Yes pito I am currently designing the circuit for a temperature controlled soldering station and i want to get everything sorted out. I guess i am going to go with the lm324 since its cheaper. I will get it to swing to +5V by supplying it with more than 7 volts. By the way the lm324 contains 4 opamps. Do you guys know the package if there is one that contains one instead of all 4 opamps?

Thank you very very much for the help people. It turns out that i only need the opamp to swing its output from 0 to +5 volts (no negative voltage involved). Is though the lm324 able to go as low as 0 volts and as high as +5 volts while being connected to +5volts and ground? I guess if not then i would have to raise the 5 volts and supply it with a negative voltage as well, or go for another opamp or implement your suggestions by using a DC to DC converter or a second center tapped transformer.

Jackrae can you be a bit more explanatory please. I don't seem to understand what you are saying.

Mike thanks for the help man.

Should I go for a rail to rail opamp or a single supply one?

How do you supply a regular opamp (lm324) with -12V if you do not have a center tapped transformer? (ie regular  240 to 12 volt stepdown transformer with a single secondary coil)

Thank you in advance
Regards Voidugu

Thank you very much for the help people

I am trying to measure the voltage drop across an RTD. Its resistance will vary from 50 to 100 ohms. A resistor of 75 ohms connected in series with the RTD, creating a voltage divider,connected to 5V DC gives a voltage range from 2V (R RTD = 50 ohms) to 2.857V (R RTD = 100 ohms) across the RTD. If this was fed directly to the ADC of the atmega328 then the resolution would be extremely poor.

My idea is that with a suitable amplifier circuit, this voltage range can be amplified around 5.83 times (5 /(2.857 - 2)= 5.83)to allow me to utilize the full potential of the atmega's ADC. The problem is that i am a complete novice with opamps since this is the first time using them.

Could you please help me by providing a suitable circuit or by providing me with some information on how to implement this?  

Thank you in advance.
Regards Voidugu

Thanks for the help magician

Magician could you please explain to me the term OPA and the table you posted below?
Thanx again

I have a soldering pencil which has an integrated PTC RTD built in. The temperature of the RTD can be represented using this equation y = 0.0056x^2+4.9382X-239.43 where x is the resistance of the RTD and y is the temperature in Celsius. The resistance would vary from 50 to 100 ohms (its a relatively low resistance for an RTD).

Could you please provide me with a way to measure the resistance of the RTD giving the greatest resolution to the arduino measuring it? (ie use all 1024 voltage levels measured with the arduino's analogRead() )

Thank you in advance.

Thanx for the help Keith

Keith i think you got the equation wrong. Check the link out.
http://www.electronics-tutorials.ws/diode/diode_6.html
"Vripple = (Iload)/fc
Where: I is the DC load current in amps, ƒ is the frequency of the ripple or twice the input frequency in Hertz, and C is the capacitance in Farads.

The main advantages of a full-wave bridge rectifier is that it has a smaller AC ripple value for a given load and a smaller reservoir or smoothing capacitor than an equivalent half-wave rectifier. Therefore, the fundamental frequency of the ripple voltage is twice that of the AC supply frequency (100Hz) where for the half-wave rectifier it is exactly equal to the supply frequency (50Hz)."

I am kind of confused now with the coefficient of f. You said that f is the frequency of the power line. If the frequency of the AC waveform is 50 hertz then shouldn't the frequency of the fully rectified waveform be 100 hertz? Why is it 200 as you suggest? Please explain.

Keith i am aware of this equation: C = I/(2FV){where C is the capacitance in farads, I is the current in amperes, F is the frequency of the wave and V is the voltage ripple] , which is essentially the same as your equation. I used 2FV though and you used 4FV. Why is that? 

Thank you both of you for your help btw.

Lets say i have a transformer putting out 11.4 volts AC at a max of 10 amperes. I connect it to a bridge rectifier and therefore the voltage drops down to 10 volts dc (2 x diodes of 0.7 forward voltage drop each). I have a heating element of 2 ohms whose resistance stays constant with temperature (lets just assume that) . If i connect a smoothing capacitor bank on the output of the bridge rectifier and connect the resistor in parallel will more power be delivered to the resistor when compared to:
>the same resistor connect to the AC 11.4 volts?
>the same resistor connected to the 10 volts dc without the smoothing capacitors?

Thank you very much for the help
Regards Void