question about sn7407n

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qsd

New Member
Hi,
I built classic Tait style PIC programmer. Everything was ok until I decided to make development board from it. The problem is than then I turn off Vdd to the 7407 IC, it seems that it still can sink current, because MCLR PNP transistor is open. Is this possible(with good design and IC)? Or maybe I have bad IC? Looking at the schematic of one gate of IC, I don't see a way how can that last transistor be open without Vcc... Input isn't connected to anything, and there are pullup on output...

**broken link removed**

Thank you
 

Any leakage current through the 7407 output pin will turn the MCLR transistor ON, removing the Vdd supply to the 7407 is likey to leave it in an unknown condition - why not simply leave it powered up?.

Or better still, use a switch to completely isolate the programmer and target device - as the Dontronics SIMMStick programmers do.
 
Thanks for answer.

I thought that if I would turn off the IC, there autimatically won't be any signal from PC , so anything left from programmer would be only pull ups on RB6 and RB7... And I would need only simple small DPDT switch...
And now I got one more question... Would it work if I turn off both Vdd and ground? And would it be ok if I turn off only ground for crystal oscillator(F1100E)?
Thats because I don't have space on my PCB for any other switches

Thank you
 

Try leaving the 7407 powered up, your PIC programmer software should leave the output pins in the correct conditions - that way you know what the pins are doing, by disconnecting the power you leave the pins in an unknown state.
 
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