Name: Vacuum Tubes and Semiconductors
Uploaded: 2019-07-19
Duration: 8 min 14 s
Description: Vacuum tubes were the first devices where the electronic discharge was realised. Further work on them gave birth to semiconductor electronics. Let's discuss...

Question

What do you mean by faithful amplification? How will you achieve it in a vacuum tube amplifier? Derive an expression for voltage gain of a triode amplifier.

Accepted Answer

Faithful amplification is the amplification of a signal that increases its amplitude without changing its shape. In faithful amplification, the shape of the output remains as the same of the input, but its magnitude increases.

Faithful amplification can be achieved using a triode amplifier.

A triode is a vacuum tube device that contains three diodes- cathode, anode and grid.

The above diagram represents the circuit of a triode amplifier. High tension battery $E_{p p}$ provides a positive voltage to the plate whereas the low tension battery $E_{g g}$ provides negative voltage to the grid. The grid in the circuit is biased with negative potential otherwise the circuit will act as a rectifier. Suppose $e_{g}$ is the weak signal that is to be amplified.

When there is no input signal, then $e_{g}$ will be zero and the grid potential will be equal to the potential provided by the battery $E_{g g}$ . However, when there is an input AC signal $e_{g}$ in the circuit, then its voltage fluctuates from positive to negative in one half of its time period, the grid voltage also fluctuates from positive to negative as well. This leads to the variation in the plate current. When this plate current $I_{p}$ passes through the load resistance $R_{L}$ , it causes an output AC voltage which is greater than the input AC voltage. In this way, faithful amplification is achieved in a vacuum tube amplifier.

Voltage gain of a triode amplifier-

Voltage gain or voltage amplification is defined as the ratio of output voltage to the input voltage.

$A_{v} = \frac{Output voltage}{Input voltage}$

We know that the output voltage, $v_{o} = ∆ I_{p} \times R_{L}$

$A_{v} = \frac{∆ I_{p} \times R_{L}}{v_{i}} - - - (i)$ where $v_{i}$ is the input voltage, $∆ I_{p}$ is a small change in plate current and $R_{L}$ is the load resistance.

We know that,

$∆ I_{p} = \frac{∆ E_{p}}{r_{p} + R_{L}}$

$∆ I_{p} = \frac{μ v_{i}}{r_{p} + R_{L}}$ (As $μ = \frac{∆ E_{p}}{v_{i}}$ ) where $μ$ is the amplification factor and $r_{p}$ is the plate resistance.

Putting this value in equation $(i)$ , we get,

$A_{v} = \frac{μ \times v_{i} \times R_{L}}{(r_{p} + R_{L}) \times v_{i}} A_{v} = \frac{μ}{1 + \frac{r_{p}}{R_{L}}}$

What do you mean by faithful amplification? How will you achieve it in a vacuum tube amplifier? Derive an expression for voltage gain of a triode amplifier.

Important Questions on Electron Emission

Learn and Prepare for any exam you want !