How does 2N4401 drive 59mA stepper motor

A while ago I asked a question about driving a stepper motor using a 2N5550 and was advised that it wasn’t really capable/a good idea. I received a good answer but would like a bit more information really so that I can make these decisions for myself in the future.

2N5550
On the 2N5550 transistors PDF it says it can drive up to 600mA (DC) under the maximum ratings (am I interpreting this correctly?):

Ic Max rating

If this is correct, why does the DC Current gain graph only show up to 100mA, is this just because it can handle 600mA but would be almost pointless to actually use it for that because of the low gain at that level?

So this brings me to the 2N4401. The **broken link removed** also says the max collector current can be 600mA (again, am I interpreting this correctly?).

Now the current gain graph on this transistor shows it going up to 300mA however it says “pulsed” in the title of the graph and there doesn’t seem to be a graph for continuous so I’m a little confused why this might be? I suppose driving a stepper motor would come under the category of pulsed though?

So assuming that this is suitable for driving my 259mA stepper motor I would need to supply the base with about (259/150) 1.7mA (259 being the number of mA needed by the stepper motor and 150 being roughly the current gain (derived from the graph)?), is this correct?
I understand that from my other question I should really use a MOSFET as they are better suited to the job, I just have a handful of these 2N4401s kicking about that I thought I could test my code and theory with.
Thanks in advance and I’d appreciate a response to even the little questions I raised like the “am I interpreting this correctly?” ones, just to reassure me that I’m on the right lines (or not).

Hi,

On the 2N5550 transistors PDF it says it can drive up to 600mA (DC) under the maximum ratings (am I interpreting this correctly?):

Click to expand...

The "Absolute maximum ratings" are the edge to kill the device (immediately). Thus your application needs in any case below these ratings. Leave a suitable margin. Mind that at powerup or powerdown your circuit may not work correctly, Short spikes of overvoltage or overcurrent may occur, even in these situations you need to be within "absolute maximum ratings".

Generally a transitor is killed by:
* overvoltage
* overcurrent
* overheat of the silicon = a function of dissipated power

To answer your question:
--> Yes, it can drive 600mA DC.... but only if V_CE is low enough, that it does not overheat.

*******

If this is correct, why does the DC Current gain graph only show up to 100mA,

Click to expand...

"Gain" is only needed with linear operation of the bjt. But with linear operation V_CE usually is relatively high.
With high V_CE you run into overheat. --> it makes for the graph to go ti higher current.

Higher current can be applied for switching operation only, where the bjt is saturated and V_CE is close to zero. Means I_b is much higher than calculated via h_fe. --> No need for knowing exact h_fe in this area.

Now the current gain graph on this transistor shows it going up to 300mA however it says “pulsed”

Click to expand...

To keep power dissipation low....
* with "DC" --> power dissipation P_d = V_CE x I_C (ignoring I_B)
* with "pulsed" --> power dissipation P_d = V_CE x I_C x dutyCycle (ignoring I_B)
This means there are times with high power dissipation (ON) and low power dissipation (OFF). For a duty cycle of 10% for exampke you have 90% of the time no power dissipation. Thus during "ON" time you may dissipate about 10 times of the DC limit. (In detail there may apply other limitations)

I suppose driving a stepper motor would come under the category of pulsed though?

Click to expand...

Only you can know. You may drive it DC style with high power dissipation, or you may drive it pulsed. Your choice.

So assuming that this is suitable for driving my 259mA stepper motor I would need to supply the base with about (259/150) 1.7mA (259 being the number of mA needed by the stepper motor and 150 being roughly the current gain (derived from the graph)?), is this correct?

Click to expand...

No. Your description seems you talk about pulsed operation. But at pulsed = switching operation, you don't calculate with h_fe. You typically drive the base with about 1/10 of the expected collector current. (Need more details)

I understand that from my other question I should really use a MOSFET as they are better suited to the job, I just have a handful of these 2N4401s kicking about that I thought I could test my code and theory with.

Click to expand...

MOSFETS have two benefits:
* Low r_ds_on = low power dissipation even at high switching current
* low gate current = easy to drive.

But honestly: I will build a dicrete stepper motor driver only for learning purpose. I would never design it for a real device.
Especially for your values of voltage and current there are a lot of ready to buy stepper motor drivers available. They have a lot of features you don't have (or it needs extremely high effort).
Some important are: overcurrent protection, overtemperature protection, easy to control, overvoltage protection (stepper EMV)...

Klaus