Electronic – Varistor S10K30 for 24-40VDC input

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I was put in charge of redesigning a motor control circuit which accepts 24-40V DC on its input.

As a side note: As far as i know, the existing circuit has been working fine for several years. The redesign aims to make the production cheaper and replace all obsolete THT parts with SMD and optimizing some other stuff.

The current circuit is protected by a 3A fuse and a varistor of type S10K30. Looking at the datasheet (see below), the "30" obviously refers to the \$ V_{RMS} \$ rating.

What bothers me is, that \$ V_{dc} \$ is 38V. I would've chosen the next lager type with a continuous DC voltage rating of at least >40V (more like 44V for +/-10% supply tolerance).

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I am trying to understand if this varistor type was a poor choice by the engineer, or if I am missing something (I never had to choose a varistor before).

The data-sheet further specifies \$ V_v=47V \$, but I am not sure how to interpret this parameter.

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What does the (1 mA) exactly mean? Does the varistor sink about 1 mA @ 47V? Is this the reason why the original developer was fine with this particular type, because he did not care if the varistor permanently consumes

$$ P = 47V * 1mA = 0,047W $$

Is this bad practice or somehow justified?
How to properly choose the correct varistor for this application?

Best Answer

The data sheets are telling you this: -

  1. The device will conduct 1mA somewhere between 47-10% and 47+10% at 25ºC
  2. At 5A the voltage across the device will be no more than 93V (this is a pulse test)
  3. The max impulse current (based on 8/20us pulse) is 500A at 85ºC
  4. At 85ºC ambient watch out because the operating voltage level drops to 38Vdc

You also need to look at the VI curve for this device: -

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The above will tell you what to expect for two conditions. The left hand area marked A is for quiescent conditions (leakage current) and at 40Vdc there may be a current of about 500uA flowing. This is OK for the design unless you are operating super low power. The right hand area is the protection side of the graph - note how the device has a staggered curve where left and right cross-over. This is giving you the 93V max at 5A figure and you can use the curve to predict what the worst case terminal voltage will be clamped at (say) for 1A. To me this looks like 82V.

My conclusion is that it is OK for a 40V supply unless you are operating at a high ambient temperature - this could give a clue about where/how you should mount the device on the PCB - If the PCB is vertical, maybe it's at a low part of the board where temperatures are bound to be lower. Maybe it's near to a fan. You need to make these decisions.

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