Solution to exercise in the book 'The Art of electronics' 3rd edition

[ajaykumarunni@gmail.com]

Hi All,

I am brushing up my knowledge in Electronics and I am following the book 'The Art of electronics' 3rd edition. Could someone please help me to find out the solution for the following exercise from the same book.

Exercise 1.9 , Page 10 The very high internal resistance of digital multimeters, in their voltage-measuring ranges, can be used to measure
extremely low currents (even though the DMM may not offer a low current range explicitly). Suppose, for example, you
want to measure the small current that flows through a 1000MΩ “leakage” resistance (that term is used to describe a small current
that ideally should be absent entirely, for example through the insulation of an underground cable). You have available a standard DMM, whose 2V dc range has 10MΩ internal resistance, and you have available a dc source of +10V. How can you use what you’ve got to measure accurately the leakage resistance?

[Ice-Tea]

You mention "when you have a DC source of +10V". You will understand better if you consider a very small current source (you are trying to measure current, right?).

[ajaykumarunni@gmail.com]

Thanks Ice-Tea, I am sorry, English is not my native language, I did not understand what you explained.
That is the complete exercise given in the book.
I would like to know answer to the part  "You have available a standard DMM, whose 2V dc range has 10MΩ internal resistance, and you have available a dc source of +10V. How can you use what you’ve got to measure accurately the leakage resistance?"

[MiDi]

That is quite simple:

10V to DMM +, DMM - to DUT +, DUT - to 10V source GND.

Then you can measure the voltage drop of internal 10MOhm in voltage range and with simple ohms law you calculate the current/resistance of DUT.

Unless you measured the value of the input resistance of DMM, the results are not expected to be accurate.

There are a couple of options to measure leakage current/resistance, couple are discussed in Measuring capacitor leakage

[Mrniceguy0o0]

DUT boards are used in automated integrated circuit testing where the term DUT stands for device under test, referring to the circuit being tested. A DUT board is a printed circuit board, and is the interface between the integrated circuit and a test head, which in turn attaches to automatic test equipment (ATE).

Just some fyi

[rstofer]

You are simply forming a voltage divider with a 1000 MOhm resistor in series with a 10 MOhm resistor.  The division is not exactly 100:1 but for my purposes here, it is close enough.  If the 1000 MOhm resistor is correct, I will read 10V / 100 or 0.1V on the meter.  Just put this number aside, it will be used to cross-check the calculated value.

For a more exact solution:

Assume R1 = 1,000,000,000 Ohms call it 1*10⁹ Ohms or 100*10⁷ Ohms
Assume R2 =     10,000,000  Ohms call it 1*10⁷ Ohms then add the resistors to get 101*10⁷ Ohms
Calculate I as (10V / 101*10⁷ Ohms = 9.9*10^-9 Amps
Calculate V (reading on the voltmeter) as I * R2= 9.9*10^-9 * 1*10⁷ = 9.9*10^-2 = 99 * 10^-3 = 0.099 Volts
Pretty close to 0.1V

[Shock]

Put in even simpler terms (I hope). You put a multimeter set to DC volts in circuit in series with the component under test. The multimeter will measure the voltage drop across it's own internal resistance (the only path current can travel is through the multimeter), by using the displayed voltage and internal resistance of the meter you can use ohms law to calculate the leakage current (the missing part of the equation).

https://www.eevblog.com/forum/beginners/is-there-any-practical-or-quick-n-dirty-method-to-measure-capacitor-leakage/