# Anyone can explain this website circuit to me?

http://www.electronics-lab.com/projects/...
On positive of input, what use for the chooice of different path of resistor?
On negative of input, what use for the adjustable resistor ?
Can explain whole circuit to me?

The two resistors (one selected by the range switch, one adjustable) form a potential divider that presents a fairly constant voltage between pins 30 and 31 of the A/D chip over a range of input voltages from 200mV to 2kV.

At its mid point, the adjustable resistor will have a value of about 1Kohm (the adjustment is necessary to calibrate the meter). This means that on the 2V range the voltage presented to the chip will be half the input voltage (because we have a potential divider with two equal resistors), which shows that the basic range of the chip is 1V. The purpose of the range switch and its associated resistors is to increase this range to cover 200mV – 2kV.

The whole circuit is basically an A/D converter followed by 7-segment display drivers for 3-and-a-half digits (i.e. the "thousands" position only ever shows 0 or 1). The test button on the left sets all outputs high to check that all the displays are working.
At the core of this whole circuit is the IC "7107".
It is a analog-to-digital converter (ADC) and display driver, all in one integrated circuit. If you want to know how this one functions in detail then you need to read its data sheet http://www.intersil.com/data/fn/fn3082.p...
The important path to look at (for your questions) is pins 30 & 31 of the IC. All the other components are necessary to make the IC itself work, and should be followed strictly, but have nothing to do with the actual measurement.
Now I don't quite understand your question, because the circuit diagram http://www.electronics-lab.com/projects/... shows it all..?
The 5 resistors (10 Ohm . 10MOhm) define the voltage range to be measured, and the 2kOhm potentiometer is for calibration. And you do not have to worry about negative voltage because the IC automatically detects this and through its pin 20 puts a minus sign infront of the display .
The reson why the voltage ranges are all defined with a "2" in front is that in your diagram, the display "thousands" can only show "0" and "1", so your highest display will always be 1999 (with a decimal point somewhere in-between where appropriate).
The resistors are set up to be an attenuator. The switch selects the amount of attenuation. It looks to me like the adjustable resistor is intended to be adjusted to 1111.11 ohm.

Note that the adjustable resistor is across the differential inputs. Any of the fixed resistors (when selected) are in series with the input voltage. This sets up a voltage divider. The 7107 input voltage will be a fraction of the voltage at the input probes. I'd say the one typo is that the 1k resistor should have been 10K.

so when
Rselc = 10 ohm Vmes = Vin (200 mv range)
Rselc = 10K ohm Vmes = Vin / 10 (2 v range)
Rselc = 100K ohm Vmes = Vin / 100 ( 20 v range)
Rselc = 1 M Vmes = Vin / 1000 ( 200 v range)
Rselc = 10 M Vmes = Vin / 10000 ( 2000 V range)

The answers post by the user, for information only, FunQA.com does not guarantee the right.