TQFP programming

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