I attach a circuit diagram.



I'd like to know, can one simply replace TIP31C with 2N3055 and expect this to work?

Also, regarding the diode, is there such a thing as a 10A diode? Can I take 10 1A diodes in parallel as a substitude?

Reference to a resource where one can look this sort of thing up would be much appreciated.

 

Hi,This is one inefficient way of controling motor speed....Rite, although the 2N3055 has a rated Ic of 15 amps, it would be pushing it to control that 10 amp motor......For continus opperation from this circuit, you wolud need a couple (maybe 3) 2N3055's in parallel, with an emiter resistor (0.2 ohms) on each device, along with a LARGE heatsink.....At slower speeds, the transistors will dissipate a lot of heat.... If you want to control a motor of this size, you would be better of looking into PWM (pulse width modulation) control, as in this type of control the output device is on or off, therefore keeping dissipation down to a minimum....Daniel.

 

not sure of the specs of the driver transistor, but it must be capable of supplying the extra base current required for the 3055's as well.

That 1N4148 will be toast as well

 

what is the wattage of1K&10K

 

Yes that's rite.... You will need to drive the 3055's with a TIP41, and then drive that with a BFY51.... Don't foreget the voltage drop through the devices....Diodes in parallel...."not a good idea" they will never be matched exactly, and thay will have very slightly differing forward voltages, causing extra dissipation from some of the diodes...Daniel.

 

what is the wattage of1K&10K

12V / 10,000\(\Omega\) = 1.2mA. 12V * 1.2mA = 14.4mW

Hobbybuilder, have you thought about using pulse width modulation (PWM)?

 

Diodes are typically rated to peak inverse voltages rather than current(10A diode). However, u need to look at the peak power the diode can dissipate- (current squared times the bulk resistance of the diode).Say, if the bulk resistance of the diode is 7 ohm,then for 10A, power diss = 10*10*7= 700W. A peek into the datasheet of the diode wud give u the above rating. If the diode icannot handle that power, it is likely to break down.

 

Yes that's rite.... You will need to drive the 3055's with a TIP41, and then drive that with a BFY51.... Don't foreget the voltage drop through the devices....Diodes in parallel...."not a good idea" they will never be matched exactly, and thay will have very slightly differing forward voltages, causing extra dissipation from some of the diodes...Daniel.

Actually, it's way worse than that. As one diode receives more current, its voltage drop will reduce which will cause it to pull more of the current. The other one will just sit there pulling less and less current until the first blows. Then the second one is all alone, and will blow soon after.

 

The max saturation voltage loss of a 2N3055 transistor when it conducts 10A is 3.0V when its base current is a whopping 3.3A. Then the TIP41 driver transistor has a max saturation voltage loss of about 1V when its base current is 300mA, then the BFY51 pre-driver transistor has a max saturation voltage loss of 1V when its base current is 20ma.
You still need an additional transistor to supply the 20mA.

The max voltage to the motor will be only a few volts.