Re: [Usbpicprog-technical] USB electrical characteristics

[Francesco M. <f18...@ya...>]

Hi,

Frans Schreuder ha scritto:
> Sorry for not responding, I just opened these mailing lists and 
> discovered that I forgot to subscribe myself!
no problem ;)

> Don't worry too much about the current consumption of usbpicprog, it 
> will draw 50mA max. It can also work from 4.2V.
very good.

> What you do have to worry about is connecting a target board which is 
> powered by usbpicprog. If that one draws more than about 300mA, give it 
> his own powersupply of 5V.
ok; but what about Vpp?
The charge pump can deliver only a very small current; I've simulated it (using 
a square wave on Vpump1 and Vpump2 at about 20kHz) and absorbing 1.5mA makes Vpp 
drop to 9V....

OTOH I couldn't find any specs about the minimal current which this pin should 
be able to provide... I've found references to a LVP (low-voltage programming) 
mode, where Vpp is used only as a voltage reference (if I got it right); while 
in non-LVP mode, Vpp is used as source for the EEPROM programming.
I may be wrong however...


> Because of the diode in the VDD connection, it is allowed to use a 5V 
> power supply (but don't use a lower voltage than 4.5V when usbpicprog is 
> trying to power up the target board)
sure; I guess that powering the external circuit with < 4.5V would turn on that 
diode, and provoke a large current flow through the PMOS, leading to its 
possible destruction...

Thanks,
Francesco


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