I have a QM300HA-24 transistor module...it has a C (collector) B (Base) E (Emitter) and then its got a BX pin...What the heck is that pin for.????Do i just use the Base.....As for load testing....i attach a 1ohm 300 watt resistor to the emitter and ground it and pull 13 or so amps....but way does approx 20vdc get fed back into the output of my voltage regulator thats putting out approx 13.5 volts???

 

The BX terminal connects directly to the base of the output transistor. You can use it for driving the output transistor directly, or you could use it with a pull-down resistor to help ensure that the transistor is in cutoff when the B terminal is pulled low.

Usually with an N-ch Darlington, you want the emitter grounded, and the collector sinking current from the load.

You should use a current limiting resistor when sourcing current to the base. Find a datasheet for the transistor to determine maximum base current, and stay below that.

 

this qm300 transistor is my only transistor....i have a lm338 voltage regulator driving its base with 14.4vdc or so...max base current is 16a i think........Correct me if Im wrong here===>so i can drive either the B or the BX with my lm338 volt reg...if i drive the B then i can ground the BX with a pull down resistor to prevent the 19-23vdc from being fed back into my volt regulators output which causes that 19-23v to show up on the E which is bad for my application. My E is supposed to have 12-14vdc and 20V is bad

 

Post a schematic of your circuit.

I really don't know what you are trying to do.

There is a datasheet available for your transistor here: http://www.datasheetcatalog.com/datasheets_pdf/Q/M/3/0/QM300HA-24.shtml

 

Here it is pretty simple...oh also there is an internal diode that goes from the emitter to the collector of that qm300 module..its not shown in the picture...for some reason when i connect the load resistor the output on the emitter shoots up to 20 or so volts....when the bx is left un connected...output voltages are not exact it just for information they are in the ballpark

 

If your input to the bridge is 24VAC, you will have approximately 34VDC-36VDC across your output filter capacitor with no load.

You need to limit the current to the base of the transistor. Gain of the transistor is 75 when saturated, so Ib=Ic/75.
That means you need 1/75th of the collector current flowing through the base as the desired collector current.

Since your load is a 0.15 Ohm resistor, and your supply is around 34v, what current do you expect through the load?

You need to have the transistor's emitter connected to ground.
Then connect the load resistor between the collector and the filter capacitor.

34v/0.15 Ohms = 226.7 Amperes through your load.
226.7/75 = 3.0222... Amperes current is needed through the transistor base.
Vbe will be about 2v when the transistor is saturated.
So, (34v-2v)/3A = 32/3 = 10.66 Ohms for a base resistor. You could round that down to 10 Ohms.
Since power = voltage x current, or voltage squared / resistance, power rating for the 10 Ohm resistor is 3,072 Watts.
Double it for reliability, so a 6,000 Watt resistor.

Alternatively, you can use your LM338 as a current regulator to limit the base current.
When used as a current regulator, you connect the VOUT terminal to the ADJ terminal using a resistor (call it R1), and take the current from the ADJ terminal.
Iout = 1.25/R1
Conversely,
R1 = 1.25/DesiredCurrent
So,
R1 = 1.25/3 = 0.416 Ohms
You will need a very large heat sink on your LM338. More like a big block of "dry ice".

Oh, you'll be dissipating about 7.7kW power in your load. It may disappear with a loud bang and bright flash when you power it up. Better have it in a very large container of water.

 

.15 was a guess...maybe .25 or .5 for load testing...the most i ever expect to pull thru the emitter was 1200 watts or so maybe a 100 amps. Well i was using the 13.5v off the emitter to power several 12v related items and maybe a car charger too....i have ZERO experience in running the load off the collector side...ive always used the emitter leg of transistors in the past to power various things...

 

This is more like how to connect it up:

The Darlington sinks current from the load instead of sourcing current.