So, I am working on a small project and forgot to order PNP transistors to switch the power to two common-anode 7-segment displays. Like this, for example:

enter image description here

All I have in hand at the moment is a series of NPN transistors but I can't think of a proper way to use them in a similar way as above.

I thought to use a simple HEX inverter like so:

enter image description here

But the supply needs to pass at least ~150mA or so for 7 segments and the 74LS04 I have is not rated for such current drive.

Worse case I can do the above and multiplex the 7 segments to stay within the current ratings for the HEX inverter, but I was first trying to come up with a solution using an NPN transistor alone.

Any suggestions?

  • \$\begingroup\$ Do you want to switch power between displays or do you want to switch both displays on or off at the same time? \$\endgroup\$
    – EM Fields
    Jul 26, 2014 at 16:36
  • \$\begingroup\$ I was going to make this twice to drive one high and one low, but I did not think about making a single circuit that does both. \$\endgroup\$
    – sherrellbc
    Jul 26, 2014 at 16:59
  • \$\begingroup\$ I'm still confused... If your displays are common cathode and you're driving the segments high to illuminate them, why would you not want to use an NPN to switch the common cathodes to GND when it's time to turn the display ON? \$\endgroup\$
    – EM Fields
    Jul 26, 2014 at 17:52
  • \$\begingroup\$ My mistake, thanks for catching that. The device is actually common anode and I drive the segments by supply low cathode voltage. \$\endgroup\$
    – sherrellbc
    Jul 26, 2014 at 18:03
  • \$\begingroup\$ OK. What are you using to drive the segments? \$\endgroup\$
    – EM Fields
    Jul 26, 2014 at 18:15

1 Answer 1


You could use an emitter follower and drive it from a 5V input. It will drop one Vbe drop, about 0.7V, so the power dissipation at 150mA would be about 100mW, which should be okay for most transistors over a reasonable temperature range. Use a pull-up resistor on the base if you don't have something like an HCMOS driver.


simulate this circuit – Schematic created using CircuitLab

The 100R base resistor is sort-of optional, but in some cases the emitter follower can oscillate at VHF frequencies if you give a really stiff base voltage, and the resistor will prevent that.

  • \$\begingroup\$ I originally thought of this circuit but did not think it would work. Actually, it's a direct replacement of the PNP with an NPN from the circuit above. I suppose my experience with transistors has mostly been with either the emitter grounded (NPN) or tied to Vcc (PNP). Why do we observe a Vbe drop as opposed to a Vce (~200mV) drop? \$\endgroup\$
    – sherrellbc
    Jul 26, 2014 at 17:06
  • \$\begingroup\$ It's not quite a direct replacement- it doesn't invert. If you want a Vce drop you have to drive it with more than 5V or whatever supplies the collectors to drive the LEDs (completely feasible- for example a series diode or two in the collector(s)) you still need the resistor (it won't oscillate when its saturated but the base current needs to be limited). \$\endgroup\$ Jul 26, 2014 at 17:15

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