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High Current (60A) High Voltage (60V)

Lord_grezington

May 3, 2013
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Hello Electronics Point

I have a new dilemma:

Fuse, I cant find a Resettable fuse which can hold around 55A and 60Vdc. I have looked at just putting some or these guys in parallel (http://www.mouser.com/ds/2/54/mfrx-48102.pdf) but although it decreases the resistance and makes it look very efficient, having 8 of these things in parallel just does not seem right, sort of non professional. I have thought of using P-channel fets to attempt to minimise the current (using discrete components). does anyone have any other good ideas on how to put some safety on this without going through a mC?

I should note as well, I have sensing the current via a sense resistor and going through am opamp into a mC, but this us to operate as a secondary failsafe in case something goes wrong.

Does anyone have any good ideas? cost is not massively critical however should be kept to a minimum (Ie, no more than a few £)

Thanks
 
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(*steve*)

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Perhaps the first step is to tell us why you need this functionality. There may be a better way.
 

Lord_grezington

May 3, 2013
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I want to fuse the design just in case I get a short or something goes wrong. A second protection if something happens to the mC,

I have attached a schematic (very rough) of an idea I have had using a relay.

R1 = pull down to ensure no current flows when the fuse (FS1) blows.
FS1 = PRT resettable fuse
R2 = Resistor (value dependant on relay coil resistance + Fuse Resistance).
RL1 = The relay which is not yet specified

Just need to know if the theory will work. I think I am going to get into issues with the tolerance of the coil resistance. Any ideas are welcome.
 

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KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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I don't think you can do it with your relay idea, but the circuit you linked to is a nice idea.

What you need to do is (a) detect the current flowing in the circuit, (b) apply some kind of comparison and/or time delay, and (c) break the circuit and keep the switch open until it's reset.

The usual way to measure the current in the circuit is by inserting a small-value resistance in series with it, and measuring the voltage across it. This resistance is called a shunt. The circuit you linked to uses the MOSFET as the current shunt; the current is converted to a voltage by the Rds(on) (ON-resistance) of the MOSFET itself. The potentiometer adjusts the gate bias to give the desired resistance. According to Ohm's Law, V = I R, the voltage across the MOSFET is equal to the current flowing, multiplied by the MOSFET's ON-resistance. Once this voltage reaches about 1.3V, the transistor turns ON and switches the MOSFET OFF. As soon as the MOSFET switches OFF, the voltage on its drain increases quickly and this holds the transistor ON until the power is removed.

Another way to measure the current in the circuit is using the Hall effect. Allegro Microsystems (http://www.allegromicro.com) make several ICs for this.

There are also MOSFETs available with built-in current sensing featurs, such as the NXP BUK7909-75AIE (see http://www.digikey.com/product-detail/en/BUK7909-75AIE,127/568-9742-5-ND/1163555). These are all N-channel, so it's simpler to put them in the negative rail. If you can't break the negative rail, you can put them in the positive rail, but you need a little voltage generator to supply the gate bias voltage and power the control circuitry.

Depending on the current sensing method, you will probably need some amplification to bring the current signal up to a usable level. You can use a comparator with a reference voltage, or just some discrete components, to detect when the limit is exceeded. An R-C circuit is normally enough to provide a delay (if you want one) and can be arranged so that the delay is shorter if the overload is larger, as is normal for a non-electronic fuse.

As for breaking the circuit, the obvious choice is a MOSFET. It needs a very low ON-resistance if you want to feed 55A through it continuously; it will dissipate about 3W for every milliohm of internal resistance! For the BUK7909-75 I mentioned above, that's 27W continuous power dissipation under normal operation! It will need significant heatsinking. Also, it will drop about 0.5V.

You need to make sure it's switched quickly, because at 60V and 55A, assuming a resistive load, the maximum instantaneous MOSFET power dissipation during switch-off will be over 800W!

Edit: I agree with Steve. You should be telling us everything you can about the application. That's the way to get the best advice quickly, without wasting everyone's time unnecessarily.
 
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Lord_grezington

May 3, 2013
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Hi Kris

Thanks for your post...

I was using some Mosfets for PWMing power anyway, I was hoping to use something standard without too much complexity and would save as much as possible if something went wrong.

I have now decided to mount into a standard 60A domestic consumer unit and use a circuit breaker.

Does anyone know anything about circuit breakers? Ie, how do they compare to fuses (when used in DC not AC). Will a 60A circuit breaker still "break" when 60A dc goes though it?

Thanks
 

GonzoEngineer

Dec 2, 2011
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LEM current sensors are cheap, and very reliable. I use them everywhere, from 100A
to 2,500A.

Feed the signal to a comparator, use the output to control the series switch.
 

(*steve*)

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Lord_grezington, when it comes to high power stuff, GonzoEngineer is the man to listen to.
 
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