Zener reverse current pulse - can the zener take this?

The data sheet doesnt specify forward voltage drop at 1A so hard to tell. If from extrapolating its as much as 2v then peak power will only be 2W, The pulse power derating curve applies to zener mode wich I gues is more fragile due to 2nd breakdown effects wich are far less significant when forward biased.

BTW at 1ua it may not to be in regulation.

Colin =^.^=

err, I mistook the 1A for forward current in the normal diode sense, (reverse of the normal zener current wich itself is reverse current) If it is zener current at 1A its way off the chart to give the voltage, and probably unlikly to survive more than a few times.

What is the impedance of the circuit driving the 1Amp ?

Colin =^.^=

Although an absorbed energy calculation makes sense from the point of view of thermal mass and temperature rise, in practice transient power handling is expressed as power for a time duration. E.g., 20 watts for 2ms, if we assume the zener stays at 20V for 1A, or 30W if we assume an internal 10-ohm resistance at one amp.

The P6KE is one of the smallest series of TVS diodes, with packages the same size as some 1/2W zener diodes, is rated at 600 watts for 1ms to the 50% decay point. This means it can handle about 20x more energy than your zener in a similar time frame. Although TVS are made with metal slugs against both sides of the die, to absorb more heat, I'd guess that you're likely to be safe with your ordinary zener diode.

It's possible to measure the actual instantaneous die temperature from the voltage measurements, using a detailed model of the zener diode under test, taking advantage of the positive tempco (0.086%/°C at 20V) of the zener breakdown voltage. You'll need to take short pulse measurements of the test zener's internal series resistance above 50mA. I've posted details on how to do this, with good results, in the past here on s.e.d. (IIRC, I observed up to 200°C rise in 50us multi-kW pulses, which I repeated tens of thousands of times, without any apparent component degradation.)

Generally, short die-temperature jumps to 150 or 175°C are considered acceptable for power semiconductors, as a maximum rating. A lower level, say 125°C, would provide a safety margin.

I wish all manufacturers specified such details in their datasheets! Thanks for the right pointer. Mike

Well, consider the downside, and whether it would be wise to invest another 7 cents and put a 1N4001 in parallel with it.

And did you really mean one microamp? I though most zeners were noisy and unspecified and unhappy at very low current. I doubt if the Spice model for the zener is a good model at microamp currents.

Also, if you ARE using it at a microamp, think about the possibly howwibble thermal tail as the zener cools off from that pulse!

Well, consider the downside, and whether it would be wise to invest another 7 cents and put a 1N4001 in parallel with it.

And did you really mean one microamp? I though most zeners were noisy and unspecified and unhappy at very low current. I doubt if the Spice model for the zener is a good model at microamp currents.

Also, if you ARE using it at a microamp, think about the possibly howwibble thermal tail as the zener cools off from that pulse!

I'm sorry, I meant to write more like this,

The P6KE, one of the smallest series of TVS diodes, with packages the same size as some 1/2W zener diodes, is rated at 600 watts for 1ms to the 50% decay point. This means it's rated to handle about 20x more energy than you're giving your zener, in a similar time frame.

Note the curve specifies a rectangular power waveform. The single-pulse curve says about 14 watts for 2ms, which could be adjusted to 28 watts peak for a pulse decaying in 2ms. (Note: 2mS is 2 milliSiemens.) I'd say the zener will be marginally happy. Mike could use a 600-watt P6KE20A TVS, which is only a little larger, but has a 20x safety margin. It would have about 400pF of capacitance, compared to about 20pF for a 1N5250 (hmm, that implies the P6KE20 has 20x higher die area). But, when dealing with high leakage and capacitance, there are a few ways to have your cake and eat it too.

Mike, what are you working on?

That would reduce the peak reverse voltage from 20V to < 1V, probably not what he wants.

Even a small TVS would have no trouble with a 20W peak pulse.

Best regards, Spehro Pefhany

If you're running it at 1 microamp (unlikely to work), you could just put a 100K resistor in series with the zener, that should lower the peak current considerably.

If you're running it at 1 milliamp, you could still put say a 100 ohm resistor in series. That would decrease the regulation a little bit but still provide considerable surge protection.

Still, as non-repetitive maximum spec at 25°C initial Tj, it's not the sort of margin that makes one really comfortable, eh?

Best regards, Spehro Pefhany

Take a look at page 7 figure 10; extrapolate to 1000mA for a crude guesstimate.

Central Semi makes diodes with a test current specified in the microamp region, and some even have a low noise spec combined with that. Presumedly that means that Centeral knows how to make zeners that are not noisy (ie: do not have negative resistance and other avalanche breakdown effects as being dominant).

The 20V zener protects 1n4007 if points a and b were interchanged by the "user" while performing the wiring on field.

The inductor is a solenoid.

1n4007 is surge absorber upon switch turn-off.

Switch is a relay.

Switch 12V _/ a | |------ -o/ o---- -||-------- | | | | === | GND | |------------+ | | | - | ^ 1N4007 | | C| | 40mH 12Ohms C| | C| | | | | | | | | V 20V Zener | z | | b |------------| | | | === GND (created by AACircuit v1.28.6 beta 04/19/05

It would have been much clearer if I had said that the zener protects

Mike

I don't know where this peak zener current figure of 1A has come from.

Based on the circuit you've shown, solenoid current will be 100mA at switch off and this is the peak current that will commutate into your series diode-zener. This current will then decay linearly to zero at a rate given by:

dI/dt = V/L

For V = 20V and L = 40mH, this gives dV/dt = 500 A/s, so the current will reach zero in nominally 200us, during which time the rms dissipation in the zener will have been just over one watt.

For any realistic switching duty cycle, you don't have a dissipation problem with this zener.

It is 12volts / 12Ohms = 1A I assume you mistook the 'O' for Ohms as a zero.