# TVS diode protection

Discussion in 'Electronic Design' started by Jamie Morken, Jan 29, 2004.

1. ### Jamie MorkenGuest

Hi all,

Does anyone know of any TVS diodes that have 5uA or lower leakage current at
5V and turn on a bit above that? I am trying to make a circuit to protect
the analog input of the microcontroller (AVR) I am using. Digikey has a 5V
one with 800uA leakage, but that drops my (weak) signal by 700mV.

cheers,
Jamie Morken

2. ### N. ThorntonGuest

I'm not clear on your circuit or what youre trying to do. If you post
it you may get more assistance. Diodes can be connected to a 0.6v rail
formed by a diode and resistor, this is one approach to fixing diode
ofsets.

Regards, NT

3. ### Jamie MorkenGuest

Hi,

Thanks, I posted a drawing of the circuit:
http://www.rocketresearch.org/new/temp/r-divider-with-TVS.jpg

The TVS is there to protect the uC ADC input from high voltage transients
that may come in from the input signal. The problem I am having is the
measured voltage is drawn down by the TVS leakage current (before it
clamps). I'd like to be able to have the TVS have under 5uA at 5V and then
clamp at 5.5V or so (the micro is specced at 5.5V for max input voltage).
Maybe I should be using a zener for this but I've heard that TVS are the
fastest voltage suppressors. Thanks!

cheers,
Jamie Morken

4. ### Chris CarlenGuest

You should put a capacitor on the output of the voltage divider, to at
least create a simple low pass filter which will slow down transients,
as well as provide an anti-aliasing filter for the sampling system. Use
the parallel combination of the divider resistors to figure the C based
on what cutoff frequency fc you want from fc=1/(2 pi R C)

Then, use just a zener diode with low leakage of 5uA such as 1N5231B,
which will be fast enough.

Note that the leakage of 5uA will only create an error of about 1 count
in the 10bit conversion. I suppose this is still to be avoided, but the
A/D in the CPU isn't that great anyway, especially unless you make every
effort to obtain maximum accuracy.

To get no leakage issue at all, use a pair of reverse biased fast signal
diodes like 1N4148 going from the A/D input to the power rails, but then
you have to think about whether this is just static protection, or
overvoltage protection. Static protection will be accomplished with
just the diodes and good bypassing on your rails. But if you apply
positive overvoltage to a dual diode protection network, you run the
risk of it pulling your supply rail voltage up. Thus, you can put the
TVS on the *supply*. Then you are protected from ESD, overvoltage at
the input of both polarities, and transients fed into the system from
the power supply. Just choose the TVS carefully to turn on before you
overvoltage your components, and not too soon so it doesn't sit there
partially conducting all the time. A more cautious approach is to use
an SCR crowbar for slow overvoltage protection of the power, and a TVS
that you are sure won't conduct at the normal supply voltage.

It can drive one nuts trying to figure out what level of protection to

Good day!

--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA

5. ### Jamie MorkenGuest

Hi,

Thanks for all the info! I checked the datasheet for the 1N5231B and it has
5uA at 2Volts, and doesn't specify
the leakage above that (its zener voltage is 5.1V)
I think I would go for the 1N5232B a 5.6V zener that has 5uA at 3Volts. Not
sure what its leakage would be at 5Volts though (not in the datasheet)

cheers,
Jamie Morken