Short Circuit Protection
I need help in trying to protect a low power
transistor output pulse. The transistor is a BJT with
max output current of 0.025A and Vsat typ 0.135v to
0.400v.
I need to provide some sort of current short circuit
protection that will limit the current to 0.060A.
Under normal operation the transistor must be able to
sink/source .020A.
Any help would be appreciated.
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transistor output pulse. The transistor is a BJT with
max output current of 0.025A and Vsat typ 0.135v to
0.400v.
I need to provide some sort of current short circuit
protection that will limit the current to 0.060A.
Under normal operation the transistor must be able to
sink/source .020A.
Any help would be appreciated.
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Listen to your Yahoo! Mail messages from any phone.
http://phone.yahoo.com
Comments
If your BJT transistor has a max current of 0.025A, it won't help to limit
the current to 0.060A, as the transistor will be reduced to a miniature
smoking pile of rubble long before 0.060A is reached (2.4 times the rated
current capacity of the device!)
You left out a couple of important pieces of information:
1) What is the Rl? (R sub ell, resistance of the load)
2) What is the Vcc? (voltage supplied to the collector)
Let's take a look at a few variations on the theme of Ohm's Law:
1) I = V / R (I=current, V=voltage, R=resistance)
2) R = V / I
3) V = I x R
I'm going to make the assumption that your Vcc is +5 volts, and that your Rl
is 50 ohms.
Note that when your transistor is biased on, it will have some resistance as
well - that's what causes the heat buildup when you exceed the devices'
current limits, and resultant meltdown. Applying variation 1 of Ohm's Law:
I = V / R
I = 5v / 50 ohms
0.1A = 5v / 50 ohms
Oops, our transistor is now a melted puddle of silicon. Let's recalculate
using variation 2 of Ohm's Law:
R = V / I
Rtot = 5v / 0.025A (Vcc / max desired current )
200 = 5v / 0.025A
Now we know that in order to not exceed 0.025 amps at Vcc of 5v, we'll need
a total Rl of 200 ohms. I've assumed that your Rl is 50 ohms; therefore
you'll need to add 150 ohms to the circuit somewhere. Adding the extra
resistance will mean that your voltage across the load will be very
significantly decreased. This might be OK if your load is inductive (eg: a
relay's coil). If not, I suggest examining a different type of device that
is capable of handling more output current.
Why not take a look at UNL2003's/NTE2013's? (7-channel inverting Darlington
pairs, 16-pin DIP). Parallax Inc. sells UNL2003's on their website at very
competitive prices (I paid a buck more for a locally-purchased NTE2013; I
was in a hurry!) If you're controlling a circuit that uses more than 5v,
there are other variations on the UNL2003's to handle more voltage.
You might also be interested in ECG2018/SK9079/NTE2018's; these are 18-pin
DIP 8-channel inverting Darlington array drivers for 5v TTL/CMOS; capable of
sinking about 600mA. It's so much easier to use standard DIP packages than
trying to build circuits from discrete components. If you need to
current-limit these IC's, you could always put a resistor on the ground pin.
Hope this helps...
Steve
> John DeFeo wrote:
> I need help in trying to protect a low power transistor output pulse. The
transistor is a BJT with max. output current of 0.025A and Vsat typ 0.135v
to 0.400v.
> I need to provide some sort of current short circuit protection that will
limit the current to 0.060A.
> Under normal operation the transistor must be able to sink/source .020A.
> Any help would be appreciated.