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I want to use 1-inch & 2-inch 7-segment display in a digital clock based circuit(six 2-inch display in a row & six 1-inch display in another row). I am using 74hc164(serial in parallel out shift register ic) for controlling 1-inch display.Data, clock and reset pin of 74hc164 is controlled by PIC microcontroller. QH pin(13th pin) of the ic is given to the data pin of the next shift ic and goes on. Since the forward voltage of 2-inch display is 9.6v, 74hc164 cannot be used(6v max).I got CD4015 with similar properties of 74hc164 and works well on 12v.

My question is, how can I connect the clock and data pins of 74hc164 to cd4015 since the operating voltage is different?.

please suggest a solution for this...

Thanks and regards, Nikhil

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    \$\begingroup\$ Personally I would say don't use either - use a constant current sink LED driver shift register like the STP16CP05 or similar and common anode LEDs - you don't care about the LED voltages then, each LED can have a different voltage if you want... \$\endgroup\$
    – Majenko
    Dec 31, 2014 at 11:21
  • \$\begingroup\$ thanks for your valuable comment... I googled through internet and I got few constant current sink led driver ic. In this design, is there any need of series resistors between segment pins and the driver ic since we can easily control the current through R-Ext pin of the ic?. \$\endgroup\$
    – Nikhil
    Jan 2, 2015 at 9:55
  • \$\begingroup\$ No, all you need is that single R-Ext pin - the LEDs are driven direct. That's the beauty of these kind of chips - they make the design so simple. \$\endgroup\$
    – Majenko
    Jan 2, 2015 at 10:30

2 Answers 2

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I agree with Majenko there are better alternatives but in answer to the question you need an interface circuit that will change a 0 -5V signal into a 0 -12V signal. This can be easily done with a few discrete components.

This is the sort of circuit (but by no means the only one) you could use. A google search will throw up many others including the use of MOSFETS, open collector gates etc.

enter image description here

Q1 and Q2 are just about any general purpose small signal type transistors - Q1 is an NPN type and Q2 is a PNP type. The resistor values are not critical so anything in the 10s of K would be suitable.

R1 limits the current to the base of Q1. An input voltage above 0.6V turns Q1 ON. This pulls current through the emitter-base of Q2 through R3 (which limits the base current of Q2). Q2 is turned ON and current flows through Q2 and R4 raising the voltage across R4 to nearly 12V.

With Q1 input around 0V, Q1 is turned OFF and no current can flow through R!,R2. The resistor R2 ensures that Q2 is fully turned OFF. With no current through Q2 there is no voltage across R4 and so the output will be 0V.

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  • \$\begingroup\$ Thanks for your comment... After posting my question I am waiting for an answer like yours. But my PIC microcontroller is running at a speed of 1/5th of a microsecond. But, is these transistors are capable for switching at this speed?. \$\endgroup\$
    – Nikhil
    Jan 2, 2015 at 10:12
  • \$\begingroup\$ @Nikhil No problem, even general purpose transistors are pretty fast. I would point out that its the only the processor that is running at that speed. The port outputs will much slower anyway. You're probably multiplexing at a few hundred Hertz for the display. \$\endgroup\$
    – JIm Dearden
    Jan 2, 2015 at 10:33
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The easiest solution is to use a CD40109 low-to-high level shifter- this is exactly what it is designed for, and it's reasonably fast. They're about 50 cents in singles.

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