Transistor Current Ic vs Power Dissipation

Question

I'm a little confused on Transistors and their power dissipation vs current.

If you look at the 2N2222A datasheet. It claims the max Vce is 40 volts.

And that the Ic max current is 1A.

It also says the power dissipation is 625mW.

So, clearly it can't do BOTH 1A & 40V and stay under 625mW.

Does that mean if I want to drive it at 40 volts, then 0.625/40 = 15.6 milliamps would be the most I could put through it?

Or say at +9V you would only get 0.625/9 = ~70mAmp.

Even at 3.3V you only get like 190milliamps.

To get a full Amp out of the transistor, you have to drive it at LESS than 1V.

What am I misunderstanding. I know I see things all over the place talking about the 2n2222 getting 1 amp out of it. But shouldn't it be burning up all the time then?

Am I misunderstanding what the power dissipation spec means? I must be.

Answer 1

Am I misunderstanding what the power dissipation spec means?

There are three basic ways to destroy a transistor: -

• Transistor not conducting but with too much voltage applied between collector and emitter. This causes break-down of the collector-base region.
• Transistor fully conducting like a closed switch but with too much current passing. Usually bonding wires fuse and go open circuit.
• The middle ground where the combination of voltage and current causes the silicon to overheat, go into thermal runaway and melt the bonding wires.

None of the three are much inter-related.

Answer 2

You understand it ok, high current only works as a pulsed switch with Vce(sat) @Ic where Vce(sat)/Ic= Rce equivalent series R of the switch and all switches alike including CMOS where it is called RdsOn. Vce(sat) is usually rated for Ic/Ib=10 not hFE.

Thus

• Vce * Ic = Pd
• and Tjcn rise = Rja* Pd = 0.2'C/mW * Pd for case TO-92 in a PN2222

Answer 3

1. Some 2N2222's are limited to \$V_{CEO}=40\:\textrm{V}\$. But, for example, On Semi's says \$V_{CEO}=50\:\textrm{V}\$. So it does vary a little. Just a note.
2. I'm not sure I've seen one listed with \$I_{C\left(max\right)}=1\:\textrm{A}\$. I think \$I_{C\left(max\right)}=800\:\textrm{mA}\$ isn't uncommon, though. That's not to disagree. That's just again to make the point that different companies may list different maximum values.
3. These are all absolute maximum ratings! That does not mean you get to operate them with the assumption they will work just fine that way. Especially, independently! There are safe operating areas which combine various parameters and you need to be aware of all the limitations as a set of simultaneous limitations.
4. Don't assume. Always go to the datasheet. For example, I assume that all TO-92 packaged devices have \$200\:\frac{^{\circ}\textrm{C}}{\textrm{W}}\$, to ambient. But looking at On Semi's datasheet just now, I see that they show a shocking figure to me that I hadn't noticed on a TO-92 datasheet prior to today: \$325\:\frac{^{\circ}\textrm{C}}{\textrm{W}}\$!! That's my ignorance, or else my bad memory. But either way, it says "read the datasheet" even when you think you know things.

The common theme here is to read the datasheet and look at all the various different ways you can imagine to look at things. And then do that again, when you are in a different state of mind, later. And talk to people, as they often see things you can't see or can't think of. And then read the datasheet again. Plus, mentally play around a bit, too. In time, you know what things you've examined before and what combinations of details are more often needed to be in view than others. Be open to that process, as well.

But as an example now: A BJT transistor will be operated as a switch (\$V_{CE}\ll 1\:\textrm{V}\$) or as part of some kind of amplifier circuit, where it operates somewhere in its active region with \$V_{CE}\ge 1\:\textrm{V}\$. Either way, you need to look at all the parameters. For example, to operate the device as a switch and expecting no better than \$V_{CE}=300\:\textrm{mV}\$ while sinking \$I_C=500\:\textrm{mA}\$, with a base drive of \$I_B=50\:\textrm{mA}\$ and \$V_{BE}=900\:\textrm{mV}\$, then the power dissipated will include dissipation from both base and collector currents of, \$P=900\:\textrm{mV}\cdot 50\:\textrm{mA}+300\:\textrm{mV}\cdot 500\:\textrm{mA} \approx 200\:\textrm{mW}\$. Using On Semi's thermal resistance, a temperature rise of \$\Delta T = 325\:\frac{^{\circ}\textrm{C}}{\textrm{W}}\cdot 200\:\textrm{mW} = 65\: ^{\circ}\textrm{C}\$ over ambient is expected. This may be acceptable. Keep in mind that the package may be surrounded by other packages also dissipating and that it may be even worse.

The above situation assumes continuous operation as a switch. If the duty cycle is 10% and the frequency fast enough that the BJT averages the power well, then you'd expect a tenth as much power and therefore a much smaller rise in temperature, too.

But you should always check these details out for yourself.

Answer 4

I think you are mixing specs in a way they shouldn't be mixed.

Vce is the maximum voltage that can be across the collector and emitter. Generally, this means "the transistor, when off, can hold back up to 40V". However, that is with (ideally) zero current flowing, and thus no power dissipation.

Vcesat is the more important number, when you add current to the mix. If you're going to be passing an Amp through the transistor, you almost certainly want to be driving it in saturation, so Ic * Vcesat + Ib * Vbesat = P will be the actual power dissipation.