I want to use this transistor Intelligent Power and Sensing Technologies | onsemi for a project but how do i find out the maximum gate voltage and the current it will draw. I am going to put 12V though it if thats any help.
Also just a quick question what are waveform generators used for, like you can have the waveform come up on your oscilloscope beyond that i don't know what other things can you do with them
a) Broken link.
b) The datasheet will be on the digikey page. Try reading that.
That's an BJT (bipolar junction transistor, the type invented in the fourties (1947))
It is MOS transistors who have gate.
A BJT are controlled by base CURRENT. The voltage are around 0,6 - 0,7 V.
The numbers are here:
VCE(sat) Collector-Emitter Saturation Voltage(1) IC = 3 A, IB = 375 mA 1.2V max
So when you put 375mA into the Base (Ib) (more than an Arduino can drive directly),
and have 3A of current flowing from collector to emitter (Ic),
the voltaga across the collector-emitter (Vce) will be 1.2V max.
If you look at Figure 2, for a more modest current, Ic = 1A, Vce will be ~0.25V, Vbe will be ~ 0.9V, and Ib will need to be 0.1A (1/10 Ic).
If you want to switch 12V current, a better part would be an N-channel MOSFET, such as AOI518.
Similar to this, but with lower gate capacitance:
Takes very little current to control the Gate, a 74HC595 with its meager 4-6mA output can drive it.
I have a card with 5 74HC595 controlling 32 of them, designed to sink 1A+ of current from LED strips powered by 12V. With their low On resistance (Rds), they won't even get warm:
You can work from "about 0.6 Volts at the gate", so from a 5V arduino high, there'll be a little over 4V through the right resistor from arduino digital out to the gate.
The table and chart of the datasheet show h_FE of between 10 and 100, so to switch "about 1 Amp" from your 12V line, it could need anything from 10 to 100 mA into the gate. That could be a bit much for the arduino, so your best bet would be to add a smaller npn to switch some current into the gate. Most of the "bog standard npn" bipolar transistors are good for more than 12V, and you should choose one which could do "up to 400 mA", though chances are that it will need less in your 1A application.
Next allow about 2V above ground from e of the second transistor (and about 0.7V above ground at c of the second transistor to b of the one in the datasheet), so there'll be about 10V above c of the second transistor. Start off on pinboard trying 10k-Ohm arduino digital out to b of the second transistor, expecting about 0.4mA to be drawn from the arduino. That should put 10 to 40mA into the base of your big transistor, switching 0.1 to 4 Amps. 40mA 10V would be well matched to 250 Ohms from 12V to c of the second transistor. I know that's a huge possible range so try it with a safe load for 12V such as one of your 1A 12V strings. The transistors may be somewhat temperature dependent, and expect to change resistors.
All of the more precise calculation methods are unwise given that all bjt are very temperature dependent. I'd say try it, with an extra npn to give you capability to switch the 'big' 0.1A base current which you might need.
It occurs to me that
i) 250 Ohms might be a bit small for your first try, until you know what h_fe really is
Try 10k from 12V to c of the 2nd npn, expecting the LED line to light visibly but less brightly than normal.
Measure I. Choose your R from that.
ii) it would be prudent to have <0.2 to 3> Ohms power resistors below the e of your big transistor. That way, if h_fe changes (on a cold day for example) you’d get less change to the main I.
You probaby will want to test thoroughly what your driver does over the range of expected temperatures.