Do you have a reference for that? Are you referring to the current-crowding effect that can happen with any negative tempco junction, presumably including Schottky diodes?
--
Thanks,
- Win
That should be fine. Peak forward current will be around 100 mA, which won't do much of anything.
Expect the 100 ohm resistor to get warm.
John
Yeah, reference appreciated. I've never heard this anywhere before.
John
As others have suggested, you don't get heaps of current either way. Think about wattage.
100 mA forward times 0.7 volts gives .07 watts during forward bias. During reverse bias you have 7 volts across a 100 ohm resistor for 70mA, which all goes through the zener if you don't have a load. That's 70mA times 5 volts or 0.35 watts. The average of .07 and .35 is about a fifth of a watt max dissiption through the zener. So the zener is okay. With the resistor you have 100 mA half the time and 70 mA half the time. Power dissipation is I^2 R. That's 1 watt during one half cycle and half a watt during the other half cycle for average continuous power dissipation of 3/8 watt with no load. With a load the power dissipation through the resistor will increase, so if you have any kind of load you have to take that into account. All the load current will go through the resistor, so add load current to your 70 mA and calculate the half cycle power and average that with 1 watt of the other half cycle to get total power dissipation in the resistor during load. You might need a power resistor. And remember a power resistor running at its full rating could about fry bacon.
The 1N52xx 500mW zener datasheet I brought up had forward voltage specified at 200mA (a bit over 1V maximum, IIRC).
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
I want to connect a 5V, 1/2Watt zener with the anode to ground and a 100 ohm resistor to the cathode end connected to a 12VAC transformer. I should get 1/2 wave pulses clamped at 5V in the positive direction. The forward conduction of the Zener should clamp the negative cycle at around 0.7 volts. But will I exceed the forward current rating of the zener this way? Will using it this way eventually harm the zener?
TIA
Rick
Zeners make very bad diodes. Forward currents are not spread over the junction area. I'd suggest you place a Schottky in parallel with it if you want to do this.
-- -- kensmith@rahul.net forging knowledge
On Sun, 23 Oct 2005 16:20:39 GMT, "kelvin_cool_ohm" wroth:
Motorola's "garden variety" 1N53xx series of zeners is characterized for
1 Amp of current in the forward direction. You're only going to be asking your zener for .02 Amps in the forward direction. I suspect any .5 Watt zener will be fine with that.Jim
FWIW, there does not appear to be anything specific about such an effect in the Motorola TVS-Zener manual, at least based on keyword searches that I came up with.
There's quite a bit about current crowding reducing the maximum
*reverse* surge rating to much less than you would expect based upon junction temperature etc. It doesn't say there is *no* problem with forward operation at high currents, rather it just does not seem to be addressed directly other than to say that the behavior is similar to a regular diode.BTW, some of the common axial-lead TVS devices are rated for hefty forward surge currents in the 100A-200A range (1/2 60Hz cycle).
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Yes, IIRC, it is current crowding that does it.
Experience:
I managed to destroy 50W 100V zeners by passing 2A through them forwards for short pulses.
This was some time ago but I do remember it was a Mot power zener and IIRC it was their app. guy that said that this was the problem.
-- -- kensmith@rahul.net forging knowledge
In article , Winfield Hill wrote: [...]
Something interesting came up in my search:
-- -- kensmith@rahul.net forging knowledge
17V / 100R = 0.17A
-- -- kensmith@rahul.net forging knowledge
The operative phrase there is "short pulses"- this is not an effect that will appear at line frequency.
In article , Spehro Pefhany wrote: [...]
Elsewhere, I put in a like to one such device. It seems that the TVS type of zeners have had a lot more thought into such issues. Perhaps the OP could use one of them if he is still interested.
-- -- kensmith@rahul.net forging knowledge
[....]
Are you saying that the 50W zener would survive 2A if I left it on for a long time? ... or did you take the 2A number to be the average and not the value during the pulse?
-- -- kensmith@rahul.net forging knowledge
I am saying the zener would have survived if you there was limited dI/dt in the application of the 2A forward pulse. If that 50W Motorola job was not a single die, it may have been a multitude of parallel standard dies of slightly different forward characteristic, then it's possible 2A would blow it no matter the slew rate.
12V RMS is about 17V ground to peak. a 0.5W 5V zener can handle 500mW/5V=100mA current in its reverse (blocking) direction continuously more fur short durations.
in the forwards direction the most you'll see through the 100 ohm resistor from 12VAC is about 160mA there's not enough current there to overheat it and it's not wildly above the rated reverse current, so it should be OK.
in the reverse direction you've got a peak current of around 120mA which should be fine too.
Bye. Jasen
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