1. What are the benefits of using a mosfet compared to a bjt transistor?
In this particular case, the major advantage is that a voltage on the gate of a MOSFET controls the current through the drain-source. With an NPN transistor, it's a
current through the
base that controls the current from collector to emitter. Voltage control is usually easier than current control.
In the general case, the advantage of MOSFETs is that when used as a switch, they have a lower resistance than an NPN. They are more power efficient, when used as a switch. But in this case we're not using the MOSFET as a switch. We're using it as a variable resistor. So that advantage is not useful to us here.
Also, NPNs can go into thermal runaway. NPNs conduct better the hotter they get. With MOSFETs it's the opposite, the hotter they get the worse they work. So MOSFETs can't go into thermal runaway nearly as easily.
> I have a few 2N3055, is it a bad idea to use them?
First, you need to know that there is a minimum voltage drop across an NPN transistor operating in linear mode of about 0.7 volts. This will put a limit on the maximum current that the dummy load can sink. However, the limit is pretty high. Something in the neighborhood of: (1.5 * input voltage) amps.
Second, NPN transistors can go into thermal runaway as mentioned before. This probably won't be a problem with the circuit Dave used, because the opamp will automatically throttle back the NPN if it goes over-current. But just be aware that thermal runaway can happen.
Third, it depends on how you're driving the base of the 2n3055. If there's X amount of current flowing through the collector of an NPN, then you need (X / beta) flowing through the base. The 2n3055 has a beta anywhere between 20 and 70. So in the very worst case, you will need to supply X / 20 amps of current into the base. If the opamp you use to drive the base of the 2n3055 can supply that much current, then no problem. But if it can't supply that much current...
Assuming a "typical" opamp with 20mA of output current capability, that would mean your dummy load would be able to handle a maximum of 20mA * 20 = 4000 mA = 4 amps. This is a lot less than the 15A maximum that the 2n3055 can handle.
To ensure the full 15A limit can be reached, you would need an opamp that can output 15A / 20 = 750 mA. There are opamps that can supply that much current, but they're not very common and may not be cheap.
BTW, you can go back and do all of the above math with 70 instead of 20 to see what kind of base current you would need in the "perfect world" case. But I don't recommend you count on the world being perfect.
(There's a relatively simple base driving circuit that you can make with a resistor and small MOSFET that will greatly amplify the output current of your opamp. But you're such a n00b I don't want to confuse you with it just now. It looks basically like this, except with the transistors swapped:
http://www.neufeld.newton.ks.us/files/electronics/mosfet-motor/MOSFET-motor-driver.png )
> 2. How to calculate the heat dissipation for the transistor? The output voltage of the PSU to be tested (Vout) - Voltage drop across the resistor (Vr) x the set current going through the resistor?
The usual answer is "P = IV", but in this case we should probably use the equivalent but alternate form:
P = I^2*R
or
Watts = (Current through component)^2 * Resistance of component
Now, the current you know. You just decide you want 1.3 A constant current, and that's that. The constant-current load will adjust its own resistance until that constant current is reached.
The resistance you will have to calculate, based on the input voltage and constant current. Example: You have 12V and set a constant current of 400 mA (= 0.4 A). Use V = IR to calculate the equivalent resistance:
V = IR
V/I = R
12/0.4 = R
R = 30 ohms
Now use P = I^2*R:
P = 0.4^2 * 12
P = 1.92W
This is much less than the ~100 watts than the 2n3055 is rated for.
But don't get a false sense of security. Remember that the equation depends on current
squared. As the current goes up, wattage also goes up
very quickly. Another example:
Constant current = 4 A; Vin = 24 volts; R = V / I so R = 6; P = I^2*R so P = 16 * 6 = 96 watts! You're getting close to the power dissipation limit of your 2n3055.
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