I used the same circuit and changed the P Mosfets to PNP TIP147 Transistors and N Channel Mosfets to NPN TIP142 transistor and the circuits doesnt work in the simulator ...
Your losses using BJTs will be much higher than MOSFETs at these current levels, however.
BJT Loss = 0.7V x current
MOSFET Loss = 0.2ohms x I
So if you are carrying anything more than an 1A through a BJT in a regular package, you will need a heatsink or else it will easily exceed 100C or so. MOSFETs are much more efficient at low current levels ("low" as in almost anything you can and will probably build at home).
audioguru
thanks, can i change the tip mentioned to tip 147,142 to handle more amperes ?
dknguyen
yea i got it , you mean the power disspiated in mosfets will be lower than that of BJTs right ?
a question, when i subistuted the mosfets to bjts it wont work because of the current is not enought ? but the voltage is enough to turn them on right ?
more than 0.7 ? why i would then care about the current and use darlington to double the current ?
Darlingtons are like two BJTs connected together so you need even less current to turn them on. MOSFETs and BJTs turn on fundamentally differently, so one circuit that works for one won't work for the other. A MOSFET uses an initial current pulse and then pretty much draws no current after that and only needing the gate voltage to be at a certain level. A BJT continually draws a small amount of current and doesn't care too much about the voltage at it's gate as long as the current is there.
So yeah, it won't work because the current is not enough.
The BJT appears as a diode when it's turned on so that's where the 0.7V comes from (it could be higher making more losses). A MOSFET appears as a tiny resistor. AT low currents the losses from this resistor are less than a diode. At realy high currents the voltage drop across this resistor becomes higher than 0.7V and the BJT becomes more efficient, but those are for realy REALLY REALLY high currents.
If there is not enough current going into the base of a BJT switch, the BJT will limit the current going through it's collector/emitter. A darlington lets you use less base current (or allows more C-E current for the same base current).
Yeah books tend to be steeped in a bunch of equations and graphs. I don't know of a book either except for my text book which is full of numbers and equations.
Of course. But then then base current is doubled so the 1k resistors must be changed to 470 ohms/2W. The base current for the BC547 transistors is also doubled so the 10k resistors should be changed to 4.7k.
The TIP142 and TIP147 each have a max saturation voltage loss of 3V at 10A so the motor will receive a minimum voltage of only 18V and the darlingtons will heat with 60W max.