"Voltage" fuse?

Current sources have a max voltage they can put out just like voltage sources have a max current they can put out.

Heres the difference, as you increase the load the current will be the same but the voltage will go down.

Suppose your load is R then V = I*R where I is the current of your current source and V is the output voltage. As R increase so does V.

So you have

Vmax = I*Rmax

or

Rmax = Vmax/I

This means that the max resistance you can have is Rmax.

So if you can be sure that your load is always less than Rmax you can be sure that the voltage is always less than Vmax.

If not then you could do something like use a transistor to cut the current to the load when it is turned on. Or maybe a zener in parallel with the source so that at the zener voltage it will turn on and short the load.

I'm sure there are other methods and these are just some ideas... I have not tried them so I have no idea how well they would work. I'm sure someone will correct me if I'm wrong.

Jon

Oh, I'm sorry. You're right. Duh, I'm thinking backwards. What I need to detect is LOW voltage.

What I'm really concerned with is the detection of a short circuit. If the supply is limited at 2.0A, it will merrily supply two amps even to a short circuit. I want to know when this occurs and shut down the circuit. It's probably something simple.

My load is very low resistance--like an ohm or so--so the difference between a short circuit and the real load is small.

sid

If I understand what you're saying - you want a "crowbar protection circuit", this still uses a current fuse but has a crowbar thyristor (or triac for AC) and a voltage sensing circuit to fire the thyristor/triac. In its simplest form, consists of a thyristor across the supply (after the fuse) and a zener from gate to Vcc - its customary to put a damping capacitor and resistor in parallel from gate to GND to prevent spikes causing false triggering.

Use a comparator - compare VO with Vf of a diode (shottky-barrier if you need really low). How you effect shutdown depends on the design of your PSU.

Without more details of your circuit, it's not easy to be too specific.

But if you have a current-limited supply, and all you're trying to do is turn it off for undervoltage (i.e. short circuit), you could do worse than getting a 5VDC coil relay (possibly with a small ohmic value series resistor). Here's how it works (view in fixed font or M$ Notepad):

| | Start | T | +4V --- V+ | o---o--o o-o-----o---o | | | | | | | | | | | | | | || | RY1 C| | '--||--' C| | || C| | CRY1 | | | | COM | COM | o----------------o---o (created by AACircuit v1.28.6 beta 04/19/05

Most all 5V relays will pull in with 4V. Press the Start button, and the relay in energized. The contact of the relay will keep the circuit closed, and allow current to flow to the load. If you get a short circuit, and the output voltage drops to less than 1.2V, the relay will open up, disconnecting the circuit until you press the start button again. If you've got a double pole relay, you can use the second pole to turn on a fault light.

The problem, of course, is that lowering the voltage decreases the current in many fault conditions. As the current limit voltage drops, the current drops too, so the fault voltage might be only 2.5V or 3V, not low enough to reliably open the relay. Relays will usually pull shut at about 75% of rated voltage, and will guarantee to open once they've closed at about 20% of rated coil voltage.

There are a number of power supply monitor ICs which can supply a logic signal in the event of undervoltage, and work reliably down to

But from a s.e.b. perspective, another way to do this would be to use a double pole momentary contact switch, and also switch a resistor in series with the relay coil, like this:

| | || | .----||---. | | || | | | CRY1 | | | | | | T | |+4V| --- | V+ | o-o---o o---o------o--------o | | | | | T RY1 C| | --- C| | .--o o---. C| | | | | | | '------o | | | | | .-. | | R | | | | | | | | '-' | | | | '---------------o |COM | COM | o------------------o--------o | (created by AACircuit v1.28.6 beta 04/19/05

You will have to fiddle with this if you're picky about the relay open voltage. If you choose R at the same resistance as the coil, it should open if your power supply dips below 2.5V or so. But try experimenting with different values of R if you want to tweak it in. Make sure the resistor can handle the wattage, if you're using a 5V relay with a low ohm coil.

Cheers Chris

I like the relay idea where the voltage is literally completing the supply circuit. But isn't there any sort of solid-state equivalent? I have a number of these low-resistance loads on separate, parallel, constant-current lines and I need to fuse them individually and I have limited space on the PCB. Even a signal that would tell the MCU that a short has occurred would be sufficient. But I have to know if even a single load shorts.

I thought about using a transistor that switches with low base/gate current/voltage, but my load voltage is so low that it gets near the transition region of the transistors, so I don't have a clean break.

sid

google "crowbar circuit"

here's a starter for 10...

Google "foldback current limit" or "... limiting".

Good Luck! Rich

Hi, Sid. There are many relays which take up the footprint of a 16 pin DIP package that have two poles and can switch a couple of amps, especially current limited, low voltage DC. Add a 1/2 watt resistor, and the pushbutton to arm/rearm the circuit, and you're done. You can use the second pole to notify the MCU easily.

If all you want is to notify the PC about overcurrent, you first have to know if your external MCU power supply common is connected to the current-limited load common. If so, you might go with a comparator to forward the voltage drop signal to the PC.

If the load power supply and the MCU power supply commons are not connected, you may want to go with an undervoltage detector IC, made to send a reset signal to a uC in the event of low battery voltage or power dropout. These work reliably down to less than 1.5V. You might want to set this up with an H11L1 optocoupler, which only needs a couple of mA of current to send a logic signal to the MCU side. Or, you might set up a TI voltage reference to source the optocoupler.

If you want better advice, you really need to provide more information, or get a better handle on your project specs. If you want, here are some sample specs:

• Do you need a "fuse", do you need a uP-controlled power switch, do you need an undervoltage signal, or what?

• Are the MCU common and the load(s) common tied together or not?

• If so, what is the MCU voltage?

• If not, you'll have to power the detector circuit with the load voltage. What is the realistic minimum voltage you're expecting in the event of overcurrent?

Give us a hand here, and you'll get better help. Try to answer *all* of the above questions, OK?

Cheers Chris