Hi all, I would be very grateful if someone can tell me what I've done wrong here. I've got a 9V battery (or thereabouts) feeding into a L7805, which should give me a 5V output (with a couple of capacitors in the circuit - see the L7805 datasheet). I've tested this and it works.
Needing to drive a PIC (16F887) and a 8-character display (
AVAGO HCMS-2915) from this, I figured a buffer is needed, as I don't think the L7805 can actually supply anything more than a tiny amount of current. So I did this:
Code:
|\
L7805 output -> |+\_____ 5V buffered supply
|-> |-/ |
| |/ |
|-------|
where the weird triangle thing is an opamp, 1/4 of a LM348. The LM348 is supplied from the 9V battery line (+) and the ground (-) (not sure whether I really need a negative supply - it's annoying to have to use two batteries, one connected backwards
)
I then connected this buffered output to the +LED pins of the display, the PIC and the display logic being powered by the unbuffered supply. However it seems the display causes a significant voltage drop - using a multimeter I got the following readings:
battery voltage when disconnected from the circuit 7.45V
battery voltage when in the circuit 7.0V
output of the L7805 5.0 V
(PIC also receiving 5V)
voltage reaching the display
1.8V
I know the display is the problem because when I disconnected the -LED wire to remove the display from the circuit, the voltage reaching the +LED pins is 5V as expected.
I've also got some LEDs to show some outputs of the PIC. As the PIC shouldn't really be made to provide much current, I need to power them separately - would a 4016 work? (signal control = the PIC pin, in = 9V supply, out = fairly big resistor + LED) If not I also have some
"darlington arrays" but I'm puzzled as to how these will help as they don't have their own power supply.
Many thanks for your advice,
ahydra