# "Voltage" fuse?

Discussion in 'Electronic Basics' started by sid, May 14, 2007.

1. ### sidGuest

Hi guys:

Hopefully this is a no-brainer.

If you have a current source that supplies a constant amount of
current, regardless of load, how do you "voltage fuse" it? In other
words, I want the circuit to kick out when too much voltage is
present, rather than too much current, like a regular fuse. What is
the simple, cheap equivalent of a fuse in this situation?

....Oh, and it's a low voltage. I want it to blow at about 4V.

Thanks for any replies.

Sid

2. ### Jon SlaughterGuest

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

3. ### sidGuest

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

4. ### ian fieldGuest

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.

5. ### ian fieldGuest

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.

6. ### ChrisGuest

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\$

|
| 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 www.tech-chat.de)

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
1V or so.

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 www.tech-chat.de)

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

7. ### sidGuest

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

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

8. ### feeboGuest

here's a starter for 10...

http://www.kbt-dc-supplies.com/crowbar.php

9. ### Rich GriseGuest

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

Good Luck!
Rich

10. ### ChrisGuest

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