TQFP programming

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

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 · Posted: Wed May 31, 2006 4:56 pm

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 · Posted: Sat Jun 03, 2006 9:36 am

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 · Joined: 06 Sep 2003 Posts: 21708

Guest · Posted: Sat Jun 03, 2006 11:55 am

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 · Joined: 06 Sep 2003 Posts: 21708

Guest · Posted: Sun Jun 04, 2006 5:37 pm

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.

Guest · Posted: Mon Jun 05, 2006 4:23 am

Wait a sec...

If the load is exactly 3k3, then everything is fine. All the current is sourced from the programmer, and the vdd line sits at 3v3.

If the load is <3k3, the programmer will tend to pull vdd <3.3v, so the on board supply has to provide extra current into the programmer...?

If the load is >3k3, the programmer holds the vdd line > 3v3, and the on board supply, or other circuitry, may be damaged.

I'm getting myself very confused wrt the above, and I'm sure I shouldn't be!

Ttelmah · Guest

Don't think about resistances, think about currents. At 3K3, with a 3.3v supply, the board requires 1mA. If the effective resistance is lower than 3K3, the board requires more rhan 1mA, while if it is hgher, the board requires less than 1mA.
In state '1', the programmer delivers 1mA, so the supply turns off (it is regulated), and all the current comes from the programmer. This is an almost impossible situation to maintain.
In state '2' the board requires more than 1mA. The progammer does not 'pull down', but sources 1mA, reducing the amount the supply has to provide by this amount.
In state '3', the voltage will rise. However the current drawn by a PIC, will rise with increasing voltage, so the rise will not be enough to cause harm. This would potentially be a different story, if the source could provide a lot more current.

Best Wishes

Guest · Posted: Mon Jun 05, 2006 6:01 am

I can see that that would keep the power consumed the same, but surely most electronic components dont' work that way - they can't vary their impedance. If you lower the voltage across them, then the current through them falls. They have a specified operating voltage, and won't work outside it. Lowering the voltage doesn't cause the current to increase, it just causes the device to shut down...

Guest · Posted: Mon Jun 05, 2006 6:24 am

ah... a current equation sorts things out. If load is equal to 3k3, our source provides no current. Less, and it provides some. More, and it needs to sink current to maintain voltage, which it may or may not be able to do. If it can't, we have a dead converter and a supply line at high voltage.

Ben