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tperkins
02-07-2007, 11:45 PM
Since I am duplicating it here.

I have a 9.2V supply powering a Propstick. To avoid loading the on board regulator as much as possible, I am supplying dry contact switches with the 9.2V through a 5.1V zener to drop the voltage from the supply to 4.1V. I have 4.6k resistors between the switches and Prop pins A0-A7. When the switch is closed, I see 9.2V on the supply, 5.1V across the zener, and 3.98 to 3.75V on the Prop IO pins relative to Prop GND. I expected a lower voltage.

Before (Monday) I was switching the regulated 3.3V onto the pins.

What am I missing, please, and am I cooking the pins?

dira for pins 0-7 = 0

Thanks, Tom Perkins

mahjongg
02-08-2007, 12:21 AM
If I understand correctly you are using a 5V1 zener to create a 5.1 Volt signal to power some 4K7 (probably not 4K6) pullups connected to I/O ports.

I also presume you have some kind of feeding resistor between the 9 Volt power supply and the kathode of the zener, to limit the zener current.

If you try to pullup a propeller I/O pin above 3.3 Volt what happens is that the internal protection diode in the chip that has its collector connected to the 3.3Volt power supply and its anode connected to the I/O pin starts to conduct, because the external voltage that is put on the I/O pin is greater than the power supply voltage of the chip, so current from the 5.1 Volt supply flows through the 4K7 through the protection diode to the 3.3 volt power supply.
The 4K7 resistor limits the current, otherwise you would force the 3.3 volt power supply up, with it the current is limited to 5-3.3= 1.7 Volt/4700 Ohm = 0.36mA. The propeller draws about 75 mA, so that a few mA flowing in frome elsewhere won't be a problem, the voltage regulator just needs to supply of few ma less to keep its output at 3.3 Volt.
So no, you are not "cooking the pins".

The voltage on the pins is not 3.3 volt, because of the diode drop of the internal protection diode, which normaly is in the range of 0.6 volt.
The expected voltage would therefore be 3.3 + 0.6 = 3.9 volt, which is what you observe.

Mahjongg

tperkins
02-08-2007, 01:46 AM
The circuit is now:

9.2VDC-----------5.1Zener-------Switch-------4k6--------PropI/O------guts_of_prop-------PropGND------bread board mOhms resistance-------Supply GND

Where I am seeing 3.8ish VDC against the Prop Ground is right on the lead from the package.

If this is OK, cool. I am still seeing (if I recall) 3.3VDC regulated on Vss.

Thanks, Tom Perkins

Paul Baker
02-08-2007, 02:58 AM
Since you are supplying >3.3V to a I/O pin what is happening is the ESD diode to VDD is turned on. Since you have a 4k6 resistor in the path you are limiting the amount of current going through this diode, place an ammeter between the resistor and I/O pin, if it's under 1mA you should be OK, if not increase the resistor's value until it's under 1mA.

However in general you should try to avoid the situation entirely, for homebrew applications it's not a big deal. But such an occurance should be avoided in commercial products in order to reduce the likelyhood of failure in the field after years of use.

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Propeller Applications Engineer
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