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I want to design a voltmeter with minimum no of components so the pcb becomes as small as possible.For this purpose i wish to direct drive the seven segment display.

Considering the fact Typically for a standard red coloured 7-segment display, each LED segment can draw about 15 mA to illuminated correctly, so on a 5 volt digital logic circuit, the value of the current limiting resistor would be about 200Ω (5v – 2v)/15mA, or 220Ω to the nearest higher preferred value.

if I am going to use PIC16f676 ,The data sheet says PORTA & PORTC Can Source & Sink 200mA combined ,with maximum source /sink per pin 25mA

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How will i calculate source/sinking current for a seven segment display working in a multiplexed environment( say 4 seven segments ) ??

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Assuming a common-anode display, if you want 15mA per segment you'll need a segment current of 4 * 15 = 60mA per segment driver (7 or 8 needed with decimal point), so the anode current will be as much as 7 or 8 * 60mA = 420mA or 480mA total.

You can forget about driving a multiplexed display that's that crummy (or is required to be that bright, if it's a high brightness type) directly with any micro.

If you use a good (high brightness) display, live with more subdued brightness, and use transistors to drive the digits (anodes in my example) you can drive the segments directly from the micro at perhaps 100mA-150mA total safely, which will give you an average of about 3-5mA per segment.

In such a case, I suggest a common-anode display with the PICs because the outputs are asymmetric and can sink current better than they can source it. You can use BJTs with internal bias resistors or dual MOSFETs to drive the digits, so only a couple parts, plus you should have a resistor in series with each segment.

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If you are using Multiplexed Seven segment displays in Sink mode the current won't exceed 15 mA per pin for the segment port. But for the selection port if they are in source mode then the current sourcing will be at max 8X15 mA i.e 120mA per pin which will damage the pin. So its always good to use transistor drivers for driving 7 segment LEDs.

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In common Anod the power comes from external source. So the MCU port will work as a drain for leds current not as source. That will be safe for MCU .

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  • \$\begingroup\$ You are overlooking what the other two answers explicitly recognized, which is that the display is multiplexed, therefore something switchable has to drive those anodes. \$\endgroup\$ – Chris Stratton Nov 11 '14 at 23:17

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