Varistor question

Nov 25, 2005 32 Replies

I have a question about varistors. Actually two.



Question 1.



Are varistors the sort of devices that cannot be expected to conduct many pulses of electricity?



Or can I expect a varistor to happily conduct many short pulses of extremely short duration, into a resistor so that energy dissipates in a resistor and not in a varistor?



In other words, are they more like one shot devices, wearing out with every breakdown, or can they be expected to conduct a tiny amount of electricity 1000 times per second?



I realize that I cannot exceeed the power limit that the varistor is rated for, for any duration like seconds or minutes. But what about very many pulses that, cumulatively, are below the varistor's rated power on the per minute basis.



Question 2. Once a breakdown voltage of a varistor is exceeded, and it starts conducting, when would it return to a non-conductive state? When current falls below some threshold? Or when?



i

Yes, when operated within their specifications, varistors can be expected to last as long as any other solid-state device.

Varistors are not like that. They do not suddenly start conducting. They do not latch. In fact, they have a very gradual current-vs-voltage curve. Take a look as some data sheets. You will see that in order to conduct enough current to protect a circuit at

400 volts, it must also conduct quite a bit of current at 100 volts too. Even at 50 volts, the current does not drop to zero. A varistor would be very poor as a voltage regulator.

-Robert Scott Ypsilanti, Michigan

Varistors can stand a great many conduction cycles, as long as the peak and average power dissipation is kept within certain bounds. Here is an information document from Epcos that contains a wealth of info on varistors.

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Can you put some specific numbers in there? For example, when you say "quite a bit of current at 100 volts" and "enough current to protect a circuit at 400 volts" it's impossible to tell what you have in mind.

Also, the "very gradual voltage-vs-current curve" doesn't paint a proper picture. For example, looking at the datasheet at

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for the 420 V MOV. (Figure 6) Note that the scale of the graph is not linear. If it was, the curve would not look as gradual as it appears. Here are some numbers taken off the chart:

Below the 420 clamping voltage, leakage current ranges from

1 uA at 100 V to over 100 uA as V approaches the clamping voltage. (There is a discontinuity on the chart, so the maximum leakage current is not shown.)

The charts starts again at the clamping voltage, where current is 1 mA. That's 1000 times the leakage current at nominal 120 volts. And the current range shown when the voltage exceeds the clamping voltage is from 1 mA at 420 volts, to over 1 amp at 1000 volts, to about 10000 amps at around 1200 volts.

I'll put that in a table: (deltas with respect to previous) Voltage Current deltaV DeltaI 120 1uA - - 420 1mA 3.5x 1000x

1000 ~1amp 2.9x 1000x 1200 ~ 10000 amp 1.2x 10000x

That is hardly a very gradual voltage-vs-current curve.

Ed

Thanks.

Thanks. I want to use a varistor to clamp voltage spikes on a DC rail.

See my today's pictures about how I used a varistor for my tig welder inverter:

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i

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Thanks... I saved the file and will print it out tomorrow.

i

Ehhh? "Closes"? You need to think a little more about that.

...Jim Thompson

| James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.

--- Do _not_ do that.

Varistors are designed to clamp once-in-a-while spikes, not all-the-time spikes, and they _will_ wear out and fail shorted.

What you could use would be a big Zener across the rail or something to follow the rail and clamp a few volts above it. That is, if the rail moves.

If it doesn't, you could do something like this:

DC>-----+--------+--->

| | [ZENER] | | D +------G | S [R] | | | GND]----+--------+--->

and let the MOSFET take the beating instead of the Zener.

-- John Fields Professional Circuit Designer

Yes, but 1 mA is not going to protect anything. To serve its function of protecting against overvoltage, you've got to get up to some serious amps. According to the table that you found, that doesn't happen until about 1000 volts.

As long as you understand this table and what it means to protection from overvoltage, you can decide which varistor is right for you.

-Robert Scott Ypsilanti, Michigan

how about TVS diodes, they seem to work much faster than zeners ?

Real Programmers Do things like this. http://webpages.charter.net/jamie_5

--- I made a mistake; the circuit should look like this:

DC>-----+------------+--->

| | [ZENER1] | | | [R1] | | D +------+---G | | S [R2] [ZENER2] | | | | GND]----+------+-----+--->

Choose Zener 1 to be enough above the rail that it's not normally conducting but close enough to the rail so that when a spike happens it goes through R1 and into the gate of the MOSFET. Zener2 is in there so you don't exceed the MOSFET's Vgs and puncture the gate oxide. About 15V will work whether you use a logic-level MOSFET or not.

R2 is in there just to keep the gate from floating, but the gate charge has to be discharged to ground through it when the spikes go away, so it has to be small enough to let the charge bleed away quickly enough to keep from shorting the rail to ground!

Another (better) way would be to use one of Micrel's low side MOSFET drivers, like this:

DCRAIL>------+-------------------+-------------+ | | | | [R3] | [R1] | | | +-----+ | | | | | | +------+------+ | \\ D +-----+-----|------|------|----| >---G |K | |+ | |K | /MIC S [Z5V] [R2] [1µF] [0.1µF] [Z15V] | 4422 | | | | | | | | GND>---+-----+-----+------+------+-----+-------+

Select the values of R1 and R2 so that the voltage on the input of the diver is less than 0.8V when the DC rail is in its quiescent state and that the 5V zener isn't being overdriven when the line spikes.

R3 and the caps and the 15V Zener are the supply for the driver and also supply the charge current for the MOSFET gate, both derived from the DC rail, so select R3 to be able to replenish the caps and drive the Zener into conduction at its test current during the times the DC rail is stable.

-- John Fields Professional Circuit Designer

This is an interesting idea! I will think about its implication for snubbing, esp. what happens when MOSFET closes.

i
[.. a bunch of stuff ..]

You may want to have a look at Sidactors from Teccor. They really do "turn on" and go to a lower drop when fired.

-- kensmith@rahul.net forging knowledge

Hmm, that's strange, the last time i used those they didn't behave exactly like Zeners. are we sure we are talking about the same thing here?

Real Programmers Do things like this. http://webpages.charter.net/jamie_5

They\'re not supposed to. In the one case you\'re talking about reverse biasing the diode until, say, with a tame 20mA of current flowing through it there\'ll be a known voltage drop across it, plus or minus the tolerance of the Zener, which can be used as a reference or a supply. In the other case you\'re talking about putting ,say, 100 amps through it for a microsecond or so, knowing that the voltage across the diode will not be greater than the value specified for that current. Same animal, different application.

I used to work on supplies that used varistors, and they were one of the least reliable parts. However I'm fairly positive this was simply because they were being pushed so hard: they always showed deterioration due to high temperatures, ie blackened, cracked, and sometimes burned metallisation. I guess you can push them like that for occasional surges, but not all the time.

NT

they are designed to clamp on over loads that is correct how ever, on heavy over loads, they will clamp to blow the fuse and most times short them self's out permanently with out physically falling apart so that the next time you put a fuse in place! it will simply blow it!. to hold that much current back while its waiting for the fuse to blow it does get hot and destroys it self and remains shorted. that main not sounds like a very good idea to you how ever, it has protected components after it that are much more expensive..

Real Programmers Do things like this. http://webpages.charter.net/jamie_5

TVS diodes are good for that, they start their latching process early once the set voltage starts the clime on the low side. because they don't act in a linear way like a zener would when the avalanche voltage is being approach, it makes this little device a nice thing to have for a fast clamp, and when it does clamp they clamp to a very low R.. and to top it off, because of the low R in the clamp state they come in some large amp sizes in small packages. and the fast time recovery is good so that you don't see a dip below the min voltage. the only problem i have seen with these items is in a case where we used a TVS that apparently was clamping alot on a circuit, the trace on the board was generating pulses to the adjacent trace there for inducing a pulse where we didn't want it. in that project we had to redesign the board to remove the noise.

Real Programmers Do things like this. http://webpages.charter.net/jamie_5

That sounds like diode-suicide in this application.

Chris

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