Jan 31, 2018 - Nixie MultiMeter, Part 10

When I started this multimeter design series, I thought it would be fairly simple and it wouldn’t take very long. It turns out that it’s more complicated than I was expecting, and I’m not one to do things halfway, so here we are at part 10. The finish line is in sight, but it’s probably another few weeks and 10-20 posts.

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Current Sense Protection, Redeaux

I spend some time over the last few days drawing up and analyzing a potential current measurement circuit with the input protection that I disucssed in the previous post.

Current Sense Protection Circuit

For reference and searchability, the table at the bottom is a rough analysis of the circuit performace at maximum rated current and the maximum protection voltage for each range.

Range Total Resistance Voltage Drop @ Range Current Current @ 200V Sense Resistor Power @ 200V
10A 1.5m Ohm 15mV @ 10A 1.3 x 10^5 A  
100mA 150m Ohm 15mV @ 100mA 1.3 x 10^3 A  
1mA 70 Ohm 70mV @ 1mA 2.85 A 122W
10uA 1.555k Ohm 15.55 mV @ 10uA 129 mA 25W

There’s a few things to note from this circuit and my rough calculations

I’ve had a hard time finding voltage vs current curves for the PTCs that I’d identified in the previous post, so I found another PTC that has better specifications, the EPCOS B59751C0120A070.

PTC Voltage vs Current

At about 200V, the current through this PTC is about 12mA (plus or minus about 1mA). At nominal voltage, this PTC has a resistance of about 50 Ohms.

Resistor Overload Rating Voltage @ 12mA Overload Current Trip Time
15 Ohm 300V / 5 seconds 180 mV 3.07 A 0.7 sec
1.5k Ohm 200V / ? seconds 18 V 129 mA No trip

With the 1.5k Ohm resistor, the overload current isn’t high enough to trip this PTC. Based on the datasheet, the required current to trip in under 1 second is about 2A. To generate 2A, the resistance needs to be reduced to about 100 Ohm total, which would require adding a 50 Ohm resistor, and a bypass diode in parallel with the sense resistor.

If I put the addtional resistor in parallel with the sense resistor, it won’t add any voltage drop, but at overload the sense resistor will still see at least 100V. If I put the additional resistor in series with the sense resistor, it will add a small voltage drop, but at overload the sense resistor will only be subjected to the voltage drop of the diode.

For the high-current resistors, the overload ratings are:

Resistor Overload Rating
1.5m Ohm 15W / 5 seconds
150m Ohm 200V / 5 seconds

I’m not completely sure I believe the overload rating on the 150m Ohm resistor, but either way the overload rating on the 1.5m Ohm resistor definitely indicates that I will probably need more than just a fuse to prevent that resistor from overloading.

I suspect the best way to solve this is with a small inductor on the common side of the circuit. I’ve found a few 10A ceramic fuses, and it looks like both of them will open in 1mS or less when subjected to at least 200A, which implies that the inductor needs to limit the rate of change of current to 2 x 10^5 A / s or less. Since V = L * dI/dT, this implies an inductance of about 1 mH will produce the desired current limiting before the fuse blows.

I’m having a hard time coming up with a good way to model this from first principles, so I’m going to model the circuit in LTSpice and see how it responds to a 200V step on the input.

I needed component values for my measurements, so I’ve selected the following components for my model:

Fuse Simulation Circuit

Download LTSpice Model

Fuse Simulation Results

Assuming a 0-Henry current-limiting inductor, the current quickly rises to almost 18kA. The lead inductance clearly isn’t helping much here, but the added resistances of the other circuit elements helps limit the current significantly compared to my previous estimates. The I^2 * t equation for the fuse I’ve selected seems to be valid all the way to the 1mS / ~160A point, so it seems reasonable to conclude that it will hold above that current as well. At 18kA, the projected fuse time is 80 nanoseconds. Given that this is more than a few orders of magnitude away from the manufacturer’s data, I think I’ve extrapolated too far. It’s late and my ability to come up with the correct integral to describe the required fusing time doesn’t seem to be working, but it shouldn’t be too hard to come up with a conservative estimate.

Extrapolating from the existing data, a fusing time of 100uS should require about 500A, and a fusing time of 10uS should require about 1600A. I’m not willing to extrapolate more than that without experimental data. Based on this data, I can conservatively estimate that the fuse will open 1mS after the current reaches 160A, 100uS after the current reaches 500A, or 10uS after the current reaches 1600A, whichever is first. This doesn’t take into account the fact that the current is still rising, which is why these estimates are very conservative.

Based on my simulated data with a 0-Henry current limit inductor, the current reaches 1600A 21uS after the current ramp starts, which suggests the fuse will blow 31uS after the initial voltage application.

With a 1uH inductor, the fuse current reaches 1600A after 46uS, which suggests the fuse will blow at most 56uS after voltage is applied.

With a 1mH inductor, the fuse current reaches 160A after 0.8mS, and 500A at 2.6mS which suggests the fuse will blow 1.8mS or 2.7mS after voltage is applied. Since 1.8mS is sooner, this suggests that the fuse will blow within 1.8mS of voltage application, and this estimate is more believable because it is based directly on data from the fuse’s datasheet, instead of extrapolated data. Unfortunately an appropriate 1mH inductor with low resistance is difficult to find on DigiKey, and I’m a little worried about the stored energy too, since my circuit model doesn’t include opening the fuse or dissipating any energy stored in the system’s inductance.

Looking at candidate inductors on DigiKey, they seem to be rather bulky and I’m not convinced that an energy-storage device is a good choice as part of a protection circuit. I suspect that if I were able to properly simulate a fuse, when the fuse opened I’d see a very large voltage spike as the magnetic field in the inductor collapsed.

The better choice here seems to be much simpler; just use a fuse that is rated to interrupt a very high current, and with an interrupting rating of 20kA, Littlefuse’s FLU series seems to be a perfect match. It helps that they seem to be explicitly designed for multimeters, and the FLU name hints that they may have been designed to meet the requirements of a well-known multimeter manufacturer (Fluke, of course!).

The 10-Amp FLU 011 fits my requirements perfectly, although at $25 each they’re a bit pricey.

For the 150m Ohm sense resistor, at 9A (just below the fuse rating) this resistor would drop 1.35V and dissipate a little over 12W. Since this could be indefinite without the fuse blowing, this sense resistor clearly needs additional protection. Given the lower power requirement here, a high-voltage PTC seems easier to deal with than a fuse. A quick DigiKey search turns up a variety of PTCs with current ratings between 200mA (2x my range) and 1A (reasonable maximum for this sense resistor). Since I want to limit the voltage drop at the working current, this Bel Fuse 0ZRE0040FF1A seems reasonable. The 600m Ohm resistance (plus the 150m Ohm sense resistor) results in a 75mV drop at 100mA (rated current for this stage). This PTC’s hold current of 400mA is well above the range’s rating of 100mA, so there’s no worry of it tripping during normal operation. At the PTC’s trip current (900mA), the sense resistor only dissipates 0.12 W, which is well below its rated power. I’m not too worried about the maximum current at max voltage, since I expect the 10A fuse to handle those situations.

Thoughts on Auto-Ranging for current

Since this circuit uses relays for range switching, the software on the microcontroller will be able to control whether the relays operate in a break-before-make or make-before-break style when switching ranges.

Since there is only one fuse, an overvoltage condition with autoranging won’t act as a “blow all the fuses” mode, although if the meter is in a low range it may transition to a higher range and then blow the fuse for the high range. With a $25 resistor in the high range, I’m not sure this is a good idea.

If I do auto-ranging, it should be aware of the fuse state and should probably turn off if the fuse blows, so that it doesn’t cycle through all of the relays trying to find an appropriate current range for the input.

Opening or closing relays in a circuit that carries a high current can result in arcing of the relay contacts, which shortens the life of the relay and increases the relay’s contact resistance. This really only matters for currents above an amp, so for most practical purposes it only affects the 10A range relay. Maybe switching in or out of this mode should only be done when the user requests it.