The galvanometer as a meter, and the diode

Aim

To find the resistance of a galvanometer and to make it into an ammeter and a voltmeter, and to draw the characteristic of a diode.

A galvanometer is a very small meter

A galvanometer shows a deflection in proportion to the current through its coil. A good one deflects fully for a current as small as $I_g$, often $100$ to $500\ \mu$A. It has a resistance $G$ of tens or hundreds of ohm. Alone, it cannot measure a current of $1$ A or a voltage of $5$ V, because those would burn the coil. With one extra resistor, it can.

The figure of merit $k$ is the current per division, $k = \dfrac{I}{\theta}$. For a galvanometer with $I_g = 150\ \mu$A and 30 divisions, $k = 5\ \mu$A per division.

The resistance by the half-deflection method

Join the galvanometer in series with a cell $E$ and a large resistance box $R$, so that the current is almost the full-scale current. The deflection is $\theta$. Now join a shunt $S$ across the galvanometer. Change $S$ until the deflection is $\dfrac{\theta}{2}$, and do not change $R$.

With the shunt in place, half of the original current passes through the galvanometer. The working shows that

Because $R$ is large, $G \approx S$. The more exactly we want $G$, the larger we take $R$.

The readings. $E = 4.5$ V, $R = 30\,000\ \Omega$, and the full deflection of $30$ divisions. The shunt that halves the deflection is $S = 119.5\ \Omega$. So $G = \dfrac{30\,000 \times 119.5}{30\,000 - 119.5}\ \Omega$ $= 120.0\ \Omega$.

Raise $R$ from $30$ to $100$ kΩ. How does the shunt change? How close is it to $G$?

An ammeter from a galvanometer

An ammeter is joined in series with a circuit, so it must have a very small resistance. Put a small resistance $S$, a shunt, in parallel with the galvanometer. Most of the current flows through the shunt. At the full-scale current $I$ of the new ammeter, the galvanometer carries $I_g$ and the shunt carries $I - I_g$. Both have the same voltage:

The resistance of the ammeter is $\dfrac{GS}{G + S}$, which is a little less than $S$.

A worked example. $G = 120\ \Omega$ and $I_g = 150\ \mu$A. To read up to $I = 1.0$ A:

1. The shunt. $S = \dfrac{150 \times 10^{-6}\ \text{A} \times 120\ \Omega}{1.0\ \text{A} - 150 \times 10^{-6}\ \text{A}}$ $= 0.0180\ \Omega = 18.0\ \text{m}\Omega$.
2. The resistance of the ammeter. $\dfrac{120 \times 0.0180}{120 + 0.0180}\ \Omega = 0.0180\ \Omega$.

The shunt is a short, thick copper or manganin wire. A scale of $30$ divisions then reads $\dfrac{1.0\ \text{A}}{30} = 0.033$ A per division.

A voltmeter from a galvanometer

A voltmeter is joined across a part of a circuit, so it must have a very large resistance. Put a large resistance $R$ in series with the galvanometer. At the full scale $V$, the current is $I_g$:

The resistance of the voltmeter is $G + R = \dfrac{V}{I_g}$.

A worked example. For a range of $3.0$ V: $R = \dfrac{3.0\ \text{V}}{150 \times 10^{-6}\ \text{A}} - 120\ \Omega$ $= 19\,880\ \Omega$. The resistance of the voltmeter is $20\,000\ \Omega$. For $15$ V: $R = 99\,880\ \Omega$ and the voltmeter is $100\,000\ \Omega$.

Checking the meter. Join the new ammeter in series with a standard ammeter and a rheostat. Compare the readings at five currents. For a voltmeter, join it in parallel with a standard voltmeter. The difference between the two readings shows the error.

The characteristic of a diode

A diode is a p-n junction. It passes current easily when the $p$ side is joined to the positive terminal (forward bias), and hardly at all the other way (reverse bias). The characteristic is the graph of the current against the voltage.

The steps.

1. Forward bias: join a battery, a rheostat or potential divider, a milliammeter in series with the diode and a voltmeter across the diode alone. A resistor in series limits the current, so that the diode is not burnt.
2. Raise the voltage in steps of $0.05$ V from $0$ and read the current. Take more readings where the current changes fast.
3. Reverse bias: reverse the diode, use a microammeter, and raise the voltage in steps up to a few volts. A silicon diode has a leakage of only nanoampere, so a good meter is needed.

The readings. A silicon diode, forward bias, at room temperature.

The current is very small up to about $0.5$ V. Then it grows by about a factor of $3.6$ for every $0.05$ V. The straight part of the curve, extended down to the voltage axis, cuts it at the knee voltage, near $0.7$ V for silicon and near $0.3$ V for germanium.

The equation. For a junction diode, $I = I_s\left(e^{V/(nV_T)} - 1\right)$. Here $I_s$ is the reverse saturation current, $n$ is an ideality factor between $1$ and $2$, and $V_T = \dfrac{kT}{e} = 0.0259$ V at $300$ K. For $V \gg V_T$, $\ln I = \ln I_s + \dfrac{V}{nV_T}$. A graph of $\ln I$ against $V$ is a straight line, with the slope $\dfrac{1}{nV_T}$.

Two resistances. The static resistance is $\dfrac{V}{I}$. The dynamic resistance is $r_d = \dfrac{\Delta V}{\Delta I} = \dfrac{nV_T}{I}$, the slope of the characteristic at that point. At $0.70$ V and $3.5$ mA, $\dfrac{V}{I} = 200\ \Omega$ and $r_d = 11\ \Omega$. A diode is not an ohmic conductor, so the two are very different, and $r_d$ falls as the current grows.

Raise the temperature. What happens to the voltage at which the diode starts to conduct? Lower $n$ to $1.0$. What changes?

Precautions. Do not let the forward current pass the rating of the diode. Take the readings quickly, because a hot diode conducts more. Use the right polarity for the meters.

Practise with new numbers

Try it in Python

The first program makes the galvanometer into an ammeter and a voltmeter, and finds $G$ from the half-deflection readings.

The second program fits the straight line of $\ln I$ against $V$ for the diode, and finds the ideality factor, the saturation current and the resistances.

Common mistakes

• Putting the shunt in series. A shunt is in parallel with the galvanometer. A series resistor makes a voltmeter.
• Forgetting $G$ in $R = V/I_g - G$. The galvanometer already has a resistance.
• Using a small $R$ in the half-deflection method. The formula $G \approx S$ needs $R \gg G$.
• Connecting the voltmeter across the diode and the milliammeter together. It must be across the diode alone.
• Taking the dynamic resistance as $V/I$. That is the static resistance.
• Raising the forward voltage too fast. The current rises steeply and burns the diode.

Remember these

• $S = \dfrac{I_g G}{I - I_g}$ for an ammeter. $R = \dfrac{V}{I_g} - G$ for a voltmeter. $G = \dfrac{RS}{R - S}$ by half deflection.
• An ammeter has a very small resistance and a voltmeter has a very large one.
• A diode: $I = I_s\left(e^{V/(nV_T)} - 1\right)$. The knee is near $0.7$ V for silicon and $0.3$ V for germanium.
• $r_d = \dfrac{\Delta V}{\Delta I} = \dfrac{nV_T}{I}$ and $\ln I$ against $V$ is a straight line.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A galvanometer has $G = 50\ \Omega$ and $I_g = 2.0$ mA. What shunt makes it read up to $5.0$ A?

2. The same galvanometer is to read up to $10$ V. What series resistor is needed?

3. In the half-deflection method, why is $R$ taken large?

4. Why is the characteristic of a diode a curve, and not a straight line?

References

• NCERT, Physics Class XII, Chapter 4 (Moving Charges and Magnetism), the section on the moving coil galvanometer, ammeter and voltmeter, and Chapter 14 (Semiconductor Electronics), the sections on the p-n junction diode.
• NCERT, Laboratory Manual: Physics, Class XII, the experiments on the resistance of a galvanometer, its conversion to meters and the characteristics of a diode.
• OpenStax University Physics Volume 3, Chapter 9 (Condensed Matter Physics), the section on semiconductors and doping.