Ive used transistors before and I have a basic understanding of them. I understand that a small current to the base will let a large amount of current through from CE. So I modelled this in iCircuit, an app for macs, and it tells me there will be 427mA running though my 1Ω load with a 1kΩ resistor at the base of the transistor
But I want more than that to burn my 1Ω resistor. If 5V out of the mcu thru a 1k R yields: 5/1000=5mA which is close to 427mA. So if I lower R from 1k to 220Ω, I get 1.93A thru my 1Ω load. My questions:
Im worried about that 220Ω. Can it be that low? Is it supposed to protect my MCU in some way and what if its that low (220Ω), could it fail to protect my MCU?
Could I get more Amps through the load without damaging my MCU? The transistor I dont care about.
1/ A small 9v battery won't provide nearly the current you need - 9 amps? I don't think so - you need a much more powerful power source. 9 amps at 9v -> 81 watts - that's quite a lot
2/ A bipolar transistor isn't a good device for this application. It's current gain is probably only about 10. And it's saturation voltage means it'll probably get very hot, if not die.
A logic level N-channel MOSFET would be a far better choice.
Why do you want to burn up small resistors anyway? for the smoke?
Im worried about that 220Ω. Can it be that low? Is it supposed to protect my MCU in some way and what if its that low (220Ω), could it fail to protect my MCU?
The absolute maximum current rating for an output pin on the ATmega chip is 40mA, so that works out to 125 Ohms minimum.
However... If you end-up burning-out the transistor and shorting it, you could put 9V across that resistor, so you probably ought to double the resistance.
DVDdoug:
However... If you end-up burning-out the transistor and shorting it, you could put 9V across that resistor, so you probably ought to double the resistance.
outsider asked: "What size resistor, 1/8 watt, 10 watt? How much current does it take to "burn" your resistor?"
I'm asking it too. And you might want to worry about whether your resistors melt, burn, or explode on overcurrent.
No 2N2222 is going to win against 1 ohm BTW, it will fry first, then maybe fry the Arduino.
And yes, you don't understand that transistor gain, hFE, is irrelevant to operation of a transistor as a switch,
which involves the transistor being in saturation or cutoff. In saturation base current needs to be 5 to 10%
of collector current, whatever the hFE value in the datasheet.
So you have to use a darlington or a MOSFET for high currents.
The FQP30N06L should do the trick. I would but a 1k to 5k resistor in series withe the arduino output to the gate of the FET. This is not a functional requirement but will help protect the arduino in case of miswiring or some other failure.
BTW the 2N2222 has no chance passing the current you nead, the FQP30N06L will pass it without a "sweat"
Also be sure the high current does not go through the Arduino board ground. Keep the high current separated with only one wire from the mosfet source to the arduino ground.
Sorry about the drawing but I couldnt find an mcu in the ltspice library so I added it as a voltage controlled switch and connected its + to one side of the switch and its - to GND1.
Marciokoko:
Ok Im trying to remotely burn a resistor.
Are you trying to make a pyrotechnic igniter for fireworks or model rocketry?
If so there are purpose built, and safe igniter elements and safety circuits.
Tom...
This is what I have so far. The load is a LED, just because I had one on that board. I added a diode just because I used a reference mosfet diagram from a solenoid project I had.
The safest way to do it would be with an opto-isolator or a relay. (Either way would electrically-isolate the Arduino from the stuff you're burning-up.) The relay would require a transistor or MOSFET driver circuit for the coil, so you'd still need a transistor.
