Understanding a technique for increasing input impedance in a circuit

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Hassan mahmoud

Joined Jan 23, 2016
19
I have struggled to understand the following paragraph in a paper:
input biasing was accomplished with two back-to-back diodes to V bias through a 100kΩ resistor,Rb, at DC and provide a path for the amplifier’s input bias current in a similar scheme to [3], but with the addition of a second diode for protection and clamping. To mask the diode’s parasitic capacitance and conductance,Cb(2.2μF)bootstraps the input for input frequencies higher than1/2πRbCbHz, thus preserving the amplifier’s high input impedance while achieving lower noise levels than what is possible with a purely resistive bias.

And what I understand so far is that bias circuit adds an addition, current yield to drop in the input impedance. what I cannot understand is that how this capacitance Cb will preserve amp. high input impedance? Also, which diode is referred to as a second diode? and how is it used for protection and clamping?
 

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MrAl

Joined Jun 17, 2014
7,672
Hi,

Just noticed this thread because it just came up on the bottom of the screen but it's an interesting question i think anyway for any era.

I think the answer is that positive feedback can be used to increase input impedance due to the fact that the input resistor current can be reduced with the introduction of positive feedback. It of course has to be done right so that nothing oscillates, but that's usually not extremely hard to achieve.

For an intuitive view, the way this works is as follows.
Say we have a single resistor, 1k Ohms, and we hold it in the horizontal position so we have one lead on the left and one lead on the right.
Now, ground the right lead, apply 10 volts to the left lead. What is the input current. Of course Ohm's Law app[lies so we see 10/1000=0.010 amps current. Now, calculating the resistance (this is important to do as we'll see shortly) we see we have 10/0.010=1000 Ohms. We expected that. But now watch what happens next.
Now we remove the ground connection on the right lead, and apply +9v to that lead. Now we have +10v on the left lead and +9v on the right lead. So what is the current now. Well, 10 minus 9 is just 1 volt, and with 1 volt across 1k we get a mere 0.001 amps, and calculating the resistance we again get 1/0.001=1000 Ohms.
However, what is the input impedance seen by the 10v source now. Well the 10v source is still 10v, and the input current is now reduced to 0.001 amps, so we see 10/0.001=10000 Ohms (10k). So we see that the input impedance went UP by a factor of 10 times!
That's how the positive feedback works although the +9v we applied is applied automatically by the circuit itself in order to get that impedance as seen by the input and that happens though the proper design of the circuit which sometimes just means adding a resistor or two to the circuit.

So in the end positive feedback gave us an input impedance 10 times that as we saw without that feedback.
 
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