Hey,
I am currently reading 'Grobs Basic Electronics'. Chapter 8 is about analog and digital multimeters. Subchapter 8-4 talks about the loading effect of a voltmeter. There is also this (see picture) equation presented.
Can someone explain me, how the author got this equation?

 

* the circuit is a voltage divider with two resistors (R1 and R2)

 

Is this homework?

Write down the voltage divider equation.

 

You need to see the circuit for this. The voltage divider probably just reduces the voltage so the meter can measure it. You have an input voltage divider plus the resistance of the load. From there its "Ohm's law" and "Kirhov's 2 laws".

Edit:
Please post a reference to that book.
You have "Voltage +5%*Voltage".

 

This is the circuit. I understood why the voltage across R2 with the voltmeter connected is different then the normal Voltage. But i dont know how you get the equation posted above.
How do you get the "correction" part ?

Is this homework?

No I am doing electronics as a hobby and I bought this book, to learn the basics.

 

As I said, write down the equation for the voltage divider circuit, example (a).

 

Hey,
I am currently reading 'Grobs Basic Electronics'. Chapter 8 is about analog and digital multimeters. Subchapter 8-4 talks about the loading effect of a voltmeter. There is also this (see picture) equation presented.
Can someone explain me, how the author got this equation?

Hi,

Since this is not homework i can provide this information, but you should go over it if you want to understand how these solutions can be found. This is one way to find the solution...

Start with the parallel combo of Rv and R2:
Rp=Rv*R2/(Rv+R2)

calculate the divider voltage Emm with the meter attached if you want to check things later::
Emm=Ein*Rp/(Rp+R1)

Using a new variable for the divider voltage Em, calculate the current rhough R2:
iR2=Em/R2

calculate the current through Rv:
iRv=Em/Rv

calculate the total current in the bottom of the divider:
iRp=iR2+iRv=Em/R2+Em/Rv

calculate the voltage drop across R1:
vR1=iRp*R1=R1*(Em/R2+Em/Rv)

add that to Em to get Vin:
Vin=Em+vR1=Em+Em*(R1/R2+R1/Rv)

now that we know Vin, multiply that by the unloaded divider to get the unloaded voltage output Vo:
Vo=Vin*R2/(R1+R2)

we get:
Vo=(Em*(R1*R2+Rv*R2+Rv*R1))/(Rv*(R2+R1))

where the numerator expands into:
Em*R1*R2+Em*Rv*R2+Em*Rv*R1

which factors into:
Em*R1*R2+Em*Rv*(R1+R2)

and dividing by the denominator above we get:
Vo=Em*R1*R2/(Rv*(R1+R2)) + Em*Rv*(R1+R2)/(Rv*(R1+R2))

and simplifying the second term above we get:
Vo=Em*R1*R2/(Rv*(R1+R2))+Em

and that is the voltage without the meter attached.

Checking, replace Em with Emm far above and we get:
Vo=(Ein*R2)/(R1+R2)

which is certainly the unloaded voltage output.

 

This is the circuit. I understood why the voltage across R2 with the voltmeter connected is different then the normal Voltage. But i dont know how you get the equation posted above.
How do you get the "correction" part ?

No I am doing electronics as a hobby and I bought this book, to learn the basics.

If you can get the voltmeter reading as a function of the true voltage and the resistances involved, then all you need to do is solve that equation for the true voltage in terms of the voltmeter reading and those same resistances.

It's hard to tell you where you are having problems unless you show the work you've done up to this point.

 

but you should go over it if you want to understand how these solutions can be found. This is one way to find the solution...

Thanks for your answer. I could follow your steps and wrote it down by myself. But I am stuck at your last step.

Checking, replace Em with Emm far above and we get:
Vo=(Ein*R2)/(R1+R2)

This is how far I got

Vo = (Em*R1*R2)/(Rv*(R1+R2) + Em
Emm = Ein*Rp/(Rp+R1)

Vo = (Ein*Rp*R1*R2)/(Rv*(R1+R2)*(Rp+R1)) + Ein*Rp/(Rp+R1)

Vo = Ein*Rp * (R1*R2/(Rv*(R1+R2)*(Rp+R1)) + 1/(Rp+R1))

How do I have to continue to get
Vo = Ein * R2 / R1+R2
?

 

You have an awful lot of variables running around.

You have Ein, Emm, Em, and Vo.

Then you have R1, R2, Rv, and Rp.

It might help to list out exactly what each of them are and you will likely find that you have some redundancies that are causing you problems.

 

Thanks for your answer. I could follow your steps and wrote it down by myself. But I am stuck at your last step.



This is how far I got

Vo = (Em*R1*R2)/(Rv*(R1+R2) + Em
Emm = Ein*Rp/(Rp+R1)

Vo = (Ein*Rp*R1*R2)/(Rv*(R1+R2)*(Rp+R1)) + Ein*Rp/(Rp+R1)

Vo = Ein*Rp * (R1*R2/(Rv*(R1+R2)*(Rp+R1)) + 1/(Rp+R1))

How do I have to continue to get
Vo = Ein * R2 / R1+R2
?

Hello again,

Thanks for your question.

Your first equation was missing a close paren and should be:
Vo=(Em*R1*R2)/(Rv*(R2+R1))+Em

but i think you followed it through ok anyway so no big deal there.

As to your last expression:
Vo = Ein*Rp * (R1*R2/(Rv*(R1+R2)*(Rp+R1)) + 1/(Rp+R1))

recall that Rp is the parallel combo of R2 and Rv which is:
Rp=R2*Rv/(R2+Rv)

So just replace all your Rp with that last Rp and then simplify. It takes a bit of doing but it works out well

 

You need only three equations.

V = Vin x R2 / (R1 + R2)

Vmeter = Vin x Rp / ( R1 + Rp)

1/Rp = 1/R2 + 1/Rmeter

where,
V = corrected voltage
Vin = Voltage input to R1
Vmeter = measured meter reading
Rmeter = internal resistance of meter
Rp = parallel combination of R2 and Rmeter

 

In the OLD days we often used analog voltmeters, I still do. These had an interesting specification called
Ohms per Volt. My Triplet 630-PL, and most others, were 20,000Ohm/V on DC and 5,000Ohm/V on AC.
Basically this means the total meter resistance is:
50KOhm on the 2.5V DC range
5MOhm on the 250V DC range
Most modern digital meters have a fixed resistance on all ranges:
Usually 1MOHM or 10MOhm.

redrok

 

You need only three equations.

V = Vin x R2 / (R1 + R2)

Vmeter = Vin x Rp / ( R1 + Rp)

1/Rp = 1/R2 + 1/Rmeter

where,
V = corrected voltage
Vin = Voltage input to R1
Vmeter = measured meter reading
Rmeter = internal resistance of meter
Rp = parallel combination of R2 and Rmeter

Hi,

Do you mean three equations to DERIVE the required equation?

 

In the OLD days we often used analog voltmeters, I still do. These had an interesting specification called
Ohms per Volt. My Triplet 630-PL, and most others, were 20,000Ohm/V on DC and 5,000Ohm/V on AC.
Basically this means the total meter resistance is:
50KOhm on the 2.5V DC range
5MOhm on the 250V DC range
Most modern digital meters have a fixed resistance on all ranges:
Usually 1MOHM or 10MOhm.

redrok

I have a small analog Triplet too from the 1970's. Havent used it in a long time now.

 

In the OLD days we often used analog voltmeters, I still do. These had an interesting specification called
Ohms per Volt. My Triplet 630-PL, and most others, were 20,000Ohm/V on DC and 5,000Ohm/V on AC.
Basically this means the total meter resistance is:
50KOhm on the 2.5V DC range
5MOhm on the 250V DC range
Most modern digital meters have a fixed resistance on all ranges:
Usually 1MOHM or 10MOhm.

redrok

For those wondering where the total resistance numbers above come from, these meters used a d'Arsonval meter movement as the indicating mechanism and the amount of meter deflection was proportional to the amount of current flowing in the movement's coil which was, in turn, proportional to the resistance of the coil.

Electrically, if you were designing a meter based on such a movement, you therefore needed to know the resistance of the movement coil and the amount of current that would result in full deflection of the movement needle.

So what is 20,000 Ω/V?

Well, flip it over and you have 1 V / 20,000 Ω or 50 μA. This is the full-scale current, Ifs, of the movement. So if you want a meter that deflects full scale for, say, 25 V, you need the total resistance to be

Rtot = Vfs / Ifs = Vfs · (1/Ifs)

and 1/Ifs is 20 kΩ/V.

So you would need

Rtot = 25 V * 20 kΩ/V = 500 kΩ

Now, we don't know the resistance of the coil and the information given doesn't tell us. But whatever it is, we would just subtract it from 500 kΩ to determine the series resistor that we would switch into the circuit to get our 25 V scale.

 

Hi,

Do you mean three equations to DERIVE the required equation?

Yes.

The equation we are trying to derive is:

Vcorrected = Vmeter + Vmeter x R1 x R2 / Rmeter / (R1 + R2)

The equations are:

(1) Vcorrected = Vin x R2 / (R1 + R2)

(2) Vmeter = Vin x Rp / ( R1 + Rp)

(3) 1/Rp = 1/R2 + 1/Rmeter

Substitute Vin from (2) in (1)

Vcorrected = Vmeter x (R1 + Rp) x R2 / (R1 + R2) / Rp

Substitute Rp = R2 * Rmeter/ (R2 + Rmeter)

Now we have one equation with all the known values.
It takes some leg work but you can arrive at the final equation.

 

Get a digital meter and you will never need to concern yourself with loading effect.

 

Get a digital meter and you will never need to concern yourself with loading effect.

Welcome to AAC.

I had times in the past when i had to use analog for different reasons.
One i remember was that i needed to have a measurement readout all the time (24/7) and did not want to keep running batteries down or have a meter that had to be plugged into the wall. We sometimes forget that most analog meters do not require power to run.
At other times the military specification was to use certain specific model meters that were traceable to NIST and they happened to be analog so any kind of new meter analog or not was not allowed.