Hi folks,
I am about to build an H-Bridge driver for a small DC motor, I know I could probably buy one but its as much about learning as anything else.

I have bought some Darlington power transistors BD679A and BD680, NPN and PNP respectively, that have an hFE of 750 and contain a reverse bias protection diode and base/emitter resistors within the package.

In the past I have always found the minimum base current that would turn a transistor fully on by experimentation but I would like to learn how to calculate this from the data sheet.

I think I understand that a BJT is a current amplifier and that hFE is its gain in specific conditions but I defiantly don’t understand how interpret the datasheet for one.

I have looked at several posts on this subject but still don’t get it …
If I want to use a BJT or Darlington BJT as a switch (saturated right?), in an open collector configuration, where ICE will effectively be limited by the load, how do I determine the required IB, given that the load and thus ICE may change. (Different motors for example.)

How would I calculate the voltage drop across the transistor for a given IC? Should I be thinking in terms of forward bias threshold like a diode or ‘on resistance’ like a FET.

If the answer to the question above is ‘forward bias’, which I suspect is the case, is that forward voltage drop then proportional to IB when the transistor is partially on.

Any input would be much appreciated …
Thanks, Alistair

 

Use the maximum current used through the motor divided by the HFE then multiply by 2 to ensure enough current into the base of the driver transistor.

 

Look at the Fairchild data sheet for these devices. From the BD679A sheet, figure 2 shows Vces vs Ic. Pick your Ic and note your corresponding Vces. Look in the upper right corner and see Ic= 250Ib. Hfe = Ic/Ib, therefor the saturated Hfe is 250.
You will also find a graph for Vbe also.