External switch to Prop

DavidM

HI,

I have an external switch about 2 - 5 metres away from my prop device.

I wish to read the switch as an input to a prop pin.

Q1) Will the following circuit do this?
I am using a resister divider to minimise the current ( I calculate about 3.2 volts when the switch is pressed)
I am supplying 5v to the switch.

I will use some kind of timing to detect a "PROPER" switch press ( i.e say >100ms) I am not concerned with de-bouncing. just need to know if this circuit will provided enough "VOLTAGE" to the prop input pin.

Thanks

Dave M

Mike Green

Your circuit should work. The 1K resistor will limit the input current to something the input stage clamp diode can handle and the resistor to ground will provide a default logic zero state. You could increase the value of the 1K resistor to maybe 4.7K or 10K to reduce the current through the clamp diode, but 1K will certainly work.· A typical value for the resistor to ground would be 100K.· Don't worry about the voltage divider effect.· Once the clamp diode conducts (at around 3.9V), that predominates.

Post Edited (Mike Green) : 11/25/2008 2:16:31 AM GMT

DavidM

HI Mike,

Thanks for responding.

According to my calculations this should make about 0.0018 AMPS ( 1.8ma ) through the circuit when the switch is closed.

I used this calculator here

http://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm

I used..

VOLTAGE = 5
R1 = 1000
R2= 1800
VOUT = 3.2143
AMPS = 0.0018

Q1) what is the maximum current the inputs for the prop can handle?


Thanks

Dave M

Mike Green

I think it's around 20mA, but that's an absolute maximum sort of value. 2mA is very reasonable.

DavidM

Thanks Mike.


Dave M

Ken Peterson

According to the data sheet (see exerpt), you should not drive more than 500 microamps through the protection diodes.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

DavidM

Hi Ken,

Would these resistor values be SAFER?

VIN = 5V
R1 = 5K1 1%
R2 = 10K 1%
VOUT = 3.3V
AMPS = ABOUT 1ma ( i think? )

regards

Dave M

Mike Green

Remember, you don't really have a voltage divider here because the substrate diode is conducting. The voltage drop across R1 with the switch closed is 5V - 3.3V - 0.6V (for the substrate diode) = 1.1V. Anything greater than 2.2K will limit the current to 500uA or less. R2 just has to hold the I/O pin low when the switch is open and the I/O pin is a high impedance. 100K would be fine and wouldn't significantly affect the voltage at the I/O pin when the switch is closed.

Ken Peterson

Is it well known what the failure mode is as a result of too much current in the protection diodes?· Is it related to PLL failures which seem to be common?· I imagine there are a lot of propellers out there with more than 500uA going through that protection diode.

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·"I have always wished that my computer would be as easy to use as my telephone.· My wish has come true.· I no longer know how to use my telephone."

- Bjarne Stroustrup

Mike Green

The current through the protection diodes has to go somewhere and it goes into the power supply network on the chip (to Vdd or Vss). If there's a lot of current, the resistance in the power supply interconnections will cause regional voltage differences. If there are external ground loops as well, it's easy to get transient voltages in parts of the chip that exceed the maximum allowable voltage and cause gate oxide damage. The PLL seems to be the most vulnerable structure for whatever reason.

The short answer for your question is "not really". Current through protection diodes may be a contributing factor in the PLL failures, but not a cause. In other words, if you have a situation close to failure (of the PLL) and add injected currents through the protection diodes, you may push the chip into failure. If you use the diodes for clamping of signals and limit the current to approximately the range specified (and particularly limit the total current for all I/O pins' diodes to avoid shifting the supply voltage), that will not by itself cause problems.

Post Edited (Mike Green) : 11/26/2008 3:48:07 PM GMT

jazzed

Ken Peterson said...
Is it well known what the failure mode is as a result of too much current in the protection diodes? ...

My experience with switches shorter term for shorter connection lengths (and by accident) is that the logic voltage gets misinterpreted especially after prolonged exposure to the "intolerable" current. Series 2.2K resistors (as Mike mentioned) do seem to eliminate the problem. My switches are much like the .pdf minus the pull-down and with cable about 8".

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DavidM

Hi,

To JAZZED - you are saying you don't need a PULL-DOWN resistor ?? Why, Won't the input pin just float?

If the maximum CURRENT AN INPUT pin on the prop is only 0.0005 AMPS ( 1/2 milliamp, or 500uA ) then according to my calculations I will use the following resister values.

VIN = 5V
R1 = 5K1
R2 = 10K 1
VOUT = 3.3V

To MIKE, if I keep the voltage under 3.3v ( by using my suggested resistors above) then why would the diode come into effect??
I also CANNOT see how this is not a voltage divider? if this simple circuit is not a voltage devider then what is?


Regards

Dave M

DavidM

OK,

I just did a breadboard with this circuit

using 2 X 10K resistors works well

ACTUAL VALUES ( using a multimeter )

VIN = 4.97
R1 = 9960 OHMS ( actual value of 10K resistor )
R2 = 9960R OHMS
VOUT = 2.48V

the current is about 0.0002 AMPS

I understand that the prop need at least 1/2 of VDD to register a high which is 1.65V

regards

Dave M