My usual focus is in digital electronics, and my BJT knowledge is fuzzy. I'm working on a design using the Intersil ICM7245 16-segment display driver. It recommends this circuit to drive the led segments with a higher drive current.


I'm having trouble figuring out exactly how this works, namely, so I can choose compatible BJTs for both the high and low sides, assuming my LEDs are Vf = 2v, If = 20mA.

2N2219 = NPN. 2N6034 = PNP Darlington

I duplicated it in Orcad/Pspice, and the results don't seem like it should work. I think I can just choose some BJTs that match, but I want to make sure before I get the PCBs ordered.


Edit: To clarify my question, I understand that each transistor has a gain (hFE) of ~100, hence 1mA->100mA, and 14mA->1.4A. I'm just not clear on how they made sure the bases would have those currents, and how much I can change the 14ohm resistor to keep my LEDs at ~20mA.


1 Answer 1


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.

  • \$\begingroup\$ Thanks! A terrible example is a valid answer. I do have a 12v supply available, but I believe I would then need to buffer the seg output, since its voltage would have to be at least ~6.2v? Changing Vcc to 12v makes my simulation appear to work. Now I have stuff to play with. ------- I'm updating a design that drives 8 16-segment displays. It's apparently bright enough already with ~23mA 1/8 of the time. It currently uses mosfets and oodles of components (~100!) which I'm hoping to reduce the total quantity of. Looks like I'll likely be sticking with the mosfets. Thanks again. \$\endgroup\$
    – brownd
    Jan 8, 2017 at 8:40
  • \$\begingroup\$ @brownd 8x16=128 segments total! Must be high efficiency LED types if \$23\:\textrm{mA}\$ divided by 8 (about \$3\:\textrm{mA}\$ on average) is giving you what you need. That's a good thing, though. Makes the power supply easier. To support all the segments your power supply must support \$\approx 400\:\textrm{mA}\$. Nice, as that isn't at all hard to achieve. \$\endgroup\$
    – jonk
    Jan 9, 2017 at 21:08

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