# BJT common base amplifier question

#### OkashiiKen

Joined Jan 24, 2007
3
I am trying to understand an answer a question from a recent lab examining BJT amplifiers, specifically a question about common-base amplifiers.

Question:
What is the effect of not bypassing the base resistors R1 and R2 in the schematic of the common base amplifier schematic attached?

I can see that given a periodic input (sinusoidal in our case), an increase in frequency will slowly begin to bypass the resistors R1, and R2. If the Capacitor C1 were not connected to ground, the base current would be lower and the gain would be smaller. Is this correct? I am having trouble understanding exactly what is supposed to happen.

Thanks.

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#### beenthere

Joined Apr 20, 2004
15,819
Hi,

R1 & 2 make a voltage divider that biases the transistor into conduction. Common base arrangements are for impedance matching, and have gains of less than 1. The variation in voltage from the signal on the emitter will influence conduction, and can even cut off the transistor if the emitter gets within .7 volts of the base. I don't think you'll see R1 & R2 getting bypassed - although you might expand on that so we understand what you mean.

#### OkashiiKen

Joined Jan 24, 2007
3
Thanks for the quick reply. This is a followup question after a recent lab I completed. I will try and explain why I think R1 and R2 will be bypassed.... As the frequency of the input is increased the impedance of Capacitor C1 will decrease. At sufficiently high frequency, this will appear as a short, bypassing the voltage divider between R1 and R2. I understand how R1 and R2 correctly bias the BJT into conduction, and what you described mostly makes sense. If the base emitter voltage drops below the turn on voltage (0.7V) for the base to emitter pn-junction, the junction will be reverse biased. The question of bypassing the resistors R1 and R2 is from my lab manual, but I assume they meant how would the circuit react if capacitor C1 was not connected to ground. Hope that makes things more clear.

Thanks again.

#### OkashiiKen

Joined Jan 24, 2007
3
After talking to a friend of mine, he said he believes that if you remove the capacitor C1 from the circuit, the gain realized at the output would decrease. Does anyone know why this would be the case? I am having trouble understanding why. Thanks.

#### DrNick

Joined Dec 13, 2006
110
The capacitor is acting as a DC block (I think). If it is not there, your Q-point could lose stability. If the Q-point loses stability, the amplifier could go in to oscillation, or do other things (that are not amplification, which is what we don't want in an amplifier).

#### JoeJester

Joined Apr 26, 2005
4,390
OkashiiKen,

Think about what happens when that capacitor (C1) is not in the circuit. Your base potential will increase and vary in phase with the input signal on the emitter. This requires more signal to overcome the .7 volts which manifests itself in the lower gain of the circuit [Vout / Vin]. This degenerative feedback is one reason you'll see a bypass cap on the emitter resistors in common emitter circuits with the input feeding the base.

If you still have it setup in the lab, you can easily measure Vin and Vout with and without that capacitor and see it.

Attached is a simulation of your circuit.

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#### ramji

Joined Nov 6, 2007
1
Hello,

The capacitor u've mentioned is a bypass capacitor used to remove the effect of voltage series feedback appearing across the parallel combination of R1 and R2. The voltage is in a diection to oppose the voltage across the base-emitter thereby reducing the collector current and hence the output voltage.

thanx

RAMJI

#### wiltec

Joined May 12, 2008
3
Notice that the base bias is taken from the same power as the collector. Any variations in supply voltage due to collector current change will affect base bias and change the operating point , however small. The capacitor connected to the base will filter most of the variation and maintain bias point. In addition any power supply ripple (which is another form of variation) will be amplified by the base as though the circuit were a common emitter amplifier.

#### Ron H

Joined Apr 14, 2005
7,014
Hi,

R1 & 2 make a voltage divider that biases the transistor into conduction. Common base arrangements are for impedance matching, and have gains of less than 1. The variation in voltage from the signal on the emitter will influence conduction, and can even cut off the transistor if the emitter gets within .7 volts of the base. I don't think you'll see R1 & R2 getting bypassed - although you might expand on that so we understand what you mean.
Just so that no one is left with erroneous information - Common base amplifiers have basically the same voltage gain as common emitter stages. The difference (and this is BIG) is that the input resistance of the common base is very low - around 40 ohms in the circuit shown. It is uncommon, but some sources do have low impedance (e.g., a speaker used as a microphone).
One advantage of the common base stage is that it has no Miller effect (assuming the base is properly bypassed), and so the bandwidth is potentially larger that one could get with a common emitter stage. Common emitter and common base can be combined to form a cascode amplifier, which has the input resistance of a common emitter stage, but has very little Miller feedback, which would otherwise limit the bandwidth of a common emitter stage. In the cascode stage, the output of the common emitter stage is basically current (due to the low input resistance of the common base stage), and that current is, of course the input to the common base stage, which has nearly unity current gain.