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coolyota:

--- Quote from: rstofer on January 09, 2019, 03:30:16 am ---1) you need a meter - say 0-1 mA full scale
2) you need to calculate the internal resistance by building a voltage divider with, say, a 1k divider resistor and a variable voltage source 0-2V (or 10k 0-12V).  Anything that is adjustable and can provide 1 mA while including the unknown internal resistor
3) set your variable power source to just move the needle to full scale and measure the output of the voltage source
4) calculate the internal resistance from the voltage divider equations by first calculating the total resistance from your measured voltage divided by 1 mA.  Then subtract off the divider resistor and what you have left is the internal resistance
5) now that you know the internal resistance, you can compute voltage divider resistors for various ranges like 0-1V, 0-10V, 0-100V and so on.

--- End quote ---

Correct me if I understood it.

Adjust power source until the DVM reads 1mA on the display. Then determine the RT by Vin / I.
Internal resistance = RT - R (in this case I picked a 10k Ohms Resistor).



rstofer:

--- Quote from: coolyota on January 09, 2019, 08:06:13 am ---
--- Quote from: rstofer on January 09, 2019, 03:30:16 am ---1) you need a meter - say 0-1 mA full scale
2) you need to calculate the internal resistance by building a voltage divider with, say, a 1k divider resistor and a variable voltage source 0-2V (or 10k 0-12V).  Anything that is adjustable and can provide 1 mA while including the unknown internal resistor
3) set your variable power source to just move the needle to full scale and measure the output of the voltage source
4) calculate the internal resistance from the voltage divider equations by first calculating the total resistance from your measured voltage divided by 1 mA.  Then subtract off the divider resistor and what you have left is the internal resistance
5) now that you know the internal resistance, you can compute voltage divider resistors for various ranges like 0-1V, 0-10V, 0-100V and so on.

--- End quote ---

Correct me if I understood it.

Adjust power source until the DVM reads 1mA on the display. Then determine the RT by Vin / I.
Internal resistance = RT - R (in this case I picked a 10k Ohms Resistor).

--- End quote ---

Yup!  That concept is all the experiment is supposed to show.  As additional information, you can review the schematics of the popular Simpson 260 VOM starting with the original version manufactured in 1940.  This is some serious history!  This schematic doesn't show the switch but it is pretty easy to step through the voltage divider chain for the various voltage ranges.

http://www.simpson260.com/downloads/downloads.htm

Look specifically at this version and you can see the protection diodes that are discussed in a note to the instructor in the lab manual.

http://www.simpson260.com/downloads/simpson_260-7_and_7m_schematic.pdf

There are applications where VOMs are still the preferred tool.  Think about the starter in a car and the internal condition of the battery.  Suppose the battery has developed a relatively high internal resistance.  When the starter cranks the voltage drops.  This is hard to see on a DMM but it's very easy on a VOM.  The meter will read 12V, like normal, and you could swear the battery is just fine.  Hit the starter while watching the needle and everything will be explained.

rstofer:
There is one other consideration:  The voltage source has an internal resistance and we didn't account for it.  Usually, this will be a small number of Ohms but it is not zero.  It might be essentially zero when compared to a 10k divider.

The higher the applied voltage and the higher the dropping resistor, the less important this internal resistance will be.  You can calculate the voltage source internal resistance by first measuring the open circuit voltage and then measuring the voltage with a known (and low valued) resistor.  This is an application of Thevenin's Theorem, also a topic for this experiment.

Maximum power transfer (P = I * E) will occur when the load resistor is exactly equal to the internal resistance of the source.  Also a result of Thevenin's Theorem.  And it might smoke the battery, do be careful!

khatus:

I have the book The art of electronics (3rd edition) . But I don't think it's a good book for theoretical analysis.it's more related to practical

rstofer:
I think 'practical' was the authors' intent.  They tend to not use deeper math and spend more time talking about actual circuits.  I have the book but I haven't done anything with it.

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