:: High Voltage @ Input Pin? :: HELPPP

[suby786]

Hey all again,

Just wanna run something past all of you wise folks..

As some of you know, i am doing CURRENT SENSING usin ga CURRENT TRANSFORMER, and it works a treat, by passing the wire i want to measure thru the core, and passing the output thru a bridge + some C's and R's i get a voltage PROPORTIONAL to the line current and my ADC converts it fine.

Now i have 4 WIRES to be measured, and instead of having 4 CURRENT TRANSFORMERS connected to 4 AtoDs , i want to pass ALL 4 WIRES thru ONE transformer, and i have tried this and it works,

When i want to acquire channel1, i turn on ch1 and turn 2/3/4 off, i then take the ADC measurement and this all works fine, BUT under normal running conditions, i.e. when im not taking a sample, i MAY have all 4 ON... and obviosuly this will produce 4 x the voltage i would expect, I am worried about this voltage, obv it will damage the input pin? what shall i do?

Use a relay to disconnect the wires going to the input pin?

Any other ideas????

Spec:

4 wires... all can be controlled to be on/off/dim-able
1 current transformer
i have enough ADC ports and DIGI out pprts (for relay) to do the job

HELPPPPP

 

[Nigel Goodwin]

Your post is highly confusing, and it's hard to understand what your problem is?, you give no details of any of the components used, but a simple series resistor will protect a PIC analogue input.

 

[eblc1388]

BUT under normal running conditions, i.e. when im not taking a sample, i MAY have all 4 ON... and obviosuly this will produce 4 x the voltage i would expect, I am worried about this voltage, obv it will damage the input pin? what shall i do?

Pass the other two wires through the CT in opposite direction to the first two.

Consider adding input overvoltage protection on the ADC pin. This can simply be a series resistor and a diode that connects to Vcc.

 

[Pommie]

Assuming that the CT connects to a resistor to convert current to voltage, why not choose a suitable resistor (or use a divider) so your readings are in the 0-1.25V range and set Vref to 1.25. This will still give you the same resolution as before but with a nice advantage, when you are not taking readings you can set Vref to 5V and your problem is solved.

Mike.

 

[suby786]

DOes any1 have a sampel circuit to show me how to protect the input...

Sorry to have confused you, the output of the CT goes to a bridge rectifier, with a R and C in parallel BEFORE the rectifier, i then have another R and C at the output of the rectifier to get an analgoue voltage.

If i have ALL 4 WIRES ON... then i will have 4 sets of current, which could potentially be 10-12v, which obviosuly i dont want to go to the input, so i wanted to know if i could protect the cpu from this in normal conditions. When aquiring the current in EACH wire, i pulse each wire seperately and thus i dont have a problem.

I cant use 1.25v as Vref as the changes in current will be in the form of mV and 1.25 / 1024 is very small and so id get alot of variation?

I like this diode/series resistor idea... but series resistance reduces VOLTAGE at the pins and reduces CURRRENT, is this why you have put this there? this will make my ADC conversion lower aswell??

 

[eblc1388]

If i have ALL 4 WIRES ON... then i will have 4 sets of current, which could potentially be 10-12v,

Pass the other two wires through the CT in opposite direction to the first two.

.. but series resistance reduces VOLTAGE at the pins and reduces CURRRENT, is this why you have put this there? this will make my ADC conversion lower aswell??

Nope. The series resistor(say 1000 ohms) will not reduce either voltage or current to the ADC. It only limits the current passing through the protection diode SHOULD the input voltage to the series resistor goes higher than VCC by 0.6V.

The input impedance of the PIC ADC pin is very high, many orders of magnitude comparing to the value of the series resistor and thus the series resistor will not affect the conversion result IN YOUR CASE.

 

[Pommie]

I cant use 1.25v as Vref as the changes in current will be in the form of mV and 1.25 / 1024 is very small and so id get alot of variation?

If you change your resistor after the bridge to 2 resistors, ¾ and ¼ of the original value then the voltage across the smaller resistor will now be 0 - 1.25V. You can feed this to the ADC with the Vref set to Vdd/4 (1.25V) and the output of the ADC will be identical as before. No loss of resolution or increase in variation. When you don't need to take readings you set Vref back to 5V.

If you want to clamp the voltage, which is actually the simplest solution, just put a 5.1V zener across the resistor after the bridge - cathode to the +side. You don't state any values and so calculating what wattage you need isn't possible.

Mike.
Edit, I just realised I have assumed that the R and C after the bridge are in parallel. If this isn't the case then put the zener from pic pin to ground.

 

[suby786]

the R and C are in parallel after the bridge. The MINIMUM wattage @ 240V AC is 40w, AND my maximum is 1000W at the same voltage, i need this range of current, which is 0.174A to ~4A to be converted to 0-5V DC,Now my CT currently churns out 1.6V at 2.1A, so i will defo make the 4A in the 0-5v range.

I didnt realise that the INPUT to the bridge had to be > 1.4V due to the drop across both diodes, i made an oversight, so i had to turn the primary around twice for ALL 4 WIRES... to get the input to bridge to be >1.4V...

But i can reduce the brightness of the load and the voltage goes down quite nicely all the way to zero...

ive never understood zener's, i know they block current in the reverse direction, and when its reversed biased it produces a 5.1V potential across this... so if i put 8V across this resistor/zener, i will still get 5.1V out?

 

[suby786]

having 2 of the wires goin in the other direction works also. It cancels it out when there ALL ON so i get 0v which is great, but then my worst case scenario is 1 of the pairs being 100% on, and the other pair OFF. i then have cumulative 2 X 4A potentially meaning my voltage could be >6V

whats the max input voltage i can put on the pin be4 killing it?

and should i STILL use the zener, they way you hopefully will guide me upon???