Discussion in 'Electronic Basics' started by Abstract Dissonance, Jan 1, 2006.

1. ### Abstract DissonanceGuest

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,

2. ### Paul BurridgeGuest

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

3. ### Bob MonsenGuest

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 << Rin, vout ~= 0 no matter what vin is.

Another way to state the situation where Rout isn't a lot larger than
Rin is that the input loading is large, and it 'attenuates' (i.e., kills)
the signal.

--
Regards,
Bob Monsen

Strange as it may seem, the power of mathematics rests on its evasion of all
unnecessary thought, and on its wonderful saving of mental operations.
- Ernst Mach

4. ### Abstract DissonanceGuest

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,

5. ### Abstract DissonanceGuest

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.