input loading

What exactly is input loading? I am reading my way through the art of electronics and he mentions then starting on about page 44(maybe a bit earlier) but I can't find where the precisely defines it. I have an idea that it has to do with what load the input see's but thats about all... hence I need some better understand of why its important.

Thanks, AD

Reply to
Abstract Dissonance
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input impedance Input Signal -----\\/\\/\\/-----o------ Output Signal | \\ / output impedance \\ / \\ | GND

If the output impedance is high compared to the input impedance, then most of the signal, as seen at 'Output Signal' above, will get through.

If, on the other hand, the output impedance is very low, then not much of the signal will get through.

This is a simple consequence of the voltage divider relation:

vout = vin * Rout/(Rout + Rin)

Clearly, if Rout >> Rin, vout ~= vin, whereas if Rout == Rin, vout =

1/2 vin, and if Rout
Reply to
Bob Monsen

To make sure I'm clear..

C V R2 Vin ---| |----o-----/\\/\\/\\/\\-GND | | / \\ / R1 \\ / | GND

The "HP filter" is the voltage drop across R1 but I can also find it across R2 but only if R2 is infinite in resistance else it has some effect on the filtering. Hence one would want R2 to be much larger than the impedance of the HP Filter so that it does "load" it... else one has a different circuit "block" than the simple HP filter that is taught.

If R2 was 0 then R1 would be completely ignored and we'd just have a --| |---GND which would give V = 0 which is definately not what the circuit was ment to do... hence the circuit was loaded improperly.

So the reason for this stuff is that when one works with small blocks and "hooks" them up there are certain conditions for that so they will actually still be relatively distinct blocks.

i.e. say I am hooking up a low pass filter up to a high pass filter... in reality its some much more complex circuit when taken as a whole but as long as I don't "load" the first block wrong then they are relatively distinct and I can think of them as two distinct blocks.

C1 V1 R2 V2 R3 Vin ---| |----o-----/\\/\\/\\/\\---o---/\\/\\/\\/\\---GND | | | | / | \\ --- C2 / R1 --- \\ | / | | | GND GND

So, that is basicaly a high pass followed by a low pass... as long as there is no "loading"... i.e. R2 would have to be much larger than R1 so the high pass see's an infinite load(or no load?) and the low pass would have to see R3 as infinite too so it is not loaded...

but in reality, if the values were arbitrary then I'd have to worry about the whole circuit and analyze it as a whole because it might not be able to be broken down into two relatively seperate parts.

so in the first half the -3dB point is 1/(2*Pi*R1*C1) and for the second half it would be 1/(2*Pi*R2*C2)

but thats only if R2 is large which causes the low pass filter to be conditioned to some degree on R1(i.e., I can just choose R2 arbitraryily because I know about loading(I hope) and hence I will have to make sure to choose it large... which means I will have to choose C2 smaller so it will still be at the same -3dB point).

Is everything I said basicaly right?

Thanks, AD

Reply to
Abstract Dissonance

Depends on the frequency of interest. If it's audio stuff, then you may regard it as entirely resistive. If radio frequency, things become a little more tricky. You have to state your area of interest for a meaningful answer.


"What is now proved was once only imagin\'d" - William Blake
Reply to
Paul Burridge

I might be getting confused with input and output loading but I think I got it now. In chapter 2 of the Art of Electronics he mentions that transistors can be used to "nullify" the effects of loading between stages so I think I'm on the right track.


Reply to
Abstract Dissonance

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