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'[EE] Why the extra resistor in a lot of circuits I'
2007\03\12@171746 by Jake Bush

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Hello,
Wondering why I see an extra resistor in a lot of circuits. For example...

-------------------------------------------------------
      __________
  ----|                 |
      | Op Amp    |--------|
  ----|__________|     _ |__
  |                           |      |
  |                           | R1 |
  |                           | __ |
  | ________________|
                              _|__
                              |      |
                              | R2 |
         A                   | __ |
                               _ |_
                               \    /
                              Ground
-------------------------------------------------------
      __________
  ----|                 |
      | Op Amp    |--------|
  ----|__________|     _ |__
  |                           |      |
  |                           | R1 |
  |                           | __ |
  | ________________|

             B
-------------------------------------------------------

What's the difference between circuit A and B? There must be a reason
for putting the extra resistor. I know R1 and R2 make a voltage
divider. But doesn't just going from R1 back into the component do the
exact same thing?

Sorry if this is pretty remedial and the reason is just staring me in the face.
--
Jake E. Bush

"For small erections may be finished by their first architects; grand
ones, true ones, ever leave the copestone to posterity."
- From "Moby Dick" by Herman Melville

2007\03\12@173002 by Marcel Birthelmer

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On 3/12/07, Jake Bush <spam_OUTjake.e.bushTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

Jake,
in case B, there is no current flowing through R1, since the inputs of
an op-amp are in theory infinite-impedance. So basically, you've built
a follower (assuming + and - are set up right on the op-amp).
In case A, you're using the voltage divider relationship to divide the
output voltage by some value, and feed that back into the op-amp. This
causes the op-amp to try to compensate the output until the two inputs
are equal - so in effect, the output is now a constant multiple of the
other (non-feedback) input. Without R2 to ground, there's no voltage
scaling, and you get a factor of 1.
- Marcel

2007\03\12@173245 by Mike Hord

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It's only remedial if you SHOULD know the answer but don't.
Otherwise it's educational.

To find the answer, use the ideal opamp assumptions:
1.  No current will flow into either terminal of the opamp.
2.  The opamp will slew its output in an attempt to minimize
the voltage difference between the two input terminals.

OR, if you prefer:
1.  A virtual open exists for the input terminals (no current
flow).
2.  A virtual short exists for the input terminals (the voltage
at both terminals is identical).

Put a voltage at the "open" input in your drawing, and then
analyze the two circuits and see what the output will do.
After you've done that, you'll know more about opamps
than you do now.

Mike H.

On 3/12/07, Jake Bush <.....jake.e.bushKILLspamspam@spam@gmail.com> wrote:
{Quote hidden}

> -

2007\03\12@174218 by Jinx

face picon face

> In case A, you're using the voltage divider relationship to divide the
> output voltage by some value, and feed that back into the op-amp.
> This causes the op-amp to try to compensate the output until the two
> inputs are equal - so in effect, the output is now a constant multiple
> of the other (non-feedback) input. Without R2 to ground, there's
> no voltage scaling, and you get a factor of 1

Presumably Jake has the op-amp with the -ve terminal at the top. If
you have a zener as feedback (a to -ve i/p, k to o/p) with -ve to 0V
through a resistor you have a positive voltage reference, the output
following and buffering the zener. Check out NatSemi AN31

http://www.national.com/an/AN/AN-31.pdf

and others. Google op-amp circuits

2007\03\12@192746 by Harold Hallikainen

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flavicon
face
In teaching this stuff, I use the "theory of the happy op amp." The op amp
is "happy" when its two inputs are at the same voltage. So, assume they
are at the same voltage, then use Ohm's law to analyze from there. This
generally means determining the voltage at the non-inverting input, assume
the inverting input is at the same voltage (and both inputs draw no
current), then analyze from there. An example (instrumentation amplifier)
is here:
http://sujan.hallikainen.org/cuesta/et113/InstrumentationAmpAnalysis.pdf

Harold


--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2007\03\12@193824 by Marcel Birthelmer

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On 3/12/07, Harold Hallikainen <haroldspamKILLspamhallikainen.org> wrote:
> In teaching this stuff, I use the "theory of the happy op amp." The op amp
> is "happy" when its two inputs are at the same voltage. So, assume they
> are at the same voltage, then use Ohm's law to analyze from there. This
> generally means determining the voltage at the non-inverting input, assume
> the inverting input is at the same voltage (and both inputs draw no
> current), then analyze from there. An example (instrumentation amplifier)
> is here:
> http://sujan.hallikainen.org/cuesta/et113/InstrumentationAmpAnalysis.pdf
>
> Harold

Harold,
this approach is the way I was taught about op-amps also, but the one
thing that bothered me is that it did not address the reason that the
inverting and non-inverting inputs are different.
But other than that short-coming, this method of analysis works really well.
- Marcel

2007\03\12@195213 by David VanHorn

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>
>
> Harold,
> this approach is the way I was taught about op-amps also, but the one
> thing that bothered me is that it did not address the reason that the
> inverting and non-inverting inputs are different.
> But other than that short-coming, this method of analysis works really
> well.
> - Marcel


It works for the closed-loop circuits.

It gets a little interesting when you have to deal with real opamps with
phase shift, and limited output voltage and current, and so on, but it's a
good model.

2007\03\12@200328 by Harold Hallikainen

face
flavicon
face

{Quote hidden}

The op amp output is A*(Vni - Vi) where A is the open loop gain of the
chip, Vni is the voltage on the non-inverting input, and Vi is the voltage
on the inverting input. As A gets very large, to get a finite output
voltage, Vni-Vi has to approach zero. The "difference between the
inverting and non-inverting inputs" is which input voltage is subtracted
from which. To get a positive output, Vni > Ni , though this difference
may be in the millivolts or microvolts. For a negative output, Vni < Vi

Harold

--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2007\03\12@201853 by peter green

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face

> Harold,
> this approach is the way I was taught about op-amps also, but the one
> thing that bothered me is that it did not address the reason that the
> inverting and non-inverting inputs are different.
> But other than that short-coming, this method of analysis works
> really well.
you could explain that by saying it trys to get back into the "happy" state but if you get the inputs the wrong way round it will head off in the wrong direction to do that.

but imo you must emphasise that this is only a model and you must be ready when a particularlly bright student designs an unstable circuit based on the model (my electronics teacher at 6th form collage didn't and as a result my second year A level project never worked properly).

Op-amps are designed to be stable for feedback gains up to 1 but if you manage to get the feedback gain above that (say by adding another stage on the output like i did) your amplifier will most likely oscilate in the RF.

Engineers have to make simplifications to get the calculations done but they also must understand that those simplifications only apply to a limited set of circumstances.



2007\03\12@202406 by Marcel Birthelmer

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> The op amp output is A*(Vni - Vi) where A is the open loop gain of the
> chip, Vni is the voltage on the non-inverting input, and Vi is the voltage
> on the inverting input. As A gets very large, to get a finite output
> voltage, Vni-Vi has to approach zero. The "difference between the
> inverting and non-inverting inputs" is which input voltage is subtracted
> from which. To get a positive output, Vni > Ni , though this difference
> may be in the millivolts or microvolts. For a negative output, Vni < Vi
>
> Harold

Harold,
(assuming this was directed at me), of course I understand that NOW.
But in my first semester of circuits, no one mentioned that - everyone
always said "just assume they're at the same voltage at all times".
- Marcel

2007\03\13@015820 by Jake Bush

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Well as it turns out the answer to this is a bit above my head. But
that just means I have some topics to focus on and read up on. Thanks
go to all of you for answering.
--
Jake E. Bush

"For small erections may be finished by their first architects; grand
ones, true ones, ever leave the copestone to posterity."
- From "Moby Dick" by Herman Melville

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