eng1

I agree with you as far as the range is concerned. But there are combinations of the two resistors that can't switch the transistor off.

andrew12345678

Ok, well thats a real kick in the pants.
What does Open Collector mean anyway?

Will a BC548 do?
so I would take the line from pin 5 and run it via a 10K resistor to the Base of the NPN, then connect a 10K from the Emitter to the Base, do I then take the output from the collector and run that to where the buffer out put is?

__________________
"The glass is neither half-empty, nor half full, it's just twice as big as it needs to be" - Unknown

Nigel Goodwin

I would completely disagree, perhaps you would care to suggest any values that you think could do that? - are you perhaps forgetting the driver IC is supposed to be open-collector?.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

Nigel Goodwin

It means what it says, a connection JUST from the collector of a transistor, with no other component connected to it. This means it can SINK current, but not SOURCE it - which is what we're looking for in this case.

If you check my PIC tutorials (Hardware Extras) there are some examples of using transistors for it, which you could use for your buffer.

Yes, that's fine.

Yes, that's it - and connect the emitter to 0V of course.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

eng1

Hmmmm... what if you consider a small leakage current? I'd like to simulate that case... do you still think that the transistor would switch off with any resistance (base to emitter)?

As Nigel already said, o.c. means that the collector of the output transistor (of each gate) is left unconnected. In order to operate the transistor as a switch you have to add an external pull-up resistor. That's useful when interfacing different voltage levels for example.
I hope you'll have your programmer working!

andrew12345678

Bugger, bugger, bugger.

I lost the pad for the buffer out put when I was trying to desolder the pin ( I dont know how because i dont think it was heated excessively), but that doesnt matter.....

with the lost pad i connected my programmer up to the computer, still the LED remained lit, however much dimmer this time. with the collector of my ghetto inverter connected to the 10 K resistor that came off buffer 4 the LED was brighter again, but unresponsive to the programming softwares "commands" ( I have been using the chech feature)

what now?

__________________
"The glass is neither half-empty, nor half full, it's just twice as big as it needs to be" - Unknown

andrew12345678

Update, almost got it, i made a mistake with the pin, i accidently used clock pin 3, but i fixed that, now the led is dim when VPP is dissabled, and bright when its enabled (looks like i almost have it).

however it still is returning errors, and isnt programming, could this be due to the VPP voltage which now sits on 1.9V when VPP is disabled... what do you think?

__________________
"The glass is neither half-empty, nor half full, it's just twice as big as it needs to be" - Unknown

eng1

Nigel, I've edited my previous post. Can you give your comment please?
Thank you.

Nigel Goodwin

You're now moving from the realms of changing resistor values to the driver being faulty - the purpose of the resistor between base and emitter of the switching transistor is purely in case there is some slight leakage in the driver. In 99.99????% of cases you could leave that resistor out altogether (which I've seen done commercially sometimes) - however, it's good practice to have some sort of resistor there, and I would always do so.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

Nigel Goodwin

Check all the voltages around the transistors, see why it's only dropping to 1.9V?. It 'may' need a resistor adding from Vpp to 0V?, try adding a 10K or so.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

andrew12345678

Someone electro had me tie VPP to ground via a 10K, that brought it down to 1.88 and i added a further 1K so now its sitting at 1.8V, almost there, but i dont know if thats a proper solution.

__________________
"The glass is neither half-empty, nor half full, it's just twice as big as it needs to be" - Unknown

kchriste

What's the voltage on the input of the "ghetto" inverter when the VPP LED is dim? You may want to add a 10K resistor from the base of the "ghetto" inverter transistor to ground if this voltage is above 0.4V

__________________
--- The days of the digital watch are numbered. ---

Nigel Goodwin

It's already there, it was in my original suggestion for a discrete inverter.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

cesarsouza

Hi, sorry to ressurect such an old thread, but I'm facing pretty much the same problem as the original poster.

Whenever I turn my programmer on, the Vpp raises to 11.66 volts and sits there forever. I'm using a custom built tait-style programmer, using the 74HC14 chip. I've already tried changing to 74HC05 but got no luck.

Should I also try with the 74hc06 or 74hc07 chips as suggested back in this thread? Would that make a difference?


Thank you,
Cesar

eblc1388

Why would a design use 7406 or 74LS06 when so many new IC or CMOS inverters are available?

You think so too?

Because there is a reason. The circuit needs the open-collector feature of these "old" IC to properly drive the transistors. If you don't know that, don't try to change these IC types. Countless builders have fallen in the same trap.

__________________
L.Chung