I am design a PIC programmer and i am using a strp up converter to convert from 5V to the required PIC programming voltage of around
13.6V.Does someone know the ammount of current the PIC takes during programming so that i design the step up converter to provide ample programming current?
Thanks very much for your help
Joseph A. Zammit Malta
Short answer, very little. The programming current comes from the +5 supply. The VHH on MCLR is only used to switch into programmimg mode.
My own device robs power from the target PIC and uses the PWM output of a PIC16F628 to drive a voltage multiplier and get about 13 volts.
-- Luhan Monat (luhanis 'at' yahoo 'dot' com) "The future is not what it used to be..." http://members.cox.net/berniekm
Yawsa! I'll check to see if my programmer does the fuses correctly.
The current varies according to the mode you are running the chip, and what you want to program. The worst case, is if you want to erase the chip protection fuses. This is specified as requiring 10mA on the Vhh supply, and not 'only a small current to switch to programming mode' (the internal programming supply can drive just about everything else, but not this). Real chips use about 6mA to do this. This is why a lot of programmers, have problems erasing the protection fuses. The specification, is IDpp, and is in the 'memory programming requirements' section of the data sheet.
Best Wishes
Maybe read the programming specification sheet?
The F (and 16C84) chips draw very litte (< 1mA), the (other) C chips draw a lot (100mA?).
Wouter van Ooijen
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As far as I know IDpp is drawn from Vpp, as the name suggests, which is the ~ 5V supply.
My Wisp628 programmers have a 1k resistor in series with the Vpp, which would not match well with 6mA drawn. Yet they work OK.
Wouter van Ooijen
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