Higher Voltage Analog Inputs

Hi All,

Please forgive me if this topic has been covered before/

I'm attempting to build a kiln temp controller using a large (about 18awg) k-type thermocouple plugged into an AD595. to get up to the max limit of the thermocouple (which is near the shutoff temp for the kiln) i need to power the AD595 with >+12.5V

the output of the ad595 likewise goes from 0.00mV up to 12.4V (about 10mV/degC)

since the arduino analog input can only receive up to +5V is there any way i can lower the voltage of this thermocouple to the 0-5V range? i'm no EE, but is there some sort of chip that will take a 0-12.5V signal and mirror it in the 0-5V range? or is there some kind of circuit i could build?

ive read some posts on optoisolators but i think they may drop out if the input signal drops below a certain voltage. since room temp will be around 0.027V i need to be able to read atleast to zero.

thanks so much for your input. :smiley:

ive read some posts on optoisolators

No opto-coupler ! Use a op-amp circuit to convert the input. You have 0 to 12.5 V out of the temp sensor, Arduino need 0 to 5 V, so an op-amp ( may need more op-amp ) may convert the input of the sensor into a proper output for the Arduino. Your code will do the calculations to display the temperature.

That will be my option. just an idea...

Look at resistor divider:

would a voltage divider work with a varying input? also wouldn't the operating temperature cause resistance drift?

Im not sure what you mean about using an op-amp... would i bias the amp so Vo/Vi is a fraction?

would a voltage divider work with a varying input?

Yes. It's a divider... :wink: If the two resistors are equal, the voltage is divided in half.

also wouldn't the operating temperature cause resistance drift?

:smiley: Don't put the resistors inside the kiln! (Don't put the AD595 or microcontroller in there either...) BTW - Resistors don't drift that much with temperature, and if there is some tiny drift, it's the ratio of the resistors that's important, so if they both drift together and there is little net effect.

Im not sure what you mean about using an op-amp... would i bias the amp so Vo/Vi is a fraction?

The gain of a "standard" non-inverting amplifier cannot go below 1 (with zero feedback resistance). But, the gain of a "standard" inverting amplifier can go down to zero (with zero feedback resistance). You could use a 2nd inverting buffer (gain of -1) to get a positive voltage, or you could bias the inverting amp so that it works between 0 and 5V, with higher voltage being cooler.

The AD595 can supply upto +/-5mA from its (low impedance) output - so if we used a resistor divider with 10k and 18k resistors, 12.0V would be reduced to 4.29V and less than 1mA is pulled from the AD595 output. The 6.4k output resistance of that divider network is low enough to give accurate readings from the analog input.

You should go with 1% resistors for accuracy.

A voltage divider is also a good one. The use two resistors for division is not too accurate in my opinion. I wil use a Multi-turn potentiometer of 10 K to 50 K should be ok. You simply adjust the put until you reach 4.98 V when the ouput of the AD595 is 12.4 V --> I go for 13 V.

Let do some maths :

5/13 = 0.384615384 Let take a 50 K pot - multiturn ... R1 = 30 769 ohm R2 = 19 230 ohm <-- set at that value.

Let check : 13 V / 50 000 = 0.26 mA <-- ok for the AD595 19 230 ohm X 0.26 mA = 4.9998 V drop of R2. Close enough.

All check out... 8) Just hope the Vout of the AD595 will not exceed 13 V.

great information everyone i think i have a clear path forward.

thanks!! :smiley: