Chapter 3.4
Ohmmeters
• How can we measure resistance? An apparent simple way is to measure the current I,
through a resistor and the voltage V, across it. We can then use Ohm’s Law to
calculate V/I.
A
DC
V
Rx
• If both meters have an accuracy of ±x%, what is the error in this method? What about
loading? Recall that the above method assumes an ideal ammeter and voltmeter
• What are the internal resistances of the ideal ammeter and voltmeter? Answer: 0 and
  respectively
• We have non ideal meters, therefore the actual meter resistances would have to be
considered.
– Hence our interest in loading
• We also showed that the error is actually its smallest value at FSD.
– ± 2x% is in fact the minimum error.
• Can we create an instrument that can directly read resistance and avoid these
concerns?
• Consider the following circuit arrangement:
A
Rz
Rm
DC
RX
• Note that we have a DC power source in this case.
• If the voltage of the source is V Volts and the circulating current is I Amps then:
• If Rx is zero, i.e, we short circuit the probes and adjust Rz to give a reading of IFSD on
the meter,
V  IRtotal  I ( Rz  Rm  Rx )
• We can say:
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
1
© 2001
I  I FSD ;
V  I FSD ( Rz  Rm );
Rz 
V
I FSD
 Rm
• We call Rz the zero ohms adjust.
• At this point IFSD is passing through the ammeter.
• We mark this as the 0 Ohms position.
0
PMMC SCALE
50A
• With the probes open, what is the circulating current?
– 0 Amps
– This implies  resistance
• We mark this as the  Ohms position:

0
0A
PMMC SCALE
50A
• How do we determine the intermediate positions?
• Since ammeter scale is linear, we can chose other points, 1/5, 2/5, 3/5, 4/5 FSD etc
and calculate the corresponding resistance.
Example
If IFSD is 50A, Rm is 3k and V is 1.5V,calculate the resistance for the positions
previously identified.
Solution
What is Rz?
For this position, I = 50 A, therefore:
Rz 
V
I FSD
 Rm 
1.5
 3000  27k
50 10 6
For the 4/5 position, I = 40 A, therefore:
V  IRtotal  I ( Rz  Rm  Rx )
1.5
 37.5k
40 10 6
Rx  37.5  30  7.5k
( Rz  Rm  Rx ) 
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
2
© 2001
For the 3/5 position, I = 30 A, therefore:
V  IRtotal  I ( Rz  Rm  Rx )
1.5
 50k
30 10 6
Rx  50  30  20k
( Rz  Rm  Rx ) 
Similarly for the 2/5 and 1/5 positions we get that Rx is 45k and 120k respectively.
Our ohmmeter scale now looks like:
45 k
120 k

20 k
20A
30A
10A
7.5 k
40A
PMMC SCALE
0A
0
50A
Comments
• Is scale linear or not?
– Scale is non-linear
• Crowded at the high resistance end
– Where is 200k for example
• Called a back-off scale since zero is to the right.
• This type of meter is called a series-type ohmmeter.
• The ohmmeter has a power source. What happens if the source voltage changes? Let
us analyse this using the previous example, but the battery voltage is now 1.4V. What
is the true reading for the 20k position on the ohmmeter?
Scenario 1 - We don’t re-zero the meter.
• The 20k position is associated with 30A through the meter movement.
Rz  Rm  30k;
Rx 
•
•
•
•
•
1.4V
 30k  16.7 k
30A
What is indicated as 20k on the meter is actually 16.7k in reality.
The ohmmeter is over reading.
Scenario 2 - We re-zero the meter
The new value for Rz is:
Rz 
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
1.4V
 3k  25k
50A
3
© 2001
Rz  Rm  28k;
1.4V
Rx 
 28k  18.7 k
30 A
•
•
•
•
Here, what is indicated as 20k on the meter is actually 18.7k in reality.
The ohmmeter is again over reading.
It is clear that the source voltage is critical to the performance of the ohmmeter.
Commercial meters would have some type of voltage regulator to avoid this.
Alternative series ohmmeter
• Consider the following circuit.
R1
A
Rz
RX
Rm
DC
• Again Rz is the zeroing resistor and an additional resistor, R1, now acts as the current
limiter. Note that Rz is now in parallel with the movement. There are 2 values to be
calculated: Rz and R1.
• In order to design this ohmmeter, we need to decide ahead of time what resistance
represents a 1/2 scale reading. We call that value Rh. Thus by setting Rh = Rx we can
draw an equivalent circuit as follows:
RE
R1
A
Rz
Rh
Rm
DC
• We can say:
RE  R1  Rm RZ
• If we short the probes, we adjust Rz to make IFSD flow through the meter.
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
4
© 2001
• The total circuit resistance is now RE
• When we make Rx= Rh, IFSD/2 now flows through the meter. Remember that we
decided that Rh caused a half scale deflection.
• Therefore RE= Rh for a current of IFSD/2
V
;
2 Rh
I 
• Therefore , to produce FSD, the total current is
IT 
V
Rh
• Returning to the circuit:
IT
Im
IZ
A
RE
R1
Rz
Rh
Rm
DC
V
at I m  I FSD
Rh
I Z  IT  I m ;
V
IZ 
 I FSD
Rh
IT 
• Thus:
RZ I Z  Rm I FSD ;




Rm I FSD  Rm I FSD 
RZ 

V

IZ
 R  I FSD 
 h

• Since
Rh  RE  R1  RZ Rm ;
R1  Rh  RZ Rm ;
• We can now calculate R1.
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
5
© 2001
Example
• You are given an ohmmeter of the following design, with Rh = 2000:
R1
50
1mA
Rz
A
RX
Rm
3V
• Calculate
(1)
(2)
(3)
Rz and R1 ;
The value of Rz to compensate for a 10% drop in battery voltage; and
The scale error when measuring 2000  with a low battery as in (2).
Solution
(1)
V
 I FSD
Rh
3
IZ 
 1 10 3  0.5mA
2000
IZ 
RZ 
Rm I FSD 50  1mA

 100
IZ
0.5mA
R1  Rh  RZ Rm
100  50
R1  2000 
 1.97 k
150
(2) With the drop in battery voltage, the new Iz is:
V
IZ 
 I FSD
Rh
IZ 
2.7
 110 3  0.35mA
2000
Therefore the new Rz is:
Rm I FSD 50 1mA

 143
IZ
0.35mA
The new value of Rh can now be calculated:
RZ 
(4)
Rh  R1  RZ Rm ;
Rh  1.97 k 
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
143  50
 2.01k
193
6
© 2001
Therefore, when the scale indicates 2000, the true reading is really 2.01k.
Is this under or over reading?
2.0  2.01
% error (under) 
 100  0.5%
2.01
• Note this arrangement has a better tolerance to small voltage drops in the DC source
than the previous design.
EE11A Handouts 106757476
Prepared by: Mr. Fasil Muddeen
7
© 2001