Electrical – Small signal equivalent circuit BJT question

The BJT power dissipation is minimal - any current it takes comes through R2 which is 15kohms. Supply is 3.3V and if half appeared across the BJT (max power transfer) it would dissipate 182uw!

The easiest way to picture the action of the BJT and D1 is by considering that the transistor cannot turn on too much because if it did, D1 won't have enough voltage across it to conduct current to the BJT's base and turn the transistor on hence, assume that the base is only very slightly forward biased at about 0.4V and therefore D1 must have 2.4V on its cathode.

And here's where you are going to hit problems with the low supply voltage....

On the original example it ran from 9V and you could assume the collector voltage would be at 2.4V + (say) 2V higher (due to R1 volt-drop). With 4.4V on the collector you can calculate collector current (\$\frac{9v - 4.4v}{15k\Omega}\$ = 307uA). And from this you could make a reasonable assumption about base current being 30 times lower\$^1\$ at 10uA. 10uA through a 220kohm resistor produces 2.2V across it and this isn't far away from the 2V I assumed by sticking my finger in the air.

But, on the reduced supply voltage design there isn't the headroom to assume the collector is 2V higher - you can barely assume 0.5V and this is making me think that the base current will be more like 2uA with a collector current of more like 60uA. This 60uA will drop 0.9V across the 15k resistor and clearly this doesn't figure because it would only leave 2.4V on the collector and not enough.

Therefore base current must be even smaller and I have doubts about the noise it will produce if any BUT good luck.

\$^1\$ Quite a few BJTs have poorer current gain at quite small collector currents so assuming 30 for \$H_{FE}\$ isn't unreasonable.

Looks like a terrible example circuit.

Just taking them at their word, at \$100\:\textrm{mA}\$, the LED is going to drop more than \$2\:\textrm{V}\$ already. The \$14\:\Omega\$ resistor will drop another \$1.4\:\textrm{V}\$. That's \$3.4\:\textrm{V}\$ (and more) and the transistors aren't even accounted for, yet.

The 2N2219 is being driven as part of a Darlington (part inside the IC, part outside), so the net drop must be at least two \$V_{BE}\$, plus a little for the internal resistor. At \$100\:\textrm{mA}\$ out the emitter, call it at least \$1.5\:\textrm{V}\$ there. (I don't see a high side boost circuit for the internal BJT base drive in their block diagram.) And the 2N6034 is a Darlington. So it doesn't saturate, either, and will require a \$V_{CE}\approx 1.5-2.0\:\textrm{V}\$ (The guarantee at \$I_C=2\:\textrm{A}\$ and \$I_B=8\:\textrm{mA}\$ seems to be \$V_{CE}= 2.0\:\textrm{V}\$ -- see this datasheet.)

They got one thing right. You can expect to achieve \$\beta\approx 100\$ in the 2N2219, since it's \$V_{CE}>1\:\textrm{V}\$. And the 2N6034, being a Darlington, can also achieve \$\beta\approx 100\$, too.

But there's no headroom left over from \$3.3\:\textrm{V}\$ after you subtract away the transistor drops. I don't know why it would work as the diagram says it can. (You'd probably need a \$6\:\textrm{V}\$ rail to make that thing work at their max pulsed current rating.)

Regarding your own use, keep in mind that since you will likely be multiplexing the LEDs, you will want to multiply up the drive current. If you are multiplexing x4, for example, then a \$20\:\textrm{mA}\$ continuous equivalent becomes an \$80\:\textrm{mA}\$ pulsed drive to operate the LED at 25% duty cycle. So your \$V_F=2\:\textrm{V}\$ probably isn't right and must be adjusted upwards a bit and also accounted for.