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Testing TVS diode

Rajinder

Jan 30, 2016
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Hi all,
I have selected a SA20A to try out and test for my own understanding.

I am thinking of the following:

1. Measure VBR by placing resistor in series say 10K at 10V so we get 1mA. Would this give me the break down voltage.

2. The reverse voltage, should not affect any operation of a circuit. I was going to put this across a 12V supply that powers a regulator and opamp circuit and ensure that circuit works fine.

3. Could I check the clamping voltage which is around 30V, by ramping up the voltage so it is greater than 30V? Would my voltage need to change in a certain time period i.e. milli or micro seconds. How could I check this out.

Have I left out any other checks?

Any help would be appreciated.


Quote
 

Harald Kapp

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Look in the datasheet. Often the test circuit is described by the manufacturer. If not for every single diode, then for the series.
 

(*steve*)

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I had a bit of fun characterizing a TVS diode. Essentially, I tested it the same way you would test a zener. Connect it up to a current limited voltage source and measure the voltage across it at various currents

One thing I suggested on the devices I tested is that they are very temperature sensitive, with the breakdown voltage increasing with temperature.

Since these are typically high voltage devices, their power dispassion is quite high and it can be challenging to get repeatable readings if you're doing it by hand. Typically this isn't a huge problem because these are not precision voltage devices.
 

Rajinder

Jan 30, 2016
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I had a bit of fun characterizing a TVS diode. Essentially, I tested it the same way you would test a zener. Connect it up to a current limited voltage source and measure the voltage across it at various currents

One thing I suggested on the devices I tested is that they are very temperature sensitive, with the breakdown voltage increasing with temperature.

Since these are typically high voltage devices, their power dispassion is quite high and it can be challenging to get repeatable readings if you're doing it by hand. Typically this isn't a huge problem because these are not precision voltage devices.
Hi,
Thanks for the information. I am trying to work out the VBR, of the device I have selected. It has a VBR of 22V. I have increased the voltage on the device to 24V, with a series limit resistor. Should I not see an increase in the current?
 

Harald Kapp

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Learn how to read a datasheet:
upload_2021-3-26_10-5-37.png
The breakdown voltage can be up to 24.5 V. At room temperature. As @(*steve*) noted, this voltage is also temperature dependend.
At the breakdown voltage the increase in current is only small:
upload_2021-3-26_10-7-19.png
Only when you increase the voltage above VBR will the current rise exponentially.
 

Rajinder

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Hi, this is what I can't see, I had the voltage set to 30V, with a 1K series resistor. The voltage across the TVS was 22.8V, should I reduce the resistor value.
Also how can I get this to clamp? The datasheet says VC @ IPP max, which is 15.7A. I was thinking of 34V with a 10R 10 W resistor. So I get 34-32.4 across the resistor i.e. 1.6V @ 3.4A = 5.44W well inside my resistor power rating. Should that be a good test? Or will I get the TVS

Lastly how could I test or simulate a transient condition?

Sorry for all the questions, thanks in advance.
 

Harald Kapp

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I had the voltage set to 30V, with a 1K series resistor. The voltage across the TVS was 22.8V, should I reduce the resistor value.
No, that's o.k. 22.8 V is totally within specs. Current should be (30 V - 22.8 V) / 1 kΩ = 7.2 mA.
If you want to qualify the TVS, you should really use a power supply with constant current output (aka current limited output) and measure at several V/I points, then plot the graph. No single measurement will fully qualify the TVS diode.
Lastly how could I test or simulate a transient condition?
Apply the test voltage via a switch that you can control electronically, e.g. a power MOSFET.. The use the switch to create a pulse. You will probably need an oscilloscope to measure the voltage waveform if the pulses are very short (which they should be as TVS diodes can take a high pulse load but will easily burn if power is applied for too long).
 

Rajinder

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No, that's o.k. 22.8 V is totally within specs. Current should be (30 V - 22.8 V) / 1 kΩ = 7.2 mA.
If you want to qualify the TVS, you should really use a power supply with constant current output (aka current limited output) and measure at several V/I points, then plot the graph. No single measurement will fully qualify the TVS diode.

Apply the test voltage via a switch that you can control electronically, e.g. a power MOSFET.. The use the switch to create a pulse. You will probably need an oscilloscope to measure the voltage waveform if the pulses are very short (which they should be as TVS diodes can take a high pulse load but will easily burn if power is applied for too long).
Thanks for the help. I am still not sure how I can test the clamping effect? My understanding was that a voltage spike >32.4V would clamp the output. I am not sure how It can do this.
 

(*steve*)

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Well, a spike above 23V will be clamped to about 23 volts. It won't be clamped to zero volts.
 

Harald Kapp

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Apply > 32.4 V, use a resistor to limit current.
Or, has has been said before, use a power supply with built-in current limiter.
 

Rajinder

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Hi
I have done exactly as you suggested. 40V with a 1K resistor, the voltage across the TVS was 23.9V which is the breakdown voltage. Does the 40V have to be a fast pulse?
 

Harald Kapp

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Does the 40V have to be a fast pulse?
No.
(40 V - 23.89 V) / 1 kΩ = 16 mA. The datasheet states the clamping voltage at a peak current of 15.7 A. No wonder your diode doesn't reach the 32.4 V you expect. You really must follow the datasheet's parameters. To achieve 15.7 A at 32.4 V clamping voltage and a 40 V source you need R = (40 V - 32.4 V) / 15.7 A = 0,48 Ω. And a very powerful source as 40 V × 15.7 A = 628 W. And you may apply that power only for a short duration (pulse), otherwise the TVS will burn out.
 

Rajinder

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No.
(40 V - 23.89 V) / 1 kΩ = 16 mA. The datasheet states the clamping voltage at a peak current of 15.7 A. No wonder your diode doesn't reach the 32.4 V you expect. You really must follow the datasheet's parameters. To achieve 15.7 A at 32.4 V clamping voltage and a 40 V source you need R = (40 V - 32.4 V) / 15.7 A = 0,48 Ω. And a very powerful source as 40 V × 15.7 A = 628 W. And you may apply that power only for a short duration (pulse), otherwise the TVS will burn out.

Hi, Thanks. Is there anyway I can simulate this is PSPICE or Simetrix? Thanks in advance.
 

Harald Kapp

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If the simulator has the right model or you know how to load a manufacturer's model into the simulator: yes.
 
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