...
> 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
need really low). How you effect shutdown depends on the design of your PSU.
On May 14, 4:40 pm, sid <sylveste
@yahoo.com> wrote:
> 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
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 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
-----------------------------------------------Reply-----------------------------------------------
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