# Q meter or RLC Meter or Quality Meter Working, Construction, & calculation

(Last Updated On: September 12, 2021)

## Q meter: Q meter- The “Q” meter is also known as RLC meter, LCR Meter, or Quality meter which was developed in 1934 by William D. Loughlin. This is used to measure the quality factor of coils and the resistance, capacitance and inductance of an electric circuit at radio frequency. Quality factor is the ratio of inductive reactance to the effective resistance of the coil.

Q = XL / R
Now we know that we are measuring the inductive reactance and effective resistance at the radio frequency so inductive reactance will be equal to:

XL= ω° L

Q=(ω° L)/R

So this is defines as the quality factor of the coil. This is quality factor is also known as storage factor. So if we want to find the storage factor of any element which are storing energy like inductor, capacitor. So for them we are defining the storage factor it is the ratio of power dissipated to the power stored in the element.

Q= Power dissipated / Power stored in the element

The storage factor defines us the goodness and performance of the coil. It is kind figure of merit of the coil. Now a point will arise in our mind that why we are measuring at radio frequency why we are not measuring directly. The difference is that the value of effective resistance in a dc circuit and in ac circuit are different. So if we are using the same circuit for measuring the resistance or effective resistance in a dc circuit or ac circuit it will give us different value because the value of the resistance it depends upon on the frequency in the ac circuit. Instead of measuring resistance directly we will measure the resistance indirectly by first measuring the quality factor.

Q= (ω° L)/R

R=(ω° L)/Q

## Working Principle of the Q or Quality meter or RLC meter:

The Q meter or the RLC meter is based upon well-known characteristics of series resonant R, L, C circuit. The circuit through which we can understand the working of the Q meter is: In this circuit we have resistor, inductor and capacitor which are connected with the AC source in series. The voltage across the resistor and inductor is collectively represented by EL and the voltage across the capacitor is represented by Ec, an the current flowing in the circuit is Io. Now this is series resonance circuit so at the resonance frequency we have condition that:

If at resonant frequency the inductive reactance and the capacitive reactance are equal.

Xc=XL—— (A)

Now we know that:

XC=1/(2πfo C) —- (1)

XL= 2πfo L —- (2)

Now put the equation 1 and 2 in equation 1 we will get:

XC= XL

1/(2πfo C)= 2πfo L

By rearranging the equation we will get

fo2= 1/(4π2 LC)

fo= 1/(2π √LC)

So this is the value of the resonant frequency.

Now if we draw a phasor diagram for this circuit we know that the voltage drop across the resistor and source voltage are in same phase and inductor voltage will leading the current by 90° and in the case of capacitor the voltage will lag the current by 90°. The phasor diagram will show the voltage drop through each element and the current flowing through it.
So if we take current as reference phasor Io the voltage drop across the resistance will be in same phase and inductor lead the current and capacitor lag the current. Input voltage = E
Output voltage = EC
Current at resonance will be = Io= E/R
The ratio of the output voltage to input voltage will be equal to the ratio of the voltage drop across the capacitor which are Io XC and Io R which is the voltage drop across the resistor.

EC/E= (Io XC)/(Io R)

At resonance frequency we know that

XC= XL

So we can write it as:

EC/E= (Io XC)/(Io R) = (Io XL)/(Io R)

And we know that:

XL= ωo L
EC/E= (Io XC)/(Io R) = (Io XL)/(Io R) = (Io ωo L)/(Io R)

Now cutting the Io we will get:

EC/E= (ωo L)/R —– (3)

Now we know that the Quality factor can be given as:

Q=(ω° L)/R —— (4)

Both the equation 3 and equation 4 are equal. So we can also write the quality factor as:

Q = EC/E

EC=Q E

As EC is the voltage across the capacitor which is our output voltage so we can write it as:

Eo=Q E

So this equation show us that the output voltage will be Q times the input voltage. If we keep the voltage constant we can read the value of quality factor directly from the output voltage or if we read the voltage across the capacitor by connecting voltmeter and keep the input constant it will be gives us the output voltage which will be equal to the quality factor. This is the principal of working of instrument Q meter.

## Construction of the Q meter or RLC meter circuit:

Q meter or the RLC meter consists of the following components:

• Thermocouple voltmeter
• Electronic voltmeter
• Shunt resistance
• Capacitor

This is the circuit for the RLC meter. The coil whose quality factor we want to measure is connected between T1 and T2. Electronic voltmeter is connected parallel to the variable capacitor which will measure the voltage drop of the capacitor. So the principal is saying that we have series resonance circuit and in this circuit we are measuring the output voltage. Due to the shunt resistance emf will be induce in the circuit which will be measured through thermocouple voltmeter which is the input voltage. This input voltage will cause the current to flow in the circuit and due to that current at radio frequency, the series resonant circuit will be resonated and we will have:

XC = KL

We will adjust the frequency to this variable RF oscillator the frequency is adjusted so that we are getting the above condition and this condition we will get:

Eo = QE

## Measurement of the Quality factor:

1. First we will adjust the RF oscillator at the desired frequency
2. The tuning capacitor is adjusted to get the maximum voltage at the output
3. Under the resonance condition we have:
Eo=QE
4. The voltmeter across capacitor is calibrated to read the value of Q factor directly
5. The Q meter is also known as directly reading instrument.
6. In the given circuit the measured value of the Q factor is different from the true value of the Q factor. This will be due to the distributed capacitance of the coil and the shunt resistance.

Correction for shunt resistance:

As the shunt resistance is connected in series with the RLC circuit. So this shunt resistance will also be added to the effective resistance of the capacitor.

Q= (ω° L)/(R+ Rsh )

As we have study the circuit and we know that the shunt resistance is very low. If the value of the shunt resistance is comparable to the effective resistance it will affect the reading.

The true value of the quality factor can be given as:

Q= (ω° L)/R

The true value is equal to:

Qtrue= Q(measured ) ×( 1+ Rsh/R )

As we know that the Q(measured ) is less than the Qtrue value. If Q is very low then Qtrue– Q(measured ) is negligible for low Q coils.

For high Q coils the error will be more.

Correction for distributed capacitance:

The coil which is under the test between the terminal T1 and T2 and it also has distributed capacitance. Some capacitance will also be in the coil. This distributed capacitor will be added with the output capacitor. So the measured value of the Q factor will be smaller than the true value of the Q factor.

Qtrue = XL/R = XC/R

Qmeasured= Xc+Xcd/R

Qtrue = Qmeasured * (1 + Cd/c)

## Application of the Q meter or the RLC meter:

• It can measure the quality factor of the coil
• It can measure the effective resistance, inductance and capacitance
• Also we can measure the bandwidth and the frequency with the Q meter
• The characteristics can be measured
• It is used for the measurement of the bandwidth

## Important Questions answer related to the Q meter:

Now we will discuss several questions which will helpful in examination and gate test. These questions occur in gate test.

• A very low loss coil tested with Q meter and distributed capacitance of coil is found to be 820 pF. Resonance occurred at angular frequency ω = rad / sec. With a capacitance of 9.18 nF. The inductance of the coil?

The solution for this question is:

As we know that  • A Q meter having an insertion resistance 0.02 Ω is used to measure the resistance of the coil. Resonance occurs at ω =106 rad / sec with capacitance of 40 pF. The inductance of the coil is ?

The solution for this question will be: • A coil is tested with a series type Q meter resonance at particular frequency is obtained with a capacitance of 110 pF. When the frequency is double of the capacitance required for is 20 pF. The distributed capacitance of the coil in pF is ?

As we know that:

C1 = 110pF

C2= 20pF

Cd = C1 – n2C2 / n2 – 1

Where  A high Q coil having Cd is tested with the Q meter. First resonance at ω1= 106 rad / sec obtained with C1=990 pF. The second resonance ω2=2 × 106 rad / sec is obtained with 240 pF capacitance. The value of inductance in mH of the coil is ?

The solution to this question is:  