In this opamp circuit, how do I determine when the diodes tu

M

MRW

Guest
Hello everyone! Please teach me how to figure this out. I don't know
how.

This is the circuit in question:
Pic 1: http://i9.tinypic.com/4xny1ph.jpg
Pic 2: http://i9.tinypic.com/4xxdxts.jpg

How do I determine what Vin value turns on/off the diodes (D3 & D4)? I
have trouble deriving a general equation.

Thanks again!

M
 
"MRW" <mr.whatever@gmail.com> wrote in message
news:1189796564.781987.218400@y42g2000hsy.googlegroups.com...
Hello everyone! Please teach me how to figure this out. I don't know
how.

This is the circuit in question:
Pic 1: http://i9.tinypic.com/4xny1ph.jpg
Pic 2: http://i9.tinypic.com/4xxdxts.jpg

How do I determine what Vin value turns on/off the diodes (D3 & D4)? I
have trouble deriving a general equation.
Click to expand...
The circuit is fairly easy to analyze if you ignore (i.e. remove) C1 and you
treat the op amp as being ideal An ideal op amp has infinite gain and
infinite input impedance. Thus an ideal op amp has no input currents.
This follows from the infinite input impedance. If the output is not saturated
then the + and - inputs have the same voltage. This follows from the infinite
gain of an ideal op amp. Any non zero difference in the input voltages
will drive the op amp input saturation. Usually op amps are used with
negative feedback so that part of the output voltage is feed back into the
negative input. The output voltage goes up or down until the voltage at
the negative input matches the positive input. Your circuit has negative
feedback either via D3 or via D4 and R18 depending upon the values of
Vin and V2.

Analysis:

The voltage at the + input to the op amp is V2. This is due to there
being no voltage drop across R19 since an ideal op amp has no
input currents.

The voltage at the - input will also be V2 since the negative feedback
will try to force the op amp's two input to match.

The current through R14 is (Vin - V2) / R14.

Since there is no input current into the - input to the "ideal" op amp then the
feed back from the output must provide a path for the current through R14.

There are two possible feedback paths from the output of the op amp
to its negative input. Fortunately it is an either/or situation. Due to the
arrangement of the diodes, only one diode will be conducting at a time.
This makes analysis easier.

One path is simply through D3. Current can only flow through this diode
from right to left. Since this current must match the current through R14,
this only occurs when Vin is less than V2.

The second feedback path is through D4 and R18. The current in this
path only flows when Vin is greater than V2.
 
"MRW" <mr.whatever@gmail.com> wrote in message
news:1189796564.781987.218400@y42g2000hsy.googlegroups.com...
Hello everyone! Please teach me how to figure this out. I don't know
how.

This is the circuit in question:
Pic 1: http://i9.tinypic.com/4xny1ph.jpg
Pic 2: http://i9.tinypic.com/4xxdxts.jpg

How do I determine what Vin value turns on/off the diodes (D3 & D4)? I
have trouble deriving a general equation.

Thanks again!

M
Click to expand...
Where do the wires go off the top of the diagram and to the right? We need
the whole circuit.

Is it an AGC threshold amplifier?

You posted this to s.e.b back in January
http://tinypic.com/view/?pic=2vafpua
 
"Andrew Holme" <andrew@nospam.com> wrote in message
news:fcf3dr$gnq$1$8300dec7@news.demon.co.uk...
"MRW" <mr.whatever@gmail.com> wrote in message
news:1189796564.781987.218400@y42g2000hsy.googlegroups.com...
Hello everyone! Please teach me how to figure this out. I don't know
how.

This is the circuit in question:
Pic 1: http://i9.tinypic.com/4xny1ph.jpg
Pic 2: http://i9.tinypic.com/4xxdxts.jpg

How do I determine what Vin value turns on/off the diodes (D3 & D4)? I
have trouble deriving a general equation.

Thanks again!

M


Where do the wires go off the top of the diagram and to the right? We
need the whole circuit.

Is it an AGC threshold amplifier?

You posted this to s.e.b back in January
http://tinypic.com/view/?pic=2vafpua
Click to expand...
C1 was 10n (C7) in the original circuit.

I will assume this is the same circuit. So the output of this circuit
controls the gain of a voltage-controlled amplifier; the input comes from a
log response RMS level detector; and the current flowing down the wire which
goes off the top of the page represents the threshold at which AGC begins to
act. V2 was ground in the original circuit. It looks like you're
re-designing it to work from a single supply. I'm going to talk in terms of
a split supply circuit. So V2 is "ground."

When Vin is less than the threshold, VOUT rises. C1 limits the rate at
which it can rise. It stops rising when it reaches one diode drop above
ground. At this point it is clamped from rising further by D3. D4 is
reverse biased all this time, and the output (via R18) is zero. This sets
the VCA to maximum gain.

If Vin rises above the threshold, VOUT falls. The rate at which it falls is
limited by C1. As soon as the output voltage slews to more than one diode
drop below ground, D4 starts to conduct and the whole stage now functions as
an inverting amplifier with a gain of -R18/R14. The response is still
"smoothed" by C1. The resultant negative output controls the VCA gain, so
as to produce a constant output level.
 
On Sep 14, 6:54 pm, "Andrew Holme" <and...@nospam.com> wrote:
"MRW" <mr.whate...@gmail.com> wrote in message

news:1189796564.781987.218400@y42g2000hsy.googlegroups.com...

Hello everyone! Please teach me how to figure this out. I don't know
how.

This is the circuit in question:
Pic 1:http://i9.tinypic.com/4xny1ph.jpg
Pic 2:http://i9.tinypic.com/4xxdxts.jpg

How do I determine what Vin value turns on/off the diodes (D3 & D4)? I
have trouble deriving a general equation.

Thanks again!

M

Where do the wires go off the top of the diagram and to the right? We need
the whole circuit.

Is it an AGC threshold amplifier?

You posted this to s.e.b back in Januaryhttp://tinypic.com/view/?pic=2vafpua
Click to expand...
Hi Andrew,

I don't consider the top part that much of a big deal. It's pretty
much a variable voltage source that sets the threshold point either
lower or higher. I do have my equations for when D3 is conducting and
D4 is open, and when D4 is conducting and D3 is open. But I cannot
properly predict when D3 or D4 actually starts conducting. That is my
problem.

Do you think if I use the standard diode equation (I = Is * (e^(qV/kT)
- 1)) I can get closer to my answer? I was thinking that e, q, k, T,
and Is are assumed values, so I just need to set V = V2-VO and work
from there. :/

Thanks!
 
"MRW" <mr.whatever@gmail.com> wrote in message
I don't consider the top part that much of a big deal. It's pretty
much a variable voltage source that sets the threshold point either
lower or higher.
Click to expand...
I suggest thinking of it as a variable current source, since it feeds
current into the opamp summing node, which is always held at a fixed
potential.

When posting schematics, it is better to post too much rather than too
little. Giving context makes it a lot easier for people to understand the
circuit and there is less risk of cropping something important.

I do have my equations for when D3 is conducting and
D4 is open, and when D4 is conducting and D3 is open. But I cannot
properly predict when D3 or D4 actually starts conducting. That is my
problem.

Do you think if I use the standard diode equation (I = Is * (e^(qV/kT)
- 1)) I can get closer to my answer?
Click to expand...
Forget about the diode equation. Diodes conduct when the forward voltage
across them exceeds 0.7V - that's all you need to know for this problem.
It's a good enough approximation. The state of the diodes only depends on
VOUT. This voltage changes slowly. The timing depends on C1. These are
the equations you need:

Q = CV
Q = It

Rgds,
Andrew.
 
On Sep 15, 4:43 am, "Andrew Holme" <and...@nospam.com> wrote:
"MRW" <mr.whate...@gmail.com> wrote in message
I don't consider the top part that much of a big deal. It's pretty
much a variable voltage source that sets the threshold point either
lower or higher.

I suggest thinking of it as a variable current source, since it feeds
current into the opamp summing node, which is always held at a fixed
potential.

When posting schematics, it is better to post too much rather than too
little. Giving context makes it a lot easier for people to understand the
circuit and there is less risk of cropping something important.

I do have my equations for when D3 is conducting and

D4 is open, and when D4 is conducting and D3 is open. But I cannot
properly predict when D3 or D4 actually starts conducting. That is my
problem.

Do you think if I use the standard diode equation (I = Is * (e^(qV/kT)
- 1)) I can get closer to my answer?

Forget about the diode equation. Diodes conduct when the forward voltage
across them exceeds 0.7V - that's all you need to know for this problem.
It's a good enough approximation. The state of the diodes only depends on
VOUT. This voltage changes slowly. The timing depends on C1. These are
the equations you need:

Q = CV
Q = It

Rgds,
Andrew.
Click to expand...
Thank you Andrew! I will investigate further. I really want to
understand this. I've been looking at the same circuit for months, but
I have not completely understood it. Diodes and capacitors scare me
for some reason.
 
On Sep 16, 2:38?am, MRW <mr.whate...@gmail.com> wrote:
Diodes and capacitors scare me
for some reason.
===================================
Click to expand...
I'll have to remind you to cover your eyes if you ever visit my lab.
 
On Sep 16, 8:04 am, BobG <bobgard...@aol.com> wrote:
On Sep 16, 2:38?am, MRW <mr.whate...@gmail.com> wrote:> Diodes and capacitors scare me
for some reason.

===================================
I'll have to remind you to cover your eyes if you ever visit my lab.
Click to expand...
Hi BobG,

I don't mean it in that sense. I mean whenever I am shown a circuit
with anything other than a simple resistor network then I get
intimidated and my mind goes blank. It normally takes me a few minutes
to recover and work along the equations, but most often, I get stuck
with problems that have capacitors, inductors, diodes, transistors,
etc. It normally takes me a few days of flipping thru book's pages or
asking questions just to get a sense of it.
 
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