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Voltage gain of an amplifier is defined as

"magnitude of output voltage / magnitude of input voltage"

OR

"Vout(r.m.s)/Vin(r.m.s)"

OR

"change in output voltage/change in input voltage"

In the figure given below they say that the by using bypass capacitor C2 which is by passing R(E) increases the gain because now the input signal by passes R(E) and so the signal faces less resistance than if R(E) is not by passed.

But what i am not understanding is that when by pass capacitor is used then magnitude of the input signal is increased but at the same time magnitude of the out put signal will also increase.

And

If magnitude of input signal decreases then the magnitude of out put signal also decreases and as gain is the ratio of output to input signal so the gain will not remain constant whether R(E) is by passed or not ? as both output voltage and input voltage are increasing and decreasing simultaneously then how the voltage gain is increased by bypassing the R(E) ?enter image description here

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    \$\begingroup\$ Pardon? How do you get the idea that C2 increases the input signal? It doesn't. The gain is influenced by the ratio of the impedance on the collector and the impedance on the emitter. C2 lowers the emitter impedance, which makes the ratio higher - you get more gain. Not making this an answer because I don't have the math to explain this properly. \$\endgroup\$ – JRE Aug 19 '17 at 7:00
  • \$\begingroup\$ C2 increases the input signal because by using C2 we will have less voltage drop now as the resistance will be less now and the magnitude of the signal across the input will be larger then. \$\endgroup\$ – Alex Aug 19 '17 at 7:04
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    \$\begingroup\$ C2 does NOT change the input signal. You are wrong, but I'm going to have to let someone else explain it. \$\endgroup\$ – JRE Aug 19 '17 at 7:33
  • \$\begingroup\$ @JRE if i do ac analysis then will the presence of capacitor C2 will not affect the gain ? in books its written that presence of C2 ( by passing the R(E) ) will increase the gain. yes please explain the phenomenon. i am confused on it. \$\endgroup\$ – Alex Aug 19 '17 at 8:04
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    \$\begingroup\$ Of course C2 influences the AC gain. It just doesn't do it the way you are trying to explain it. \$\endgroup\$ – JRE Aug 19 '17 at 9:09
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Without the bypass capacitor:

Irrespective of the resistor from collector to Vcc, the signal voltage at the emitter is a little bit smaller than the signal voltage at the base - this is because that part of the circuit acts like an emitter follower.

Because emitter current can be said to equal collector current (with only a small error due to base current), any signal current flowing through RE also flows through RC (or RL in your example).

The implication of this is that the voltage gain of this circuit is simply RC/RE.

With a bypass capacitor:

Signal emitter current is increased due to the lower impedance of the capacitor and therefore collector signal current is increased and this develops more signal voltage across RC hence, gain increases.

The general case now is that gain = RC/XE.

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  • \$\begingroup\$ thanks for the answer. i want to understand that in terms of the definitions i wrote. RC/RE is not apealing to me. can you please explain it in terms of the definitions i wrote ? \$\endgroup\$ – Alex Aug 19 '17 at 8:35
  • \$\begingroup\$ question is that if gain is the ratio of out put to input voltage or change in output voltage to input voltage or peak to peak out put voltage to in put voltage then how the by pass capacitor parallel to R(E) increases the gain ? \$\endgroup\$ – Alex Aug 19 '17 at 8:38
  • \$\begingroup\$ The gain is the ratio of input to output voltage. When you put in C2, the output voltage will increase. Gain is measured by comparing the input and output voltage. Gain is designed by setting the ratio of Rc and Xe. \$\endgroup\$ – JRE Aug 19 '17 at 9:10
  • \$\begingroup\$ @Alex the main part of your question makes the incorrect assumption that the input voltage increases. This is not true. Regarding a defintion of gain all of the versions you state are fine but it doesn't get you to the heart of the problem and many engineers will use Rc/Re for gain in this type of circuit (for the reasons I've given). \$\endgroup\$ – Andy aka Aug 19 '17 at 9:13
  • \$\begingroup\$ @Andyaka thanks for correction. so the input voltage will not change but since its an ac voltage it will cause variations in input current. when there is by pass capacitor more current will flow and when there is no by pass capacitor less current will flow. when more current will flow Ic will be more and there will be more voltage across RC and because of that voltage gain which is Vout/Vin will be more. now am i correct ? \$\endgroup\$ – Alex Aug 19 '17 at 10:53
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These diagrams try to show in simple why the bypass capacitor increases the voltage gain.

Without the bypass capacitor:

The input signal is divided between Vbe junction and Re resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

Hence not the whole input signal is seen by the base-emitter junction.

The voltage across \$Vre\$ is

$$Vre = V_{in}\frac{r_e}{R_E+r_e} = 0.1 V_{in}$$

Tha output voltage is :

$$ V_{OUT} = - \alpha*I_E*R_C$$

And

$$ I_E = \frac{Vre}{r_e} = 0.01Vin$$

So the output voltage is

$$V_{out} =- \alpha*I_E*R_C = -1 *0.01Vin*1k\Omega = -10*Vin $$

Therefore, the voltage gain is:

$$A_V = \frac{V_{out}}{V_{in}} = - \frac{R_C}{r_e+R_E} * \alpha = - \frac{R_C}{r_e+R_E}*\frac{\beta}{\beta+1} \approx -\frac{R_C}{r_e+R_E} \approx- 10V/V$$

With a bypass capacitor:

schematic

simulate this circuit

As you can see now the whole the input signal is present across base-emitter junction. And the is why the voltage gain is larger.

$$V_{out} = -\alpha*I_E *R_C$$

$$V_{in} = I_E*r_e $$

And tha voltage gain is:

$$ A_V = \frac{V_{out}}{V_{in}} = \frac{-\alpha*I_E *R_C}{I_E*r_e} = -\frac{R_C}{r_e}\alpha \approx -\frac{R_C}{r_e} \approx -100V/V $$

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Putting a capacitor in parallel with a resistor doesn't just change the resistance of the resistor. At DC the capacitors look like open circuit, so there will be no effect on RE. Hence your bias levels will remain the same.

The frequency of your input signal will determine how much impedance C2 presents to the circuit. At DC it will present quite a bit and have negligible effect, at MHz the impedance will be probably not much (depending on the value of the capacitor).

If you do an AC analysis on the circuit in whatever your favourite spice simulator is you will find that the gain of the amplifier will change with frequency.

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  • \$\begingroup\$ if i do ac analysis then will the presence of capacitor C2 will not affect the gain ? in books its written that presence of C2 ( by passing the R(E) ) will increase the gain. \$\endgroup\$ – Alex Aug 19 '17 at 8:02
  • \$\begingroup\$ Yes, its presence will affect the gain in a frequency dependant manner, as I mentioned in my answer. As the frequency rises, the impedance of C2 will decrease, which will cause the combined impedance of RE and C2 to decrease, in turn increasing Vout. But it will not, as you have mentioned, increase the magnitude of the input. C2 does not completely or immediately bypass RE. \$\endgroup\$ – JCollins Aug 19 '17 at 8:09
  • \$\begingroup\$ what if the frequency of the source does not change. say its fixed at 1 Mhz or 2 khz etc? then will presence of capacitor will not change gain ? \$\endgroup\$ – Alex Aug 19 '17 at 8:12
  • \$\begingroup\$ yes, it will. the amount by which it changes will depend on the frequency of the input. if it's 1MHz, the gain will change more because of C2 than at 2kHz. \$\endgroup\$ – JCollins Aug 19 '17 at 8:25
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    \$\begingroup\$ You can't work with capacitors in amplifiers without considering frequency. The capacitor when AC signals are involved is essentially a frequency dependent resistor. So, if you really must disregard frequency, imagine the capacitor is just another resistor. The parallel resistance will be lower, less resistance, means more current. More current means bigger change in Vout --> bigger gain. Magnitude of Vin is fixed. \$\endgroup\$ – JCollins Aug 19 '17 at 8:55
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Me thinks the explanations are way over complicated. The emitter resister is there to provide bias for the transistor to operate. Resistors resist, right? But you want less resistance to the AC signal. The bypass (and by the way, it is not by pass, it is bypass) capacitor is there to what? BYPASS. What is being bypassed? The AC signal. The bypass capacitor is a low impedance path for the AC signal and BYPASSES the resistance of the emitter resistor and therefore increases the output signal. All the fancy mathematics notwithstanding.

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