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Why is input impedance is higher in Common-Emitter configuration as compared to Common-Base configuration as signal is applied on Base-Emitter junction in both Amplifiers. Not in both configurations the resistances r'e and r'b will be considered?

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Although in both the common base amplifier and in the common emitter amplifier (non-swamped: with a nonexistent or bypassed emitter resistor), the signal appears to be coupled to ground only across the BE diode. So intuitively, you might expect the same low impedance in both.

However, in the common base amplifier, there is a dynamic action going on which lowers the input impedance.

The action in the common base amplifier can be understood like this. Suppose that the input AC signal swings a little bit low. This increases the \$V_{BE}\$ voltage. The collector-emitter current is very sensitive to \$V_{BE}\$ so a small increase \$V_{BE}\$ causes a large increase in the current. But in this common base amplifier, the collector-emitter current goes through the input. So, the input pulls down the voltage slightly, and a lot more current pours out!

Since the current flowing out of the emitter reacts sensitively to the input, it means that the input stage behaves rather like a voltage source. The opposite (a current source) would not react to voltage changes at all: the same current would flow regardless of our attempts to shake the voltage this way and that.

And, of course, voltage sources are what? Low impedances!

So in summary, the common base amplifier has a kind of active voltage source behavior, according to how the collector-emitter current responds to changes in the applied input voltage, and this gives it a lower impedance than you would expect just from looking at it statically.

This behavior is absent in the common emitter, because the input interacts with the base current, not the collector emitter current. In the common emitter, we even get feedback-driven impedance raising action if we add an unbypassed emitter resistor.

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  • \$\begingroup\$ "The collector-emitter current is very sensitive to VBE so a small increase VBE causes a large increase in the current." Upvoting because of THIS sentence. \$\endgroup\$ – LvW Dec 21 '14 at 9:47
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In common emitter, the current flowing into the base is \$(\beta+1)\$ times smaller than the current flowing into the emitter in the common base. If the voltage amplitude is identical, but the current is lower, then the resistance is higher with the same factor \$(\beta+1)\$. This is derived from \$i_e=i_c+i_b\$ and \$i_c = \beta \cdot i_b\$

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ How can we show it using h-parameters of Bjt transistor? \$\endgroup\$ – Ali Khan Apr 1 '13 at 15:06
  • \$\begingroup\$ You can replace \$\beta\$ with \$h_{fe}\$. It's been too long since I used them to be of more use at this moment. \$\endgroup\$ – jippie Apr 1 '13 at 15:16
  • \$\begingroup\$ The input resistance at the base node (emitter grounded) is r,in=hie and at the emitter node (base grounded) r,in=1/gm=hie/hfe (gm:transconductance). \$\endgroup\$ – LvW Dec 21 '14 at 12:13
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Another way of thinking about it:

Common-base The "input" is connected to the emitter and, the "input" has to supply the signal current that goes through the collector load. Let's say collector signal is 5mA peak to peak.

Common emitter The input supplies a fraction of the collector's 5mA peak to peak because BJTs have current gain in this configuration - typically in the order of 100 but can be as low as 5 and as high as 500.

If you looked at the signal current flowing in the base when it is a common base configuration you will see 5mAp-p/100 = 50uAp-p

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  • \$\begingroup\$ @ Andy can you more elaborate on your answer? \$\endgroup\$ – Ali Khan Apr 2 '13 at 11:42
  • \$\begingroup\$ @AliKhan I think I've said enough that needs to be said without repeating. Is there one particular area you don't understand? \$\endgroup\$ – Andy aka Apr 2 '13 at 12:03
  • \$\begingroup\$ Can you define it in h-parametrs? \$\endgroup\$ – Ali Khan Apr 2 '13 at 12:14
  • \$\begingroup\$ @AliKhan - your question asked about input impedance being higher in common emitter and I have answered the question by stating that the current taken from the input in common-base is higher. Ohms law therefore tells us that Zin is lower on common-base. \$\endgroup\$ – Andy aka Apr 2 '13 at 12:18
  • \$\begingroup\$ But you did't define it convincingly. I am asking the reason to prove. \$\endgroup\$ – Ali Khan Apr 2 '13 at 12:25

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