TQFP programming

[barkerben]

Question number 2...

I have a TQFP package. I intend to program it over a serial link using a bootloader, but to do this I first have to get the bootloader onto it!

I'm assuming the easiest way to do this would be to try to temporarily clip wires onto the pins to program it - might be a bit tricky, but could be done. Once soldered in place, it obviously can't be removed again, and dedicated adapters are extremely expensive...

Ben

[rwyoung]

Add a pin header nearby for the programming. Could be pins, pads or plated holes. Then you need a programmer with the approprite mating portion.
_________________
Rob Young
The Screw-Up Fairy may just visit you but he has crashed on my couch for the last month!

[Guest]

The problem is the ICSP works at 5v, wheras most of the circuit runs at 3.3v,
so I think I need to get the bootlaoder on before I put the PIC in place on the board - then I can just program it over the serial bus

[kender]

[rwyoung]

[treitmey]

If the board isn't made yet, I thought the correct way to do this is make a header for programming,.. and a jumper to remove the high programming voltage from the rest of the circuit.

Remove the jumper and program while keeping the programming voltage from the rest of the circuit.

Then put the jumper in place and the PIC is feed from the other normal low voltage supply.

[FirstSteps]

Is this a one-off or for production?

I've done this before. You'll need to bend the pins of interest slightly down, flip the chip and solder fine wires to the pins. The fine wire can be got from an ordinary multiconductor wire. Strip the insulation and take one of the hair-fine strands.

To solder to them you'll need a rosin flux (not the "no-clean stuff") from Radioshack.

After programming, unsolder the wires and clean them up with a little desoldering braid.

Let me know if you need a better description and photo of it being done. I do this sort of fine work all the time. No special skill is needed, just the ability to follow instructions and a few basic materials.

[kender]

[PCM programmer]

You don't need a jumper.

These instructions are for ICSP programming with the Microchip ICD2
when the project board runs at +3.3v.

1. Run your project board under its own +3.3v power. Its power supply
may be a wall-transformer that goes to a +3.3v linear regulator,
or it could be a battery which runs a DC-DC converter which has
a +3.3v output. Or it could be connected to a bench power supply.

2. The MCLR pin on the PIC should go to +3.3v through a 10K resistor.
(If you're using the CCS ICD, they recommend a 47K resistor).
Don't use any diodes or other components on the MCLR pin.
The 10K resistor permits the MCLR function to work properly during
normal operation of the board. During programming, the resistor
limits the current going from the programmer into the Vdd supply
on the board. During programming, my ICD2 puts out about 13.3v
on MCLR. So the voltage drop across the resistor is 13.3 -3.3v =10v.
The current is 10v/10K = 1 ma. So the ICD2 is sourcing 1 ma
into the Vdd rail on the project board. This won't affect the board
because the voltage regulator (linear or DC-DC converter) will
continue to hold the project board's Vdd at +3.3v.

3. Configure MPLAB so the ICD2 knows that your board runs under
its own power supply. Go to the Programmer / Settings menu
in MPLAB. In the window that pops up, select the Power tab.
There is a check box that says "Power target circuit from MPLAB
ICD2 (5v Vdd)". Make sure that box is not selected.

4. Turn on power to your project board.

5. Plug the ICD2 into the ICSP connector on your project.
The +3.3v Vdd voltage on your project board must be connected
to the Vdd pin on the ICD2 connector. The ICD2 uses the Vdd
voltage on your board to set the proper voltage for its i/o buffers
which talk to your board.

6. Go to the Programmer menu and click on "Connect". You should
see messages in the Output window showing that ICD2 connected OK.

7. Then click the Program button to program the PIC on your board.


Caution: Do not set the Code Protect fuses, as you can only erase them
if the board is running at +5v Vdd. You can't erase them on a board
which is powered at +3.3v.

[Guest]

Hi,

Thanks for all the replies!

Unfortunately the PCB is already made - it looks like I could have used ICSP after all if I'd put a pin header in. I had assumed wrongly that the 3.3v was a majr problem. In this case, it is a one off project - so I think I'll have to take the suggestion of trying to solder the other programming pins verrry carefully - either that or see if anyone I know has a TQFP adapter, although that seems quite unlikely :-)

It may be a while till I have a chance to do this, but I'll let you know what happens,

Cheers,

Ben

[Guest]

By the way PCM programmer, why do you suggest not putting a protection diode between MCLR and Vdd - would that not prevent the power supply having to source current? I'm using a MAX710, and can find nothing on its tolerance of this.

I'm guessing the problem is that the MCLR pin draws so little current, that the voltage drp across the diode is a problem?

Cheers,

Ben

[PCM programmer]

[Guest]

Ok - thanks for that. I still can't find anything on the MAX710 datasheet about its ability to sink current, but a fraction of a milliamp doesn't seem much... do you think it will be ok? I take your point about static discharge with a diode...

Higher resitors will reduce the current further, but increase its response time - but since MCLR isn't a data line I don't suppose response time really matters.

[PCM programmer]

[Guest]

Ah I see... If the circuit load is exactly 3k3, then the programmer ends up providing the power for the entire circuit and the on board supply supplies no current. For lower loads, the balance is changed between current from the 2 sources. For Higher loads, assuming infinite input impedance to the voltage source, the voltage on the line has to rise.

Wrt ESD, is the following correct:

The MCLR pin is not tied internally to Vdd

Thus, if isolated, a static charge can build up on this pin

Eventually this may be enough to cause brakdown within the chip - bang.

By providing a bleed path, there can no longer be a differential across the circuit due to static

A PCb is likely not to be earthed, so the whole circuit can still be charged up relative to earth by the static, but there can be no relative differences that could lead to breakdown.