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I am referring MOV characteristics attached as shown below:

MOV Characteristics

As per it's characteristics,

"The conceptual sketch above shows the current I through an MOV is as a function of the voltage V across the MOV. Below the threshold voltage Vth, we see leakage current. Above Vth, the MOV switches to high conductance mode and the current increases significantly with only a small increase in voltage. (The incremental resistance in high conductance mode is the slope of the VI curve, or ΔV/ΔI"

As per above quote "Below the threshold voltage Vth, we see leakage current" Here, we can see in characteristics that leakage current shown and it's very close to zero.

But if we see voltage at this point, its also less till its reaches zero. At point its zero and voltage across MOV is zero. Since MOV is connected across load, at this point load is shorted and can harm our load.

Am I correct in my opinion ? How it maintains const. voltage to load when MOV is shorted for short instance?

Thank you.

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The MOV doesn't provide power to the circuit (that's why it's I-V curve is only in quadrants I and III of the graph). So if the voltage across it is 1 V or 3 V or 0 V, it's not because of the MOV, it's because of whatever power source (voltage source or current source) is being used to power the circuit.

At point its zero and voltage across MOV is zero. Since MOV is connected across load, at this point load is shorted and can harm our load.

If you drove the voltage across the MOV with a 0-V source, then you shorted out the load with that source, not with the MOV.

If you just connect the MOV across a load with no voltage source then the leakage current will allow any charge in the load to dissipate. This is the same as connecting a high-value resistor across the load. It's not likely to damage the load because the discharge current is going to be fairly small.

How it maintains const. voltage to load when MOV is shorted for short instance?

Say you applied a 9 V battery across the MOV (assuming you chose an appropriate MOV for a 9 V circuit) and some load. The MOV would only draw a small leakage current from the battery, allowing most of the battery's available current to supply the load.

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If the voltage across the MOV is zero then the current through it is also zero. If the current through it is zero then it cannot be shorting the load. Therefore either the voltage is not zero or it is not shorting the load.

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