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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

So on the 2N5550 transistors data sheet 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?

hFE

2N4401

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

Ic Max rating

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?

hFe pulsed

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).

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    \$\begingroup\$ I don't see a flaw in your reasoning, just know that you want to have the bjts in the saturation region, so aim for a higer base current. Check the datasheet, some 20mA should be enough. great question though, +1 \$\endgroup\$ Aug 15, 2014 at 9:58

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For switching purposes you should be using the saturation voltage guarantees rather than the DC current gain. For example, the 2N4401 is guaranteed at Ic/Ib = 10, so a base current of ~26mA will typically result in 220mV or so drop at 260mA (eyeballing the below plot).

enter image description here

The reason they say "pulsed" is that they are not accounting for the self-heating of the transistor. 26mA is quite a bit of base current, and the 220mV (typical!) drop will result in about 10 or 20 degrees C heating for the transistor (depending on the package).

If this is a hobbyist thing, you can probably get away with forced \$\beta\$ = 20-30 or so, so a base current of more like 9mA rather than the \$\beta\$ = 10 shown in the plot. I usually use \$\beta\$ = 20 at Ic <= 100mA, which might be a bit on the high side for a microcontroller to supply directly in your case (13mA of base current).

Edit: A "forced beta" of, say, 20 can be achieved as follows. Suppose that the collector current is limited the load resistance to say 200mA. No matter how well the transistor turns on it will never exceed about 200mA (even a dead short). You use a base resistor so the base current is 20mA. The ratio of Ic/Ib is now 20, and we can call that a "forced beta". The gain of the transistor at 200mA would be more like 150, but that's only when Vce is high enough that the gain determines the collector current. The datasheet specifies it at Vce = 5V, which would be a disaster for a switching application.

enter image description here

You can "force" the ratio of Ic/Ib to be as low as you like, but the upper limit is constrained by the transistor characteristics. When you're designing, ideally you'd like to eliminate the exact transistor characteristics, so that the circuit functions provided the transistor meets the minimum/maximum guarantees. The gain should never determine the collector current in a saturated switching application.

A small MOSFET with logic-level gate might be a better choice when you get much above about 100mA. As you can see, the performance starts to go a bit sour above approximately that current, even on the 2N4401.

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    \$\begingroup\$ Thanks for the clear explanation Spehro! The only thing I think I'm unsure of is how can I "get away with forced β = 20-30", I mean, if the gain (β) is 10 how can I "force" it to be something else? Surely that's a fixed characteristic of the transistor.. isn't it? \$\endgroup\$
    – OdinX
    Aug 15, 2014 at 11:47
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    \$\begingroup\$ Excellent question, I'll edit my answer to cover that. \$\endgroup\$ Aug 15, 2014 at 11:56
  • \$\begingroup\$ Thank you for the update. You mention that if Ic is limited to 200mA and you put 20mA into the base β is now 20, isn't it still 10 (200mA/20mA = 10)? I understand most of what you've written I think. However, sorry to be a pain, but I didn't understand what you were saying about Vce in the last two sentences before the image. Specifically, "when Vce is high enough that the gain determines the collector current" and why would it "be a disaster for a switching application"? \$\endgroup\$
    – OdinX
    Aug 15, 2014 at 12:27
  • \$\begingroup\$ Yes, forced beta is 10 in my example. I'm saying if the transistor is operated in the region with Vce = 5V Ic = 260mA then the transistor will get very hot and probably die, plus there won't be much voltage across the load so nothing will work and the transistor will fail. That's pretty much a disaster. ;-) \$\endgroup\$ Aug 15, 2014 at 12:56
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    \$\begingroup\$ Okay, it will work, it just isn't the best. If you want to give it, say, 15mA or 20mA of base current, you can use another transistor to drive the base (say a PNP), and all should be well. A micro might not be able to reliably source 15mA-20mA with a predictable minimum output voltage so there is danger of not saturating the transistor and causing overheating. The transistor should be fine. Its also bad karma to switch inductive loads that are too close to the maximum current capability without verifying the (missing) Safe Operating Area curves but that's whole 'nuther discussion. \$\endgroup\$ Aug 15, 2014 at 13:49

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