Using Op-Amp to drive Amp Gauge with Audio Signal

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)

  1. 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?

  2. 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?

  1. 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

Reply to
Tom
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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.

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Regards,

John Popelish
Reply to
John Popelish

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

Reply to
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.

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Regards,

John Popelish
Reply to
John Popelish

Thanks again John,

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

Tom

Reply to
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.

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Tim Wescott
Wescott Design Services
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Tim Wescott

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

Reply to
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.

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Regards,

John Popelish
Reply to
John Popelish

Awesome,

Thanks for the help!

Tom

Reply to
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 ->

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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?

Reply to
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.

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Regards,

John Popelish
Reply to
John Popelish

Thanks John,

I will give it a try.

Tom

Reply to
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

Reply to
Tom

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

Ed

Reply to
ehsjr

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

Reply to
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

Reply to
ehsjr

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

Reply to
Tom

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