choosing a suitable transistor

For on off switching, the maximum collector current spec on data sheets is a poor guide, except that it is way too high.

Lists the maximum DC collector current at 3 amps.

For low voltage drop switching I usually look for a transistor with a gain at the load current that is at least half the peak gain shown on the DC current gain curve, at some low DC collector voltage.

The above data sheet indicates that this half peak gain current (with

4 volts collector to emitter, on page 2) is about 1.5 amps.

I also check the saturation voltage curves for a forced gain of 10 and make sure that the maximum voltage drop is below about 200 mV. This is about 1.2 amps for TIP31. So, though it has a higher current capability than you need, it is not grossly oversize, unless fast switching is also important, because the over sized die has more capacitance and stored charge than necessary.

Taking a look at 2N4401,

with a maximum collector current rating of 800 mA,

But the gain curve falls to half its peak value at about 450 mA (with

5 volts collector to emitter). Checking the saturation voltage, it is about 300 mA for a saturation voltage of .2 V and a forced gain of 10 (with 30 mA of base current).

If you can spare this much base current, this transistor is just about as small as you can go for a saturated load of 400 mA, with a reasonable base drive and saturation voltage.

A really fine saturated switch for this sort of current would be one of the high gain, low saturation voltage devices from Zetex. You haven't mentioned the off state voltage the transistor must withstand, but assuming 40 volts, take a look at ZTX1051A (Digikey sells these, but they cost as much as the TIP31).

Half peak current gain is about 5 amps and the gain is still over 200. The saturation voltage at 400 mA is only about .05 volts with a forced gain of 100 (4 mA of base current). All this in a tiny to-92 package.

I love this line.

I'm hoping I can use the 2N4401, as it's a real cheapy. My base current will be 200mA, it's coming from a 555 timer.

I read at some website

that a transistor is fine if:

1) Ic > your load current (this one has 600mA Ic, check.) 2) hFE > 5 * Lc/Supply Current (5 * 400/200 = 10)

Is this sound advice?

Thanks a lot

-sam

If you limit the base current to 40 mA with a suitable resistor, it should do fine.

Since the Ic max specs are often exaggerated, I would have said that Ic max should be 3 or more times the loads current, for good saturation voltage. The 2N4401 is good here, with an IC max of 4 times your 200 mA load current.

That is 5*Load current/chip output current. This just says that the minimum (linear, not saturated) gain is at least 5 times more than needed to boost the maximum chip output current up to the load current. This allows the gain to fall to 20 % of the linear minimum when the transistor saturates, to get a low saturation on state voltage drop.

Pretty reasonable.

Are the manufactors cheating on a regular basis ..?

Manufacturers don't cheat, they just tell the truth in different ways.

As an example, look at the data sheet for the ZTX1051 above. Under the Absolute Maximum ratings, it lists maximum power dissipation as 1 watt at 25C.

That's the highest claimed AbsMax Pd for a TO92 I've seen. While I believe in sunshine just like everyone else, I also know a TO92 package is gonna be hot as a $3 pistol at the firing range if it goes above a half watt without a heat sink, and you realistically shouldn't put more than 1/4W steady state on a transistor in that package without a heat sink at a normal range of ambient temperatures without forced air if you want a reliable circuit.

But I'm sure the manufacturer's claim is definitely true. If you keep the package surface temp at 25C, it will handle 1 watt of power. To do that, you'll have to immerse the device in flowing ultra-low conductivity deionized water, but there it is.

Theta(Junction-to-Case) of a TO92 package is about 83 degrees C per watt. Your TO92 immersed in 25C deionized water will have a junction temp of (25C + 1.0W * 83C/W) = 108C, which is fine. Not even close to being realistic, but fine. Of course, Theta(Junction-to-Ambient) of a TO92 is 200C/W. And most TO92 heat sinks don't do much of anything anyway -- the mating surface of the transistor is too small, and is plastic instead of metal.

Mr. Popelish's choice for the OP is best (I gotta get me some of those on my next DigiKey order). With a Vce of 50mV, it'll barely get warm at 400mA. Assuming I had the base current available, though, I might have gone with the TIP31. It's a gumball part (low cost multiple sources), it can easily handle the current, and the Theta(JA) of a TO220 is only 60C/W or so.

Overkill is a rather strong word, especially if you're not building hundreds or thousands of 'em. If you avoid the obvious things that decrease circuit reliability, you can usually get it done and go home on time without having to worry about fixing it. I'd call it overbuilding or bulletproofing.

Cheers Chris

(snip)

No. It is just not practical to make use of transistors at the maximum possible current they can survive. That spec is a limit to be avoided, a boundary not to be crossed.

You also should not operate 1 watt resistors at 1 watt, if you expect reliability and stability from them.

Your last paragraph begs something else, entirely. At least, it appears to ask for you to be "taught to fish, rather than be given a fish to eat," so to speak.

Understanding how to set part values and otherwise using transistors for switching (and just staying within the domain of bipolar transistors [BJT's] here), involves knowing more than what you suggest above. It means understanding a little about how they work and knowing at least some of the many possible topologies -- hopefully, in fact, being able to arrive at some of them on your own imagination, in fact.

So rather than answer your more immediate questions, I'm going to point out some directions to look.

BJT's are usually "looked at" as inverting. When you use one BJT as a switch and actively drive it, the low-voltage drop of its emitter to collector is usually a condition caused by creating a significant (in the neighborhood of 0.8V) voltage difference between its base and emitter. So "pulling down" the base of a PNP tends to "pull up" its collector; or "pulling up" the base of an NPN tends to "pull down" its collector. However, there are other cases -- such as "pulling down" on its emitter (with base tied at some V) also pulling down the collector.

Here are some non-inverting voltage translation circuits (which amount to switches):

Vcc2

| | |

\\

R5 / Vcc2

\\ |

/ |

| |

| |

| |\\c

| |

| '-,

|/c Q3 |

|>e \\

| / LOAD

| \\

| /

\\ |

R4 / |

\\ gnd

/ | |

gnd

If you learn to understand these reasonably well, you will pretty much be able to do your own calculations of resistor values.

BJT's also include some effective capacitance between base and collector and base and emitter. These values are often roughly in the area of 10pF, or so, for general purpose BJTs. How these affect speed depends on the resistor values (you can often use lower values to help out on speed, at the expense of operating current), but in most cases where one millisecond is considered "fast enough," you won't need to worry about that aspect. If very short switching times are required, you will need to worry about this capacitance and provide a means to overwhelm it.

Jon

Hmm, if you saturate a 2n2222 that will work.

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