# Op-Amp calculations

Discussion in 'Homework Help' started by allcircuit, Mar 18, 2009.

1. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
0
Maybe i can get some tutorials and guidance here:
According to this picture: [here]

How am I able to get the part on Op-Amp calculated? Coz Op-Amp involve in dealing with the scaling of the current, then multiplied with the voltage ....
How to get rid of having the values of resistances and capacitances across the schematic?

Thanks

2. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
- is very unclear.

Any graphic program that can load a jpg file will probably let you delete the component values.

3. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
0
Well, I just wanna know, to built a watt meter like what is displayed as the link....
To have the current input being scaled and feed into the PIC for calculation...
What is the main criteria to have/calculate the values for the resistances or capacitance around the op-amp? Both at the inverting and non-inverting inputs?
I just want some idea to starts this major part - current shunt for watt meter.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
The circuit appears to be something to do with a motor. There is an RPM input, for instance. The op amp circuit is for an ammeter. So the whole thing can display current and/or RPM, but that doesn't give power, which is voltage (not measured) times current.

The sensing resistor (not a shunt) is .001 ohms. The voltage input to the op amp is given by E = IR, basic Ohm's law. For 10 amps, the voltage will be 10 mv.

The op amp amplifies the voltage developed over the sensing resistor. There is also a trim pot to let the circuit be calibrated, meaning that the input will be scaled to the current in a predictable manner.

This is a link into our Ebook that explains how resistors may be used to set an op amp's gain - http://www.allaboutcircuits.com/vol_3/chpt_8/5.html.

5. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
0
But from the perspective of a schematic designer....
what makes it have a value of those lumped components?
As the MCU doesnt meet those values in the code.

6. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
- is very unclear.

Can you expand on the question?

One does not design a schematic - it is a map of the circuit.

A network of components may be resolved into a lumped value, but that has no meaning for the ammeter circuit. Did you read the section of the Ebook?

Why would the MCU "meet those values in the code"? What do you mean by that?

7. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
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what i mean is:

The PIC sense the voltage and current input. So inside the PIC ASM codes, it does not implement any calculations regarding the lumped components outside the PIC. I just want to know how a circuit designer know the values for the capacitance and resistors that need to use? Using try and error method? Especially the part on op-amp using feedback, how the designer deal with it? It must have certain gain value for that? Thanks.

8. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
See my post #4. You have a fixed resistance which will develop a voltage proportional to current. The op amp gain is made with that in mind, as well as the range of current that will be considered full-scale.

The analog input to the PIC gets made into digital values by the ADC. Using the scaling in the op amp circuit, the programmer can say that a value of such-and-such is equal to a current of X amps.

With enough knowledge of electronics, it is quite possible to design everything properly ahead of time.

9. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
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How about the voltage input fed into the PIC? As shown below:

Might want to know the voltage input is how much, with calculations perhaps?

So in conclusions, do you mean the Op-Amp functioned as a scaling factor for the current measurement?

10. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
The basic equation for the voltage divider is Ohm's law - E = IR. The 10K resistance is a variable resistor, so its value can be 0 - 10,000 ohms. Total resistance is all the values added together. Total current is the applied voltage divided but total resistance.

With variable at 0, Rt = 51.7K and It = 174ua. So the current through the 4.7K resistor will drop 818.18 mv across it.

With the variable at 10K, Rt = 61.7K, It = 146ua. Voltage on the 4.7K resistor will be 685.57mv.

The calpacitors are there to pass noise off to ground, and don't affect anything. The 4.7K resistor from the resistor junction and the 100nF cap are another low pass filter to eliminate noise.

I would strongly suggest you read our Ebook to learn these basic concepts.

11. ### allcircuit Thread Starter Active Member

Jan 7, 2009
31
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Hey, been there, thank you so much. As I am not sure about the function of capacitors there and i thought capacitors and resistors need to calculate as a lumped circuit.

Here is another major part of watt meter (or I assumed it as ammeter,) which is quite tough to me.

I tried to calculate the Rt as needed without considering the capacitor (Izzit correct by the way?)

As I am surprise that the current is 90A??
Actually this part is on current scaling, the positive voltage supply has been drops to 3.3V using voltage regulator. The current input is the one fed into the current sensor.

[Full schematic here]

Jan 7, 2009
31
0