Switch anode current on with HIGH

Question

Hello. I hope the topic makes sense. I would like to add that this is for educational purposes, and that a dedicated driver IC would probably be better. What I basically want to do is drive a multiplexed 4 digit common anode 7 segment, using some sort of current switching on the anode side and a shift register on the cathode side. I have this circuit working with pnp transistors on the anode side (right on the schematic below). A low signal on pins 2,3,4 or 5 makes the current flow from the emitter to the collector, and the LED light up in the chosen digit (in a pattern controlled by the shift register). Is there any way I can replace the PNP transistors with something that reacts the same way to a low signal? First thought was of course NPN transistors, but they have to go on the cathode side, which is not practical with this display. My plan is to try driving the anodes with a counter such as the CD4022 using the latch signal from the shift register, thus potentially saving 4 pins on my arduino. (sorry about the messy schematic, it is a quick drawing for learning to work with the PCB-layouting tool in KiCAD) 
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johnbeetem · Accepted Answer

You could wire four NPN transistors as emitter followers. In this case the NPN collectors are connected to Vcc and the NPN bases are connected to your counter outputs. You probably don't need resistors, since the current through the NPN base is limited by the amplification of the NPN transistor. The emitters drive the common anodes for the four digits.

An emitter follower does not pull up as high as your PNP transistors: it is limited to the high value of your counter output minus approx 0.7V drop across the NPN B-E junction. However, if your counter outputs are near Vcc = 5V then pulling up to 4V is fine for the LEDs: just adjust the segment resistors.

If your counter outputs don't pull up high enough by themselves, you can add additional pull-up resistors at the NPN bases.

I don't think you're going to save all four pins -- you need one to synchronize the counter to a known state.

dougw · Answer

Hi Niels,

I'm not too clear exactly what the question is so I'll toss a couple of ideas to see if they apply...

If you want to reduce component count, you could use FETs on the anodes to eliminate base resistors.

If you want to change from active low to active high control signals you could use something like a PEMD6 or PUMD6 which can use an internal NPN transistor to invert the signal and drive the internal PNP transistor output. These parts even have the base resistors built-in. PUMD12 has even more resistors built-in. They cost about 39 cents.

If you want to change from active low to active high control signals and consolidate 4 drivers into a single chip, you might try a quad op-amp such as an AD8534 which can source about 100 mA before the output sags 0.4 V. I don't know if .4 V will cause a noticeable difference in brightness, but it should be considered. Cost is just under $2.

Doug

dougw · Answer

Hi Niels,

Another thought....

Could you extend the shift register by 4 bits by cascading a second chip and use these to drive your PNP transistors? - no need for a Johnson counter, signal inversion or more Arduino pins - just make sure only one of the 4 "digit bits" is low.

Doug