Question about a schematic

I was wondering if someone knows how the following circuit works:
http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF
I understand most of it, but have a few questions.
Firstly, I don't understand why we need the first diode
(the one which is connected between C and GND)
Then, why do we not use the 3 pins of the variable resistor, surely
we are meant to trim the voltage by "taking" only part of it, so for
this I would have though the two "ends" of the pot should be
connected to the diode and GND, the the output should be the
middle pin rather than the top pin ?
Tia

 

[John Popelish]

beginner@btopenworld.com wrote:

I was wondering if someone knows how the following circuit works:
http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF
I understand most of it, but have a few questions.
Firstly, I don't understand why we need the first diode
(the one which is connected between C and GND)

A capacitor cannot pass DC current. If the first diode was missing,
there would be a brief current out, and then the cap would charge up
with enough voltage to keep the other diode reverse biased over the
whole cycle. The first diode keeps the cap output side from going
more than one diode drop below ground during the negative half cycle,
so that once the wave starts to swing positive, it starts from that
voltage. This diode capacitor combination is called a voltage
doubler, because it produces and output that swings from zero to the
whole peak to peak voltage of the wave (twice the center voltage to
peak difference).

Then, why do we not use the 3 pins of the variable resistor, surely
we are meant to trim the voltage by "taking" only part of it, so for
this I would have though the two "ends" of the pot should be
connected to the diode and GND, the the output should be the
middle pin rather than the top pin ?
Tia

I might have used the pot as a voltage divider on the output, and
added a second capacitor across the while resistor, to give a variable
DC output.

Their way, the resistor and output capacitor form a voltage divider
with a variable time constant that is discharging the cap, pulse by
pulse.

The best method depends on what you want.

--
John Popelish

 

A capacitor cannot pass DC current. If the first diode was missing,
there would be a brief current out, and then the cap would charge up
with enough voltage to keep the other diode reverse biased over the
whole cycle. The first diode keeps the cap output side from going
more than one diode drop below ground during the negative half cycle,
so that once the wave starts to swing positive, it starts from that
voltage. This diode capacitor combination is called a voltage
doubler, because it produces and output that swings from zero to the
whole peak to peak voltage of the wave (twice the center voltage to
peak difference).

1.
Regarding this voltage doubler, can we say that if our power supply is
-5V to +5V than the output can be from 0V to 10V ?

2.
Can we say that the first diode is responsible for out output to start from
0V rather than from 0.6V

Then, why do we not use the 3 pins of the variable resistor, surely
we are meant to trim the voltage by "taking" only part of it, so for
this I would have though the two "ends" of the pot should be
connected to the diode and GND, the the output should be the
middle pin rather than the top pin ?
Tia

I might have used the pot as a voltage divider on the output, and
added a second capacitor across the while resistor, to give a variable
DC output.

Their way, the resistor and output capacitor form a voltage divider
with a variable time constant that is discharging the cap, pulse by
pulse.

The best method depends on what you want.

I forgot to mention this but I think you understood anyway, this circuit
is supposed to give a voltage output which is proportional to the sound
level. Given that, and to phrase is in your terms, what do we want ?

 

[John Popelish]

beginner@btopenworld.com wrote:

A capacitor cannot pass DC current. If the first diode was missing,
there would be a brief current out, and then the cap would charge up
with enough voltage to keep the other diode reverse biased over the
whole cycle. The first diode keeps the cap output side from going
more than one diode drop below ground during the negative half cycle,
so that once the wave starts to swing positive, it starts from that
voltage. This diode capacitor combination is called a voltage
doubler, because it produces and output that swings from zero to the
whole peak to peak voltage of the wave (twice the center voltage to
peak difference).

1.
Regarding this voltage doubler, can we say that if our power supply is
-5V to +5V than the output can be from 0V to 10V ?

It would if the opamp could produce a full supply output swing. But a
real amplifier can seldom do this. The output stage of a 741 op amp
can only approach within about 3 volts of the supply voltage, so you
will be lucky to get a +- 2 volt swing from it when it is powered by
+- 5. In this case, it is powered by a single 9 volt battery, so its
output can swing from about +3 to +6 volts. The output cap and diode
pair reference this swing from about -.6 to +2.4. Long after the 741
was designed, opamps have been made that have input and output stages
that work to one or both supply rails. The very common LM324 (quad)
and LM358 (dual) have inputs and outputs that function all the way to
the negative supply rail, and to within 1.5 volts of the positive
supply rail. So these can provide 7.5 volts of swing from a 9 volt
battery. There are cmos amplifiers that work with the full supply
voltage at both inputs and outputs.

2.
Can we say that the first diode is responsible for out output to start from
0V rather than from 0.6V

You could if it were a perfect diode (zero volts forward drop, no
reverse current).

Then, why do we not use the 3 pins of the variable resistor, surely
we are meant to trim the voltage by "taking" only part of it, so for
this I would have though the two "ends" of the pot should be
connected to the diode and GND, the the output should be the
middle pin rather than the top pin ?
Tia

I might have used the pot as a voltage divider on the output, and
added a second capacitor across the while resistor, to give a variable
DC output.

Their way, the resistor and output capacitor form a voltage divider
with a variable time constant that is discharging the cap, pulse by
pulse.

The best method depends on what you want.

I forgot to mention this but I think you understood anyway, this circuit
is supposed to give a voltage output which is proportional to the sound
level. Given that, and to phrase is in your terms, what do we want ?

The method shown gives a meter reading (which is the average of the
sound waveform) that gives a fast response to changing signals, but
doesn't show much imformationb about brief loud peaks. The extra
capacitor and pot approach I mentioned, acts more like a peak hold
that will register the loudest peak and fade it out slowly (depending
on the output RC time constant). Neither of these is very accurate
because of the diode voltage drops.

--
John Popelish

 

Take a look at:
http://members.tm.net/lapointe/Cockcroft_Walton.htm

Can't understand how this works, but thanks.

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

Nice link, lots of usefull circuits, no explanations as to how they work though.

What about this, eliminate the C and the first D, leave 2nd D and then
add a big C (electrolotic) from D to GND. Basically like in the very
basic power supplies which have one diode and C. Will be get a DC
voltage which is proportional to the sound level ?

 

[John Popelish]

beginner@btopenworld.com wrote:

Take a look at:
http://members.tm.net/lapointe/Cockcroft_Walton.htm

Can't understand how this works, but thanks.

Each end ot the transformer winding puts out a sine wave referenced to
the ground on the center tap of the winding. So both ends of the
winding do the same thing, but on opposite halves of the cycle.
Taking the top half...
The first series cap is just like the one on your opamp. The first
diode keeps the negative half cycle from going below ground, by making
the right side of the cap charge up positive with respect ot the left
side, so that when the wave turns around and starts positive, it
starts from ground, instead of a negative voltage. The second diode
dumps the peak positive voltage from this pushed up wave into the
right side of the first cap along the middle of the circuit (the one
that is connected to ground). That full peak to peak positive voltage
is then used as the reference point that the third diode uses as its
clamp voltage for the second capacitor in the top row, which passes
the same AC wave that went through the left one. but now the wave is
forced to swing from the positive voltage from the first center row
cap to the full peak to peak voltage more positive that that, etc.
etc.

The top and bottom rows both dump charge into and draw it from the
middle row of caps on alternate half cycles.

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

Nice link, lots of usefull circuits, no explanations as to how they work though.

What about this, eliminate the C and the first D, leave 2nd D and then
add a big C (electrolotic) from D to GND. Basically like in the very
basic power supplies which have one diode and C. Will be get a DC
voltage which is proportional to the sound level ?

That would work, using only the positive half of the sound wave. But
it will still waste the first diode drop before any positive signal
comes out. The capacitor will charge up on the biggest peaks and
discharge very slowly (set by what resistance you connect across it).

--
John Popelish

 

[John Popelish]

beginner@btopenworld.com wrote:
(snip)

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

Nice link, lots of usefull circuits, no explanations as to how they work though.

Once you have the basic black box concept of how an opamp works, most
of these are pretty obvious. Do you have a conversational
understanding of the concept of an opamp?

--
John Popelish

 

Yes, you are most likely right about that, it is so trivial they
omit explanations. I simply like explanations if only to
confirm what I think. Yes, the basic principle of the op-amp
with a ve- feedback is that the minus input follows the plus
input.

 

What is this beast? My guess is its an audio noise detector for intrusion
detection or the like, or a sound level meter (less likely).

Thanks, I sort of get it now, I mean with the capacitor and 2 diode's.
Firstly, I assume that the first diode can have any meaning only because
the op-amp has a ve- voltage and because it's o/p can go below GND.
Then, when the signal is -ve, the capacitor charges through D1. Then,
when the voltage becomes +ve, the capacitor conducts, but because
in the +ve cycle the current goes through R, it does not change it's
charge and what we get is the o/p voltage plus the volateg from the
-ve cycle which is still on the capacitor, and this is the doubling effect.
Is this about right ?

Regarding as to the purpose of this circuit, the answer is that it is some
exercise on the net to demonstrate how sound level decreases with
distance, and I think they conclude there that is follows the inverse
square law relationship. You can play with the link I provided by
subtracting the directories one by one, this will reveal all their files,
eventaully you can reach this
http://hop.concord.org/s1/mess/s1ma2dgrm.html#anchor884816
The reason I adopted this circuit is that I want do build a proper
sound level meter, so I though why not use this one (I found it
from yahoo). Do you think it will be OK ? The only other option
I believe is to use a proper audio amplifier ( preferabley in a discrete
component form) instead of an op amp, but I am not sure if we need
this.

 

[John Popelish]

beginner@btopenworld.com wrote:
(snip)

Can we actually say this: when we have +ve and -ve rails as we
normally do (with op-amps) the voltage is sampled during the
-ve cycle (when D1 is conducting), and in the +ve cycle the charge
on the capacitor is given a chance to discharge through R. If R was
too big then the circuit would get stuck on the highest peak.

I don't quite know what you mean by, "the circuit gets stuck". But if
R is infinity, then the charge that gets pumped into C by the first
diode is trapped there, and the output voltage just goes up starting
at zero, and back down during one cycle.

I think of the two diodes as having the following function. The cap
is like a bucket that is raised and lowered by the opamp output
voltage. The first diode represents the bucket going below the
surface of water, and the water spills into the bucket. The second
diode is like a tip mechanism that dumps the bucket little by little
as it is raised out of the water (emptying it through the resistor).
This is a very rough analogy, but gives the general sense of how the
doubler works.

--
John Popelish

 

[R. Steve Walz]

beginner@btopenworld.com wrote:

Take a look at:
http://members.tm.net/lapointe/Cockcroft_Walton.htm

Can't understand how this works, but thanks.
[]
Nice link, lots of usefull circuits, no explanations as to
how they work though.
-----------------

These guys are guessing.

The output stage is NOT a voltage doubler, though there is a slight
resemblance, this is merely a half-wave rectifier. The diode to ground
simply allows cap recharge.

The op-amp is merely an inverting amp and the gain is -1000. The 10K
to ground is merely the parallel of 10K and 10M for optimal common
mode biasing. The upper input is (-) and the lower is (+) (inverting
and non-inverting).

The input device looks like an electret mic which is why there is a
bias resistor from +9V and why there is a cap to isolate the offset
and pass only AC.

This is just a sound meter that relies on a -dB setting on the
voltmeter.
-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

 

Found this from yahoo:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/limiter.html#c2

Isn't it the same thing ?

 

These guys are guessing.

or maybee you are ?
read this
http://rf.rfglobalnet.com/library/ApplicationNotes/files/1/an956-4.pdf

 

[Robert Monsen]

I spent a few minutes modelling this circuit using spice.

The really odd thing about this circuit is the biasing seems wrong; the
circuit runs without a -V rail, but the V+ input is grounded. Seems like the
V+ rail should be pulled up to Vcc/2.

In addition, there doesn't seem to be any DC path for current to ground,
except through the input of the opamp. I'm not sure why, but this seems to
push the output to float up to about 2V even when the V- input is > the V+
input. I think it depends on the leakage of the V- input.

If the input amplitude is big enough, it just manages to push the V- input
below the V+ (which is at ground + some small amount) and generates a
positive spike on the output, because of the huge gain of the amp
configuration. The length of the spike is determined by how long the input
pushes V- below V+, and thus the 'duty cycle' is determined by the amplitude
of the wave.

I'm not sure why the diodes and variable resistor are there. They don't seem
to do anything important. It looks like somebody was trying to design a
charge pump. If so, it doesn't work as advertised. It actually attenuates
the output at 1k, which is probably an important frequency for a sound
detector.

The thing that bugs me about this circuit is the lack of proper biasing. I
think the duty cycle will depend on the leakage of the input, which isn't a
predictable parameter. It may be that for a given opamp, its always the
same, and so tuning the circuit will only have to be done once. However, it
may also make the output temperature sensitive.

Seems like a simple integrator would be a better solution, assuming that
this is a sound level meter...

Regards,
Bob Monsen

<beginner@btopenworld.com> wrote in message
news:bgn534$ct5$1@sparta.btinternet.com...

I was wondering if someone knows how the following circuit works:
http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF
I understand most of it, but have a few questions.
Firstly, I don't understand why we need the first diode
(the one which is connected between C and GND)
Then, why do we not use the 3 pins of the variable resistor, surely
we are meant to trim the voltage by "taking" only part of it, so for
this I would have though the two "ends" of the pot should be
connected to the diode and GND, the the output should be the
middle pin rather than the top pin ?
Tia

 

[John Popelish]

Are we looking at the same circuit?
http://hop.concord.org/s1/ext/s1epix/s1e2a.GIF

Robert Monsen wrote:

I spent a few minutes modelling this circuit using spice.

The really odd thing about this circuit is the biasing seems wrong; the
circuit runs without a -V rail, but the V+ input is grounded. Seems like the
V+ rail should be pulled up to Vcc/2.

I missed it the first glance, too, but there is a -9v supply in pin 4.

In addition, there doesn't seem to be any DC path for current to ground,
except through the input of the opamp. I'm not sure why, but this seems to
push the output to float up to about 2V even when the V- input is > the V+
input. I think it depends on the leakage of the V- input.

The + input is grounded, and the 10 meg feedback resistor pulls the -
output to ground also. This is probably too much resistance for the
bias current or a 741, but there is a path. At least the + input
should also have a 10 meg resistor to ground, instead of 10k to
balance the voltage drops caused by the input bias currents.

If the input amplitude is big enough, it just manages to push the V- input
below the V+ (which is at ground + some small amount) and generates a
positive spike on the output, because of the huge gain of the amp
configuration. The length of the spike is determined by how long the input
pushes V- below V+, and thus the 'duty cycle' is determined by the amplitude
of the wave.

It is just a linear amplifier with a gain of 1000.

I'm not sure why the diodes and variable resistor are there. They don't seem
to do anything important. It looks like somebody was trying to design a
charge pump.

A charge pump is exactly what it is. Otherwise known as a voltage
doubling rectifier.

If so, it doesn't work as advertised. It actually attenuates
the output at 1k, which is probably an important frequency for a sound
detector.

Certainly, when the pot is set to zero resistance it not only
attenuates the output but puts an AC short circuit to ground on the
opamp output. Not a particularly elegant design.

The thing that bugs me about this circuit is the lack of proper biasing. I
think the duty cycle will depend on the leakage of the input, which isn't a
predictable parameter. It may be that for a given opamp, its always the
same, and so tuning the circuit will only have to be done once. However, it
may also make the output temperature sensitive.

This is just not right.

Seems like a simple integrator would be a better solution, assuming that
this is a sound level meter...

An integrator does not measure average level but just low pass filters
a signal. I think an ideal rectifier followed by a peak hold circuit
might have been more useful, but this thing will provide a DC volt
meter reading that is roughly proportional to sound level (except for
the diode losses).
--
John Popelish