3,173
The slope is almost vertical around 5mA. Its datasheet shows that its "dynamic impedance" is 15 ohms max so if you increase its current to 20mA, then its voltage increases 15 ohms x (20mA - 5mA)= 0.225V max.
The datasheet also shows that its voltage rises a little if it gets warm. Sure it will get warm or get hot. Low voltage zeners reduce their voltage when they get warm or hot.

#### crutschow

Joined Mar 14, 2008
27,184
Here's the LTspice simulation of a 1N758, 6.8V Zener with ±50mA current.