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Virtex-4 5V tolerance

Started by Heiko Kalte July 15, 2005
Hi,
is there any difference between Virtex-II and Virtex-4 5V tolerence? I did 
not find any information about that. I want to connect 5V outputs to 
Virtex-4 inputs. I read all the articles about 174ohm resistors and 
QuickSwitch. Does all that apply to Virtex-4, too?
Regards
Heiko 


"Heiko Kalte" <kalte@csse.uwa.edu.au> schrieb im Newsbeitrag
news:db7djq$eb5$1@enyo.uwa.edu.au...
> Hi, > is there any difference between Virtex-II and Virtex-4 5V tolerence? I did > not find any information about that. I want to connect 5V outputs to > Virtex-4 inputs. I read all the articles about 174ohm resistors and > QuickSwitch. Does all that apply to Virtex-4, too? > Regards > Heiko >
you can assume pretty much the same applies, yes. Antti
Yes, it does.

Austin

Let me put my tutorial hat on:
Virtex-4 pins should not see a voltage significantly more positive than
4 V, because that would overstress the thin gate oxide in some
transistors.
As a means to avoid damage from electro-static discharge, there is a
diode between each pin and its Vcc supply connection, preventing the
pin from going more positive than Vcco + 0.7 V.

If you drive the pin with a voltage >4 V, this diode gets
forward-biased and will conduct tens of milliamps (provided Vcco is 3.3
V, which it should be if you want to be 5-V tolerant.)
Now you need something to limit the current that the 5-V output drives
into the protection diode, and through it into the 3.3-V supply.
Limiting it to <10 mA is a good idea. A resistor comes in handy.

The accuracy of that resistor is irrelevant. A higher value would slow
down the signal, since the pin represents a capacitive load. If speed
is not an issue, use 1 kilohm. But remember that this resistor might
also be in the way when the FPGA pin is an output.
With an even higher resistor, the voltage tolerance goes up.
With a 10 kilom (1W!) resistor, it becomes 100-V tolerant, if anybody
would care.  :-)
This is really Basic Electricity 101.
Peter Alfke

Note that app. notes also specify, and this is important, that you
power regulator must support inverse voltage

Imagine a case where your 3.3V network draw 50mA (for FPGA IO, 3.3
devices, ...).  Then, the regulator that supply the 3.3V will outpu
50mA

Now, take a few IO that you put 5V on them, with resistors to limi
the current to 10mA.  The way the FPGA is done, the IO pins hav
protection diodes, which will take cut excess voltage at any IO t
Vcco + 0.7V (diode at IO connected to the Vcc).  Now, if you limi
current to 10mA, with a resistor for example, the primary MO
transistor at IO pin won't draw more current (still only some pico o
nano-Amps).  So, the 10mA that goes through the IO end-up feeding th
3.3V Vcco network

In the case above, having only one IO feeded by 5V (with 10mA goin
through the IO), will feed the Vcco 10mA, while the regulator wil
now only provide 40mA.  You will still have 3.3V on Vcco.  Th
problem get if you feed more current than what is needed device
connected to Vcco.  If you get too much current through the IO, then
the 3.3V regulator will no longer supply current.  All the curren
will come from the FPGA through the protection diodes.  The very BA
thing is that the Vcco line will no longer be regulated to 3.3V, bu
will get up higher (4.3V or more)

So, you must ensure that the regulator can also sink excess current
to maintain the voltage to 3.3V