Society of Robots - Robot Forum
Electronics => Electronics => Topic started by: krich on March 03, 2008, 07:04:15 PM
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Okay, so I hooked up a circuit on my breadboard. One of my resistors started to smoke, a clear sign I screwed something up. I checked and double checked my circuit and it was hooked up correctly. The only thing I could think of is that there is too much current going through that resistor. I decided to do the math. This is where I'd like to make sure I'm doing the simple calculations right.
voltage = 16v
resistor = 100ohm
I=V/R = 18v/100ohm = 0.18A
W=V*I = 18 * 0.18 = 3.24W
So, 3.24 Watts going through a 1/4Watt resistor. No wonder. Adding additional resistance to the circuit eliminated the smoking.
For a little background, this resistor is part of a circuit going to the base of a transistor (2n3904). In "production", there will be more resistance than the 100ohms introduced by way of separation of contacts in a liquid (non-flammable, resistance yet to be calculated). Basically, this part of the circuit is a water level detector. Liquid is present, all okay. Liquid not present, sound an alarm.
A second related question. It seems reasonable to me that this circuit should protect itself from having the probe contacts shorted. I'm thinking I need to figure out how much base current the transistor needs in order to switch on and try to find a happy medium between base current, resistor value, the rated wattage of the resistor, and the expected resistance of the liquid. Am I going down the right road here?
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Yeap, you needed a resistor that can handle more watts.
The equation is:
P=IV, or P=(V^2)/R
solving . . .
P=16^2/100 = 2.56 Watts
I'm thinking I need to figure out how much base current the transistor needs in order to switch on and try to find a happy medium between base current, resistor value, the rated wattage of the resistor, and the expected resistance of the liquid. Am I going down the right road here?
Sounds fine. Or you can be lazy and use a resettable fuse :P