designing electronic voltmeter

We were asked to design digital voltage and current meter with 3 ranges
(20 100 200) v for voltages and two ranges
(2, 20)A for amperes
Note: supply voltage is 5V
maximum value for resistors of ampere meter must not exceeded 0.5 ohm


V1=Vin. (R2+R3+R4)/(R1+R2+R3+R4)
V2=Vin.(R3+R4)/(R1+R2+R3+R4)
V3=Vin.(R4)/(R1+R2+R3+R4)

Where V1=V2=V3=V4=1 V
Vin changes respectively with V1,V2,V3 as 20,100,200
So we used those equations mentioned above when we considered R1+R2+R3+R4=6000 Kohm but we get these values:
R4=R3=30Kohm
R2= -30
Note: we have no problem with current meter,that's why I haven't referred to it at the attachment
Need help!
Thanks in advance!

MODERATOR: stupidly large graphic reduced in size!.

It's very simple, it's a simple resistive attenuator (also known as a potential divider). Assuming you require 1V to the opamp, then for a 200V range the ratio between the bottom and top resistor is 1/199 (so it drops 199V across the top resistor, leaving 1V to the opamp. For 100V the ratio would be 1/99 and for 20V 1/19.9 - it's that easy!.

The input of a digital meter is normally 10Mohms, so the total resistor chain should add up to that - so to give you a start, the bottom resistor should be 10,000,000/200 = 50,000, and the chain above it should total 9,950,000 ohms.

I think that's the solution:
let's go with Kohms units
R4=10,000/200=50kohm

For range(100v):
R3+R4=10,000/100=100kohm
So R3=50

for range(20v):
R2+R3+R4=10,000/20=500

SO R2=400

And finally R1 is equal to (10,000-500)=9500

good luck :wink: