Using Op-Amp to drive Amp Gauge with Audio Signal

[Tom]

I currently have a small audio amplifier in a custom case I built. In
order for it to look cooler, I decided to find an old analog gauge to
add to the case. Well, my local surplus store had a REALLY cool
looking gauge that I HAD to have. The gauge is a 0 - 1mA analog
gauge, 130ohm internal resistance. It is a bit small in range. What
I would like for the gauge to do of course is move in step with the
audio input to the amp. I did not want to hook it to anything on the
amplification portion of the amp, as it might reduce/change some audio
quality. Someone suggested a great idea that I split out the input
signal (Note: only 1 channel would be needed) and "use an op-amp to
drive a dummy load with the gauge inline". The source of the audio is
a computer audio out jack.

This sounded like a great idea, so I started researching op-amps and
their use in circuits. I have learned quite a bit, but I am by no
means an electrical engineer, so I am turning to you for help. Here
are my current stumbling blocks.

1. What circuit to use? Voltage gain, Voltage to Current...??? (Note
I built a simple Voltage to current amp last night, The gauge went to
about .8mA and stayed there. There was no fluctuation that I could
see based on the audio input)

2. I gather that I am using the op-amp mainly to act as a high
impedance buffer only, I do not think I need any gain that I can think
of. Does this sound correct?

3. I would prefer using a single power supply (24V off of the amp
power supply). If I do this I would need to add a DC offset to the
signal. I would like to avoid this, as I do not want to get any DC
sent back to the main amp. I guess I could use a capacitor if need
be, but I would like to avoid it, so was going to use resistors to
split the 24V into +12V, and -12V to feed the op-amp so that it can
handle +/- swings. Perhaps add the DC offset after the op-amp?

4. I currently am testing this with a lm741 op-amp. Is the 741
bandwidth good enough for this (It only goes to 1kHz)? Should I be
using a different op-amp?

Any help or even a good "Right path" would reeally help me out. I am
having quite a bit of fun learning this, but a nudge in the right
direction could help.

Thanks,
Tom

 

[John Popelish]

I currently have a small audio amplifier in a custom case I built. In
order for it to look cooler, I decided to find an old analog gauge to
add to the case. Well, my local surplus store had a REALLY cool
looking gauge that I HAD to have. The gauge is a 0 - 1mA analog
gauge, 130ohm internal resistance. It is a bit small in range. What
I would like for the gauge to do of course is move in step with the
audio input to the amp. I did not want to hook it to anything on the
amplification portion of the amp, as it might reduce/change some audio
quality. Someone suggested a great idea that I split out the input
signal (Note: only 1 channel would be needed) and "use an op-amp to
drive a dummy load with the gauge inline". The source of the audio is
a computer audio out jack.

This sounded like a great idea, so I started researching op-amps and
their use in circuits. I have learned quite a bit, but I am by no
means an electrical engineer, so I am turning to you for help. Here
are my current stumbling blocks.

1. What circuit to use? Voltage gain, Voltage to Current...??? (Note
I built a simple Voltage to current amp last night, The gauge went to
about .8mA and stayed there. There was no fluctuation that I could
see based on the audio input)

2. I gather that I am using the op-amp mainly to act as a high
impedance buffer only, I do not think I need any gain that I can think
of. Does this sound correct?

3. I would prefer using a single power supply (24V off of the amp
power supply). If I do this I would need to add a DC offset to the
signal. I would like to avoid this, as I do not want to get any DC
sent back to the main amp. I guess I could use a capacitor if need
be, but I would like to avoid it, so was going to use resistors to
split the 24V into +12V, and -12V to feed the op-amp so that it can
handle +/- swings. Perhaps add the DC offset after the op-amp?

4. I currently am testing this with a lm741 op-amp. Is the 741
bandwidth good enough for this (It only goes to 1kHz)? Should I be
using a different op-amp?

Any help or even a good "Right path" would reeally help me out. I am
having quite a bit of fun learning this, but a nudge in the right
direction could help.

To deflect the meter with the average of the absolute value
of the signal (ideal rectifier), use either the inverting to
non inverting opamp configuration for a voltage gain more
than 1 (so you have a voltage divider from output to ground,
with the tap fed to the inverting input). But replace the
output to inverting input resistor with the AC terminals of
a bridge rectifier (could be made of 4 1N4148 diodes or a
packaged 1 amp or power rectifier bridge) Connect the meter
between the + and - terminals of the bridge, so that no
matter which way the opamp passes current through the
bridge, it always goes the same way through the meter.
Adjust the sensitivity by changing the grounded resistor
that connects to the inverting input.

 

[Tom]

To deflect the meter with the average of the absolute value
of the signal (ideal rectifier), use either the inverting to
non inverting opamp configuration for a voltage gain more
than 1 (so you have a voltage divider from output to ground,
with the tap fed to the inverting input). But replace the
output to inverting input resistor with the AC terminals of
a bridge rectifier (could be made of 4 1N4148 diodes or a
packaged 1 amp or power rectifier bridge) Connect the meter
between the + and - terminals of the bridge, so that no
matter which way the opamp passes current through the
bridge, it always goes the same way through the meter.
Adjust the sensitivity by changing the grounded resistor
that connects to the inverting input.

Thanks John,

Duh, a rectifier. That makes sense. A am a little confused as to
where it goes, based on your description. From what I understand, the
op-amp output goes through a voltage divider to ground. The tap on
this divider feeds the inverting input. I gather that the audio input
goes to the non-inverting op-amp input. (or does it go to the
inverting input with an inverting input resistor to set the gain?) I
am getting confused with -> "But replace the output to inverting input
resistor with the AC terminals of a bridge rectifier". Do you mean
the first resistor in the voltage divider, right before the tap? So
if I understand correctly op-amp-out -> rectifier(AC) -> tap ->
resistor -> ground. The tap feeds back to the inverting input?

Tom

 

[John Popelish]

Duh, a rectifier. That makes sense. A am a little confused as to
where it goes, based on your description. From what I understand, the
op-amp output goes through a voltage divider to ground. The tap on
this divider feeds the inverting input. I gather that the audio input
goes to the non-inverting op-amp input. (or does it go to the
inverting input with an inverting input resistor to set the gain?)

Either will work. connecting the signal to the non
inverting input will load the signal source the least, so
that might be best if you just want to borrow the signal
from some point in the amplifier with minimum chance of
changing anything. The signal will have to be ground
referenced (zero volts when no signal) not biased at some
DC, or you will have to add a capacitor to block the DC and
add a resistor to ground to bleed the cap voltage to zero on
the opamp side.

I
am getting confused with -> "But replace the output to inverting input
resistor with the AC terminals of a bridge rectifier". Do you mean
the first resistor in the voltage divider, right before the tap?

Yes. the one that normally connects between the output and
the inverting input. The current that would pass through
that resistor, instead passes through the bridge rectifier
and the meter movement. The bridge rectifier makes sure
that for both positive and negative currents, they all go
through the meter in the direction that will deflect it
above zero.

So
if I understand correctly op-amp-out -> rectifier(AC) -> tap ->
resistor -> ground. The tap feeds back to the inverting input?

Right. The opamp produces whatever output voltage it takes
to force the two inputs to have the same voltage. When the
input on the non-inverting input swings, say, positive, the
output must force current through the bridge and meter
movement and then through the grounded divider resistor,
till that resistor's voltage drop matches the input voltage.
The amount of current that takes for a given input voltage
depends on the value of that grounded resistor. For
instance, a 1k resistor will need a milliamp through it to
produce a 1 volt drop. And that milliamp will be delivered
by the output and passing through the meter before it gets
to the resistor.

 

[Tom]

Either will work. connecting the signal to the non
inverting input will load the signal source the least, so
that might be best if you just want to borrow the signal
from some point in the amplifier with minimum chance of
changing anything. The signal will have to be ground
referenced (zero volts when no signal) not biased at some
DC, or you will have to add a capacitor to block the DC and
add a resistor to ground to bleed the cap voltage to zero on
the opamp side.


Yes. the one that normally connects between the output and
the inverting input. The current that would pass through
that resistor, instead passes through the bridge rectifier
and the meter movement. The bridge rectifier makes sure
that for both positive and negative currents, they all go
through the meter in the direction that will deflect it
above zero.


Right. The opamp produces whatever output voltage it takes
to force the two inputs to have the same voltage. When the
input on the non-inverting input swings, say, positive, the
output must force current through the bridge and meter
movement and then through the grounded divider resistor,
till that resistor's voltage drop matches the input voltage.
The amount of current that takes for a given input voltage
depends on the value of that grounded resistor. For
instance, a 1k resistor will need a milliamp through it to
produce a 1 volt drop. And that milliamp will be delivered
by the output and passing through the meter before it gets
to the resistor.

Thanks again John,

Its seems to make sense. I will give it a try!

Tom

 

[Tim Wescott]

I currently have a small audio amplifier in a custom case I built. In
order for it to look cooler, I decided to find an old analog gauge to
add to the case. Well, my local surplus store had a REALLY cool
looking gauge that I HAD to have. The gauge is a 0 - 1mA analog
gauge, 130ohm internal resistance. It is a bit small in range. What
I would like for the gauge to do of course is move in step with the
audio input to the amp. I did not want to hook it to anything on the
amplification portion of the amp, as it might reduce/change some audio
quality. Someone suggested a great idea that I split out the input
signal (Note: only 1 channel would be needed) and "use an op-amp to
drive a dummy load with the gauge inline". The source of the audio is
a computer audio out jack.

This sounded like a great idea, so I started researching op-amps and
their use in circuits. I have learned quite a bit, but I am by no
means an electrical engineer, so I am turning to you for help. Here
are my current stumbling blocks.

1. What circuit to use? Voltage gain, Voltage to Current...??? (Note
I built a simple Voltage to current amp last night, The gauge went to
about .8mA and stayed there. There was no fluctuation that I could
see based on the audio input)

2. I gather that I am using the op-amp mainly to act as a high
impedance buffer only, I do not think I need any gain that I can think
of. Does this sound correct?

3. I would prefer using a single power supply (24V off of the amp
power supply). If I do this I would need to add a DC offset to the
signal. I would like to avoid this, as I do not want to get any DC
sent back to the main amp. I guess I could use a capacitor if need
be, but I would like to avoid it, so was going to use resistors to
split the 24V into +12V, and -12V to feed the op-amp so that it can
handle +/- swings. Perhaps add the DC offset after the op-amp?

4. I currently am testing this with a lm741 op-amp. Is the 741
bandwidth good enough for this (It only goes to 1kHz)? Should I be
using a different op-amp?

Any help or even a good "Right path" would reeally help me out. I am
having quite a bit of fun learning this, but a nudge in the right
direction could help.

Thanks,
Tom

Do a web search for "VU Meter". That's what you'd expect a meter
connected to the input would be reading, and you may even be able to
find one that's designed for a 1mA movement.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Tom]

Thanks again John,

Its seems to make sense. I will give it a try!

Tom

Just to let you know it worked great! I do have one concern that I
was wondering if I could get some advice on. I set up everything as
mentioned above. To get the +/-12V for the op-amp, I took my 24V
supply and split it with two resistors. The tap in the middle being
ground. I had to use that as ground for the audio ground and the final
output resistor. My concern is what problems could be encountered
having the audio ground hooked to this "floating ground" and at the
same time having the ground go to the main amp? Is there a way to
have the V- be ground without having to DC offset? Possibly the
bridge before the op-amp?

Tom

 

[John Popelish]

Just to let you know it worked great!

Excellent. Now we can move on to making the scale read in
decibels. ;-)

I do have one concern that I
was wondering if I could get some advice on. I set up everything as
mentioned above. To get the +/-12V for the op-amp, I took my 24V
supply and split it with two resistors. The tap in the middle being
ground. I had to use that as ground for the audio ground and the final
output resistor. My concern is what problems could be encountered
having the audio ground hooked to this "floating ground" and at the
same time having the ground go to the main amp? Is there a way to
have the V- be ground without having to DC offset? Possibly the
bridge before the op-amp?

The only risk I can think of is if turning the volume way up
bounces the 24 volt supply around, feeding another,
unintended signal into the circuit. But if the 24 volt
supply is stiff or regulated, that is not a problem. The
opamp has to be able to swing both ways to work with the AC
signal. So the simulated split supply is probably fine.

 

[Tom]

Excellent. Now we can move on to making the scale read in
decibels. ;-)


The only risk I can think of is if turning the volume way up
bounces the 24 volt supply around, feeding another,
unintended signal into the circuit. But if the 24 volt
supply is stiff or regulated, that is not a problem. The
opamp has to be able to swing both ways to work with the AC
signal. So the simulated split supply is probably fine.

Awesome,

Thanks for the help!

Tom

 

[Tom]

Awesome,

Thanks for the help!

Tom

I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.

Diagram at -> http://spilledwhine.com/pics/AmpCircuit.JPG

Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.

The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.

I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.

Can anyone help me out?

 

[John Popelish]

I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.

Diagram at -> http://spilledwhine.com/pics/AmpCircuit.JPG

Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.

The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.

I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.

Can anyone help me out?

I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.

 

[Tom]

I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.

Diagram at ->http://spilledwhine.com/pics/AmpCircuit.JPG

Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.

The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.

I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.

Can anyone help me out?

I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.

Thanks John,

I will give it a try.

Tom

 

[Tom]

I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.
John Popelish

Thanks John,

I will give it a try.

Tom

Hey John or anyone, if you are still there,

I have finally had a chance to start working on this again. I am a
bit confused to where you are suggesting the resistors go. I do not
have any resistors connected to the inverting input that you are
referencing. Everything else seems to make sense, just where the
resistors go. What size cap should be used for DC blocking?

Thanks,

Tom

 

[ehsjr]

Tom wrote:

I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.

Diagram at ->http://spilledwhine.com/pics/AmpCircuit.JPG

Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.

The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.

I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.

Can anyone help me out?

I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.
John Popelish

Thanks John,

I will give it a try.

Tom

Hey John or anyone, if you are still there,

I have finally had a chance to start working on this again. I am a
bit confused to where you are suggesting the resistors go. I do not
have any resistors connected to the inverting input that you are
referencing. Everything else seems to make sense, just where the
resistors go. What size cap should be used for DC blocking?

Thanks,

Tom

The R1/R2 divider junction is connected to the inverting
input through R3 in your schematic.

Ed

 

[Tom]

Tom wrote:
I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.
Diagram at ->http://spilledwhine.com/pics/AmpCircuit.JPG
Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.
The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.
I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.
Can anyone help me out?
I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.
--
Regards,
John Popelish
Thanks John,
I will give it a try.
Tom

Hey John or anyone, if you are still there,

I have finally had a chance to start working on this again. I am a
bit confused to where you are suggesting the resistors go. I do not
have any resistors connected to the inverting input that you are
referencing. Everything else seems to make sense, just where the
resistors go. What size cap should be used for DC blocking?

Tom

The R1/R2 divider junction is connected to the inverting
input through R3 in your schematic.

Ed

I am a very visual person, so let me see if I understand.

1. The resistor divider goes to the inverting input, (Thus the
inverting input gets half the total voltage)
2. The op-amp is powered with a single supply.
3. R3 goes between the the divider junction and inverting input (This
is where I am a bit confused still)
4. Remove the signal ground

Does a cap still need to be between the signal and Non-inverting
input, as I do not see where any DC offset would come from.

Thanks,

Tom

 

[ehsjr]

On Mar 5, 8:55 pm, Tom <[email protected]> wrote:

On Mar 5, 6:34 pm, John Popelish <[email protected]> wrote:

Tom wrote:

I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.

Diagram at ->http://spilledwhine.com/pics/AmpCircuit.JPG

Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.

The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.

I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.

Can anyone help me out?

I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.
John Popelish

Thanks John,

I will give it a try.

Hey John or anyone, if you are still there,

I have finally had a chance to start working on this again. I am a
bit confused to where you are suggesting the resistors go. I do not
have any resistors connected to the inverting input that you are
referencing. Everything else seems to make sense, just where the
resistors go. What size cap should be used for DC blocking?

Tom

The R1/R2 divider junction is connected to the inverting
input through R3 in your schematic.

Ed

I am a very visual person, so let me see if I understand.

1. The resistor divider goes to the inverting input, (Thus the
inverting input gets half the total voltage)
2. The op-amp is powered with a single supply.
3. R3 goes between the the divider junction and inverting input (This
is where I am a bit confused still)
4. Remove the signal ground

Does a cap still need to be between the signal and Non-inverting
input, as I do not see where any DC offset would come from.

Thanks,

Tom

Ok, since you are a visual person, here's an ascii schematic of
what you have at _now_, re-drawn to illustrate the divider that
you have connected at present to the inverting input through R3:


+-----+------------------------+
| | |
| | |\ |
| [R1] | \| +----------+
| | Signal Input---|+ \ | |
--- | R3 | }---[Br1]---+ [meter]
- B1 +---+---/\/\/--+------|- / | | |
--- | | ^ | | /| +----------+
- [R2] | | | |/ | |
| | | +----+ | |
| | | | | |
+-----+------------------------+ |
| | |
Signal Gnd-+ +----------------------+


John Popelish suggested adding a similar divider to the
non-inverting input, with a DC blocking cap in series
between the signal input and the non-inverting input of
the op amp.

Adding the divider (R4 & R5) and the blocking cap (C1)
would make it look like this:


+-----+---------------------+-------+
| | | |
| | [R4] |\ |
| [R1] | | \| +----------+
| | Signal Input-[C1]--+----|+ \ | |
--- | R3 | | }---[Br1]---+ [meter]
- B1 +---+---/\/\/--+-----------|- / | | |
--- | | ^ | | | /| +----------+
- [R2] | | | [R5] |/ | |
| | | +----+ | | |
| | | | | | |
+-----+---------------------+-------+ |
| | |
+ +---------------------------+
***

***Note that the signal ground has been removed from the
junction of R1/R2, as John suggested. He mentioned the
assumption that it is connected to the supply _somewhere_
You will have to supply that detail.

Both John and you would have to verify that the drawing
reflects what you have and John's suggested mods to be
sure it is accurate.

Finally, yes, you need the blocking cap.

Ed

 

[Tom]

Tom wrote:
Tom wrote:
I finished up the circuit for the VU meter. It worked great when I
hooked the inputs to my iPod, so this morning I began installing it in
the amp(I was supposed to be painting the house, and there was hell to
pay) Well lo and behold, it does not work. I was wondering if any one
could help me out? Below is a simple diagram I made in SPICE.
Diagram at ->http://spilledwhine.com/pics/AmpCircuit.JPG
Please note that the values are correct and the bridge is actually a
chip, but I did not find one in the library, so I just drew 4 diodes.
My meter is the between the bridge and resistor.
The problem I am having is that when I connect the signal ground to
the amp, it seems to remove my floating ground, so that the input to
the op-amp is now 27V, where as when it is hooked to the iPod, the
power is properly split +13.5/-13.5. I am sure it has to do with using
a virtual ground in this setup. I think the virtual ground is getting
shorted out to system ground.
I have been looking into using something like a MAX4242 Op-amp that
does beyond-the-rail. From the literature, it can be used for AC
without any bias. I am hoping that it could be the solution.
Can anyone help me out?
I think you can correct the problem by adding another p[air
of resistors in series, somewhat like the two that connect
to the inverting input, but connected to the non inverting
input. A pair of 100k resistors would work. Then add a DC
blocking capacitor between the signal input and the non
inverting input. Then eliminate the input signal ground
connection to the resistor divider. This assumes that one
side of the supply is already connected to signal common,
somewhere else. To simulate that, you will have to connect
your simulation source as the actual circuit connects it to
the supply.
--
Regards,
John Popelish
Thanks John,
I will give it a try.
Tom
Hey John or anyone, if you are still there,
I have finally had a chance to start working on this again. I am a
bit confused to where you are suggesting the resistors go. I do not
have any resistors connected to the inverting input that you are
referencing. Everything else seems to make sense, just where the
resistors go. What size cap should be used for DC blocking?
Thanks,
Tom
The R1/R2 divider junction is connected to the inverting
input through R3 in your schematic.
Ed

I am a very visual person, so let me see if I understand.

1. The resistor divider goes to the inverting input, (Thus the
inverting input gets half the total voltage)
2. The op-amp is powered with a single supply.
3. R3 goes between the the divider junction and inverting input (This
is where I am a bit confused still)
4. Remove the signal ground

Does a cap still need to be between the signal and Non-inverting
input, as I do not see where any DC offset would come from.

Tom

Ok, since you are a visual person, here's an ascii schematic of
what you have at _now_, re-drawn to illustrate the divider that
you have connected at present to the inverting input through R3:

+-----+------------------------+
| | |
| | |\ |
| [R1] | \| +----------+
| | Signal Input---|+ \ | |
--- | R3 | }---[Br1]---+ [meter]
- B1 +---+---/\/\/--+------|- / | | |
--- | | ^ | | /| +----------+
- [R2] | | | |/ | |
| | | +----+ | |
| | | | | |
+-----+------------------------+ |
| | |
Signal Gnd-+ +----------------------+

John Popelish suggested adding a similar divider to the
non-inverting input, with a DC blocking cap in series
between the signal input and the non-inverting input of
the op amp.

Adding the divider (R4 & R5) and the blocking cap (C1)
would make it look like this:

+-----+---------------------+-------+
| | | |
| | [R4] |\ |
| [R1] | | \| +----------+
| | Signal Input-[C1]--+----|+ \ | |
--- | R3 | | }---[Br1]---+ [meter]
- B1 +---+---/\/\/--+-----------|- / | | |
--- | | ^ | | | /| +----------+
- [R2] | | | [R5] |/ | |
| | | +----+ | | |
| | | | | | |
+-----+---------------------+-------+ |
| | |
+ +---------------------------+
***

***Note that the signal ground has been removed from the
junction of R1/R2, as John suggested. He mentioned the
assumption that it is connected to the supply _somewhere_
You will have to supply that detail.

Both John and you would have to verify that the drawing
reflects what you have and John's suggested mods to be
sure it is accurate.

Finally, yes, you need the blocking cap.

Ed

Wow! Thanks Ed, this really simplifies it for me. Good Work with the
ASCII drawings! Now it makes sence, adding half the voltage to both
sides of the opamp, thus providing the DC offset. Makes sense now.

Tom Kuhn