good day all,

this is a circuit from a pdf which i am basing some of my design concepts on.

however there are a few things i dont understand!

for example:

1> how do i choose the correct op-amp?? (the sensor/transducer measures a peak value of 500V, and 5A from an induction motor). here the LT 1058 is used, and i have no idea why it is??

2>R1 and R2 --- i have no clue how to determine the resistor values here, i know that the board that this signal conditioning unit is connected to has a current max at its ADC input pins of ~10mA


3>shouldnt there be a cap in parallel with R7 in order to perform the operation of a low pass filter cutting out frequencies > 1/2 the nyquist freq?? instead the literature says that this is done by op-amp with (R6 +P2) and C1.

4>P2 (pot) is adjusted so that the max voltage to be measured (at signal conditioning input)gives max voltage of 10V (for the DSP board max voltage). this means that this forms an analogue divider circuit. so does this mean that instead of point 3> above R6 and C1 form the cut off freq.?? but then again adjsuting P2 would mean the knee point of the LPF changes to a higher/lower value (defeating the purpose of an LPF).. indirectly as we attempt to correlate max measured voltage to max DSP board input (of 10V).



please reply asap, this is important



Thanks team!

 

1. It is unclear whose sensor you refer to. You will have to get the data sheet for an LT1058 and deduce from the parameters why it got selected. Your application is even more unclear.

2. Without your schematic and sensor, that is impossible to answer. R1 & 2 values have to do with gain in conjunction with R6 & the P2 setting, and with the P1 setting plus the sensitivity jumper there. The signal magnitude and the input resistance of the mystery ADC will determine the current.

3. If you're dealing with DC, it's very hard to get concerned with Nyquist frequency. We don't have the literature you mention available. See #4. Why do you need an LP filter? What is your application?

4. P2 varies the gain of the first op amp. C1 will integrate the signal over time. No frequency, no idea about the overall effect.

 

thanx beenthere....


my app is measuring of 3 phase AC line voltage using a voltage transducer (such as the LV 25-P), n then connecting this to the signal conditioning part of the design(the circuit presented above) so that the signal is not corrupt/distorted before it is inputted into the µC. so thus i will need 2 of the LV 25-P Chips to measure across 2 of the 3 phases of the stator terminals of the induction machine, and each one will have its own signal conditioning circuit. which then goes to the ADC of the µC for evaluation.

 

I wouldn't think you would need a lot of signal conditioning on a low impedance output running at 60 Hz. Just apply the signal on the Rm resistor to an op amp with some gain to match the ADC input and there you are.

I'm imagining you are using a uC with a 10 bit ADC. You might worry a bit if the reso were higher, but it's really not worth much concern for only 1 part in 1024. Good ground layout will take care of about everything.

 

ok thanks. i havent done electronics in a while (since im an electrical enginer) but there is 1 quick question. if i send the signal to the inverting end of the op-amp input does that mean the output signal will be inverted or negated?? for example, if the gain of the op-amp is 2...and the input signal is 5V(to the inverting end). does this mean the output is -10 or 2.5??or something like that??

i know its a basic question, but ive lost touch with op-amps for a while now

thanks in advance!!!!!!!

 

Your signal is going to be inverted, but it's AC, so it won't matter. If your ADC is unipolar, you may have to use a precision rectifier to convert to DC. The circuit in the National Semiconductor op amp collection works well. Use modern FET input op amps. I get good results with OPA2134's (they are dual op amps - the LT1013 is another good one).

Here's a link to out Ebook that will show you how to calculate the gain -http://www.allaboutcircuits.com/vol_3/chpt_8/5.html.

 

if the gain of the op-amp is 2...and the input signal is 5V(to the inverting end).

-10V
You are measuring a current proportional to the voltage so that circuit doesn't just "tack-on" to the sensor.