Interfacing of 7 segment...

[shaheen]

Which of the following circuit is correct?
I want to interface a common anode 7 segment display with microcontrolar (5v logic).
Thanks.

 

[(*steve*)]

Figure 2 is generally what you'd use because you want the same current for each segment.

Remember that the logic for the segments is reversed! They are pulled low to turn them on.

 

[alenco]

The second circuit will work better. The first will cause the brightness of the segments to dim as more segments are turned, so a '1' will be bright, but an '8' will be very dim.

 

[shaheen]

Figure 2 is generally what you'd use because you want the same current for each segment.

Remember that the logic for the segments is reversed! They are pulled low to turn them on.

Thanks.
I think 330ohm resistor is enough for supply of 5v, and if there are 12v (rather than 5v) supply 1 killo ohm resistor is enough.

 

[alenco]

Beware of connecting the display anode pin to higher than +5V, this can result in the PIC chip output pins receiving higher than 5V on the output pins, something it may not like. It can cause current flow into the pins that disrupts the PIC's power scheme or even trigger a latch-up condition that can destroy the PIC. LEDs typically have a 2 to 3 volt forward voltage drop through them, so a +12V anode means the cathodes would have 9 to 10 volts on them. The PIC output pin drivers may have the equivalent of a diode anode on the pin with it's cathode going to the +5V supply, so 9 volts from the LED cathode could produce 8 volts or so on the PICs internal +5V rails. Normally a 12V connected display would use a special driver circuit on the PIC output pins to avoid this. As simple as an NPN transistor and two resistors per LED segment. Or another IC.

 

[shaheen]

Beware of connecting the display anode pin to higher than +5V, this can result in the PIC chip output pins receiving higher than 5V on the output pins, something it may not like. It can cause current flow into the pins that disrupts the PIC's power scheme or even trigger a latch-up condition that can destroy the PIC. LEDs typically have a 2 to 3 volt forward voltage drop through them, so a +12V anode means the cathodes would have 9 to 10 volts on them. The PIC output pin drivers may have the equivalent of a diode anode on the pin with it's cathode going to the +5V supply, so 9 volts from the LED cathode could produce 8 volts or so on the PICs internal +5V rails. Normally a 12V connected display would use a special driver circuit on the PIC output pins to avoid this. As simple as an NPN transistor and two resistors per LED segment. Or another IC.

Please tell me which driver circuit or IC I use ?