ISO224: Simple and inexpensive solution for leakage measurement

Hi Javier,

a measurement range of 1nA to 10mA means a range of 1:10,000,000. This can impossibly be achieved with only one single shunt. Because of this, you should think about what total measuring range is realistic and how many individual measuring ranges (each with its own shunt) you really need to have. And you should think about how you want to switch between these shunts. Not switching between them but having several different and separate circuits, each with a very certain measuring range and shunt may also be an (inexpensive !) option.

When it comes to short-cicuit protection in a high voltage application, you urgently need a high voltage source with an output current limiting. The easiest way to achieve this is to have a current limting resistor directly at the output of high voltage source, best being already internally implemented in the high voltage power supply unit so that you cannot unintentionally bypass it.

The current limiting resistor should be chosen in such a way that the voltage drop across it caused by the maximum measuring current is under let's say 1% of the measuring high voltage. So, if your maximum measuring current is 1µA and the measuring voltage is 2kV, then a voltage drop of 20V may be allowed resulting in a current limiting resistor of

R = 20V / 1µA = 20M

In a short circuit event the short circuit current is limited to

I_sh = 2kV / 20M = 100µA

then, resulting in a heat dissipation of 200mW. And to make the 20M resistance survive 2kV without needing to buy an expensive HV-resistor, take ten 2M resistors in series. This trick is also and especially useful, when you have to deal with a higher maximum measuring current.

If you have the possibility to take the voltage drop across the current limiting resistor into calculation (µC, etc.) you can even choose a higher voltage drop across the current limiting resistor than the above 1% of measuring voltage. This can dramatically simplify the design when you have to deal with a much higher maximum measuring current than the 1µA of the example.

High voltage applications almost always contain such a pure passive short-circuit limiting resistor at the output.

Kai

Hi Kai and Alexander,

first of all thanks a lot for your valuable inputs. I have put some thought into it and instead of a shunt it seems that a logarithmic amplifier, such as the LOG114 should be able to fulfill my needs in regards to measurement range and reduce the circuit complexity. What do you think?

In this case, for the protection resistors what I am wondering is how high the input impedance of the LOG114 is and how this would affect the protection concept.

By the way, in series with the semiconductor whose leakage is measured, my plan is to have a well a support Switch that would react fast to an overcurrent or over voltage event. As an additional safety measure.

Regarding the signal acquisition path, my idea was to have directly after the LOG114s an ADC (such as the ADS124S08) to measure many Devices Under Test. Afterwards I would place an isolation barrier to go to the main uC via SPI. For the supply I would use an isolated flyback. The ADS124S08 and Isolated flyback is something I have already implemented in other project for a readout of several Thermocouples and it works nicely.

BR,

Javier

Hi Javier,

Arijav said:

By the way, in series with the semiconductor whose leakage is measured, my plan is to have a well a support Switch that would react fast to an overcurrent or over voltage event. As an additional safety measure.

Semiconductors can die within tens or hundreds of nanoseconds. In a short-circuit event a microsecond can mean "eternity". So, in any case to have to limit the short-circuit current to a value the semiconductor can withstand for longer periods. A power resistor with a high heat capacity needing a long period to heat up may profit from such an overcurrent or even overtemperature switch, but not the semiconductor. No melting fuse can protect a semiconductor against a heavy short-circuit event, even not a super fast acring fuse. So, the best remedy is the current limiting resistor at the output of high voltage source, as I meantioned earlier. In any case you have to limit the current running into the LOG114 to 10mA.

Arijav said:

first of all thanks a lot for your valuable inputs. I have put some thought into it and instead of a shunt it seems that a logarithmic amplifier, such as the LOG114 should be able to fulfill my needs in regards to measurement range and reduce the circuit complexity. What do you think?

A log amplifier is a good idea for handling huge measuring ranges. But keep in mind that due to the very high compression when taking the logarithm the resolution and the precision of the signal at the output of LOG114 may decrease compared to when using a shunt in a linearily working circuit. See the "average total error" curves in the datasheet of LOG114 on page 8.

Have you already drawn a schematic you could show us?

Kai

Hi Kai,

thanks for your valuable input. As safety switch I was thinking about a MOSFET not a relay or fuse, but the reaction time would be nevertheless in the range of 10us, therefore probably too long. As a current limiting resistor, my understanding then would be that in series to the LOG114 I would have to place a 200KOhm resistor to limit the current to 10mA in case of breakdown (@2kV). Do you see it in the same way? The issue that I see is that at 10mA which would be still a range I want to measure leakage, the Semiconductor under test would not see the 2kV, but almost all the voltage would be in the limiting resistor. The test I want to performed is an HTRB and therefore the semiconductor really needs to see within the measurement range the voltage of the high voltage power supply.

For now I have no schematic to share, I am still at concept level selecting what to put. Regarding the accuracy, you are right, I still have to check if it is sufficient, but hope so, since using several Shunts in parallel and switching them seems to increase the complexity too much. In the future I might need to be able to design a system to test in parallel up to 100 semiconductors.

Regards,

Javier

Hi Javier,

sorry, I don't want to offend you, but 2kV times 10mA gives a heat dissipation of 20W within the semiconductor. This is quite a lot and can easily become explosive. And multiplied by 100 semiconductors this gives a total power of 2kW at 2kV... Do you really want to measure up to 10mA? I would 10mA not call any longer an "usual leakage current" but more an "unlimited breakdown" current 

And do you really want to measure down to 1nA when the maximum leakage current can be up to 10mA? This would mean a range of 10,000,000:1...

Kai

Hi Kai,

no offense taken at all, I appreciate your feedback. 10mA is actually the top limit, at this point the semiconductors are dying off already, most of the time the leakage is in nA range. 20W is not critical, due to the fact that this happens at the very end and in this testbench I shall be able to test big Power Devices (f.i. 1.7kV SiC MOSFETs), some of them in big packages with big heat sinks (f.i. Pinfin structures) and if needed cooling. I know for a fact that others use this measurement range of 1nA to 10mA as well and I know specifically a University which is using the concept of a logarithmic transimpendance amplifier for the current measurement. My bid headache now is more on how to protect the measurement system including the amplifier. It might very well be that the HV Power supplies do include this current limitation and my protection switch is good enough with its 10us reaction time, but I am not sure.

Javier

Hi Javier,

in any case you must limit the current into the LOG114 to below 10mA.

Assuming a short circuit current limited to 100mA by the help of a current limiting resistor at the high voltage source, a low leakage diode clamp from the input of LOG114 to signal ground in combination with a current limting resistor could do:

javier_log114.TSC

Kai