Power Darlington base drive requirements

Thread Starter

Denesius

Joined Feb 5, 2014
106
I picked up half a dozen devices at a garage sale, with the thought of a good opportunity to make a battery spot welder - data sheet attached. So I have an Arduino circuit that has programmable pulses (5v, 100ms duration, 50ms pause in between), a bridge rectifier rated at 300A, and a big power transformer that puts out 3.8v at 400amps. I'm not really versed in IGBT/Power transistor circuit design, but I do know that IGBT are voltage driven, vs a transistor's current. So I thought I'd do a bench test on one of the bricks before putting it all together. The data sheet shows the nominal Vbe as 3.5v, max 7v - so 5v should be perfect. So I put a test load on the device and powered the base via a 5v, current limited power supply. Ibe current was a surprising 3.5 amps. The datasheet shows the gain (hFE) as 100.
Am I to understand that expecting to switch a 300A current load, I have to supply 3 amps thru the base - or am I missing something here?
If so, why the low gain on a Darlington design?
Finally, would an IGBT work better here, with direct drive off the Arduino, or will it need a driver nonetheless?
 

Attachments

dl324

Joined Mar 30, 2015
11,315
What makes you think this is an IGBT? The datasheet only mentions Darlington. Circuit below:
clipimage.jpg
Am I to understand that expecting to switch a 300A current load, I have to supply 3 amps thru the base - or am I missing something here?
clipimage.jpg
Given a typical beta of 100 at a current of 300A, yes, you would need 3A of base current.
If so, why the low gain on a Darlington design?
Beta for that transistor falls off sharply after about 100A.
Finally, would an IGBT work better here, with direct drive off the Arduino, or will it need a driver nonetheless?
We need to know more about how it will be used. Switching frequency would be important. Arduino can only source/sink 20-40mA. That might not be sufficient for high speed switching.
 

BobTPH

Joined Jun 5, 2013
2,505
Show us your schematic. It sounds like you applied 5V directly to the base. That is essentially putting 5V across two diodes to ground. You need a resistor to limit current.

Bob
 

Thread Starter

Denesius

Joined Feb 5, 2014
106
Thank you all - especially Dennis for the detailed explanation. I never thought this device was an IGBT. They were $3 each so I thought I could substitute for a much more expensive device, not realizing the base current requirement. Without a datasheet in hand, and a 3 stage Darlington design marked on the device, I hoped the current requirement would be trivial. I guessed wrong!
Any explanation of what the BX pin is for? It's not addressed in the datasheet....
 

Papabravo

Joined Feb 24, 2006
14,253
Thank you all - especially Dennis for the detailed explanation. I never thought this device was an IGBT. They were $3 each so I thought I could substitute for a much more expensive device, not realizing the base current requirement. Without a datasheet in hand, and a 3 stage Darlington design marked on the device, I hoped the current requirement would be trivial. I guessed wrong!
Any explanation of what the BX pin is for? It's not addressed in the datasheet....
BX is the second emitter in the cascade, and the base of the final transistor in the cascade. It is brought out for convenience and/or monitoring. I've never had a need to use such an animal, but it might be possible to simulate one to get some idea of what you might be dealing with.
 
Top