Re: Question about opa350 data***

Is it also an adder? i.e. if instead of 2.5V from the divider we apply
any voltage, how will that affect the formula?

....and what if after the divider (or thevenine equivalent) we have
a 60k resistor (which goes to the opamp minus) ?




Tom Bruhns wrote:
In addition to what Bill already told you, consider that the divider
between +5 and gnd, using two equal-valued resistors, is equivalent to
a 5V/2 (=2.5V) supply connected through an R/2 resistance. Then the
gain, just like other non-inverting op amp configurations, is
(practically) 1+Rf/(R/2). If R=Rf, then the gain is 1+2Rf/Rf = 3. The
2.5V Thevenin equivalent of the divider just sets a bias, and it should
be clear that if the input to the (+) terminal is 2.5V, and feedback
causes the (-) input to follow the (+) input so it's at (practically)
2.5V also. Thus there is no current in the Thevenin-equivalent R/2
resistance, and no current in Rf, and Vout = 2.5. So, Vout =
(1+2*Rf/R)*(Vin-2.5V)+2.5V. Sound like what Bill wrote? Hope so.
Sound like what you wrote for the Rf=R case: Vo = 3*Vin - 5? Hope so.
Note that yours can be re-written as (Vo-2.5) = 3*(Vin-2.5). Maybe
then the "amplifies around 2.5V" will be more obvious.

Cheers,
Tom


joakent@xxxxxxxxxxxxxx wrote:
On Oct 5, 2:48 pm, bill.slo...@xxxxxxxx wrote:
joak...@xxxxxxxxxxxxxx wrote:
I am looking at an analog opamp amplifier for a digital
scope and I was wondering if someone knows how
this opamp circuit works. I have captured the section
in question, of course there are more stages, but this
is the one which is the most interesting for me.

The circuit goes like this:

there is an op amp with a negative feedback loop through
a resistor Rf

then, the minus input is also connected to a resistor
divider which is connected from the positive 5v supply
to ground and both resistors are 1.8k, so the divider
is 2.5v.

the input goes into the plus input and the output is the
output of the opampTrivial - centres any amplifiction around 2.5V. Repost on
sci.electronics.basics nd see if you can find anybody with the patience
and spare time to provide chapter and verse.

One moment,
It is not completely trivial. Lets assume all R's are the same,
which in reality is almost the case, I have 1.8K for the resistor
divider, and Rf = 2.2K + RV=1K (final factory adjustment).
So, Kirchoff's current law says:

(5-V)/R - V/R + (Vo-V)/R =0
this simplifies to
Vo = 3V - 5
I can't see that this is amplifying around 2.5Volts ?
??


--
Bill Sloman, Nijmegen

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