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Also, the Heating Resistor will have "thermal inertia" and may stay hot for a while even when it's not receiving power from the PWM/Mosfet circuit. DMMs do a pretty good job of averaging voltage from a PWM circuit, so you could hook your DMM in voltmeter mode across the power resistor, and it should show you the resulting voltage steps as the PWM feeds the resistor. You may need to play around with the delay() statements in your code.
Hey... you could try this. Connect a potentiometer to one of the Analog input pins and use an analogRead() statement to control the delays!
Connect the two ends of the pot to +5 and Ground on the analog side of the Arduino and connect the wiper to A0. It doesn't matter what the pot value is, but 10k, 50k or 100k are good choices.
Then put a statement at the top of your loop() (or in each of the "for" loops) like this:
int PotDelay = analogRead(A0);
Then for each delay() statement in the PWM increment and decrement , instead of putting in a number of milliseconds like delay(100), use
delay(PotDelay);
and then the potentiometer setting will give you a delay from 0 to 1023 milliseconds each time the code hits the delay() statement.
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Ok one more question alsetalokin. I feel like I'm getting close to nailing this breadboard down and being able to play with PID code.
My question is this. I've gotten the LED to work great with the MOSFET circuit and PWM code. But when I replace it with the heating resistor, nothing happens.
I thought this was due to the PWM code and not allowing the heating resistor enough time to heat up. So I gave it an even simpler code of just putting pin 3 to high for 1,000,000 MS (17 mins). Resistor does not heat up. Then you plug the LED back in and it stays lit.
Is this due to the big 270 Ohm (non-heating) resistors I put on the board? Too much resistance on the current before it's fed into the heating resistor?
See picture attached
Code I used
// the setup function runs once when you press reset or power the board
void setup() {
pinMode(3, OUTPUT);
}
void loop() {
digitalWrite(3, HIGH); // turn the LED on (HIGH is the voltage level)
delay(1000000); // wait for a second
digitalWrite(3, LOW); // turn the LED off by making the voltage LOW
delay(1000); // wait for a second
}
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It works! All I did was bridge the gaps as you suggested
Cool!
But now you have 2 series resistors for your LED: one on the breadboard and one soldered to the LED itself, so your LED will be a lot dimmer than it could be... -
Ok one more question alsetalokin. I feel like I'm getting close to nailing this breadboard down and being able to play with PID code.
My question is this. I've gotten the LED to work great with the MOSFET circuit and PWM code. But when I replace it with the heating resistor, nothing happens.
I thought this was due to the PWM code and not allowing the heating resistor enough time to heat up. So I gave it an even simpler code of just putting pin 3 to high for 1,000,000 MS (17 mins). Resistor does not heat up. Then you plug the LED back in and it stays lit.
Is this due to the big 270 Ohm (non-heating) resistors I put on the board? Too much resistance on the current before it's fed into the heating resistor?
See picture attached
Code I used
// the setup function runs once when you press reset or power the board
void setup() {
pinMode(3, OUTPUT);
}
void loop() {
digitalWrite(3, HIGH); // turn the LED on (HIGH is the voltage level)
delay(1000000); // wait for a second
digitalWrite(3, LOW); // turn the LED off by making the voltage LOW
delay(1000); // wait for a second
}
Yes, as I mentioned earlier the resistor on the board is a _current limiting_ resistor for the LED. Don't use it for the Heating Resistor! Look at the Fritzing sketch I posted up above (Heating Resistor3) and you'll see how to put the HR in parallel with the LED+CLResistor, so that the current through the HR isn't limited by the LED's resistor.
The power dissipated in a resistor is P=I2R, so it's easy to see that the power dissipated in your 270 ohm resistor is more than ten times the power dissipated in the 20 ohm heating resistor, if you keep the 270 ohms in series with the HR. The current through the 270+20 ohms, if you are using the 4.5 volt supply, and neglecting the Mosfet's resistance, is I = V/R or 4.5/290 = about 15 mA. So the 20 ohm HR will only be dissipating I2R, or (0.015)2x20 = a bit over 4 mW, and the 270 ohm resistor is dissipating about 61 mW !! (if I did the math right...)
If you don't use the current-limiting resistor in series with the HR, then you should have I = V/R or 4.5/20 = 0.225 A, and the HR should be dissipating (0.225)2x20= a bit over 1 Watt. (Again neglecting the mosfet's on-state resistance, and assuming 100 percent ON from the PWM.) You should be able to feel this with your fingers, I should think. But for real heating, I'd use a higher supply voltage for the HR. -
Now... I have to wonder why you are using the Caddock 1% resistor if all you are using it for is a PID-controlled heater. An ordinary wire-wound "cement" power resistor would be more suitable, I should think, and would allow you to use more power. And would be cheaper too!
Even if you are going for a very precise heating control, the PID system should take care of that, even if you use a normal 5% or 10% tolerance cement power resistor.
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I got that circuit up and running. Works well. The parallel LED indicator is great.
What did you have in mind to use the 2nd LED for?
See pics
will post more once I get working on PID