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I'm working on an audio crossover for a friend's stereo system. I have a passive design that uses a single 330uF in series with the positive audio line. It works like a charm , but I'd like to see if I can pull off an active filter instead.

I have a MAX410, MAX492, TLC7111 (which so far has produced the worst quality), as well as an ADA4895-1 (which I'm working on putting on a board to test). So far the MAX410 has produced the best results and sound (which is what I have currently hooked up). But with one issue....

It seems the volume is about 1\16th that of the input signal. I have tried at least a half dozen schematics\configurations with the same results. And this is a big problem as I need the input and output signals to match volume. But it's seeming like this may not be possible

Here is my schematic. VCC is +5V, R2, 3, and 4 are 10k (with the exception of R1 which is 1k), and C1\C2 are 0.1u:

High Pass Op Amp Schematic

And here is a picture of my breadboard using the MAX410 (Ignore the board in the upper left corner. It's NOT being used or at all active):

enter image description here

I have also tried running the op amp with both a regular ground as well as a virtual ground by using a 1k voltage divider; same result of the output being too quiet.

The output is to speakers. Currently I'm using headphones with a 23 ohm impedance. His system is a 4 ohm impedance.

EDIT: Please ask for any information that you need on top of the information I have provided. I have given everything that I currently have.

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closed as unclear what you're asking by Matt Young, Daniel Grillo, PeterJ, nidhin, Dave Tweed Apr 17 '15 at 21:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Are you sure you've got your inverting input (pin 2) correctly connected? It looks like the orange wire is going from pin 2 to row 10 near-side, but the pair of resistors is on row 11. \$\endgroup\$ – brhans Apr 16 '15 at 20:15
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    \$\begingroup\$ And that circuit needs either a good virtual ground or a split-supply. \$\endgroup\$ – brhans Apr 16 '15 at 20:16
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    \$\begingroup\$ What are you feeding the op-amp output into? What are your op-amp supply voltages? Also, please show your exact circuit diagram not some cut and paste from the web. \$\endgroup\$ – Andy aka Apr 16 '15 at 20:17
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    \$\begingroup\$ What's the load on the op-amp i.e. what is connected to the jack and what impedance is it? Please don't BS me if you want my help - THAT IS NOT THE EXACT CIRCUIT DIAGRAM - I see regulators and supply decouplers and THAT circuit WILL NOT run from a single 5V supply. \$\endgroup\$ – Andy aka Apr 16 '15 at 20:24
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    \$\begingroup\$ So we have an ambiguous picture, schematic with no component values, mention of components that seem to have nothing to do with said schematic, and a chain of comments that adds more confusion. This is a mess, voting to close until it's fixed. \$\endgroup\$ – Matt Young Apr 16 '15 at 20:51
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The MAX410 will not drive headphones (23 ohms or 32 ohms) as you expect - it's just an op-amp (not some small power amp) and might be expected to deliver a 2.3Vp-p output signal into a 2kohm load. It's got a short circuit current of 35mA so, you might get 20mA peaks with an output voltage of maybe 2Vp-p. That's a power of about 25 milli watts.

Also, the schematic as shown in your question will not work from a single 5V supply - you need a voltage midpoint to reference your input and output signal against. As it stands your output will be pityfully low and probably show signs of being half wave rectified against the 0V (most negative) rail.

Read the data sheet and please also provide a proper "full" circuit diagram if you are interested in other questions being answered.


EDIT section following disclosure by OP of a resistor value change.

Changing R1 to 1k from 10k will also not achieve what you want (irrespective of other errors). This will produce a gain of about 11 and when you say: -

I need the input and output signals to match volume

This contradicts R1 being 1k. In fact, with it at 10k you will have a pass-band gain of 2 so this might be another issue you wish to consider. Try removing R1 and splitting R3 into two resistors (20k each) to both supply rails.

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Something seems wrong to me - you mention that you are currently using a 330 uF capacitor.

What is the impedance of your load? Are you expecting to feed something like a speaker or headphones with this or is it feeding an amplifier?

For example, many audio amplifier inputs have an input impedance of 10 KOhms. Your 330 uF capacitor would give you a high-pass filter with a break point of about 0.048 Hz. So you obviously can't be working with a high impedance (10K) input.

That means that you may be trying to have your filter directly drive your load. That isn't a good idea from many respects - cost being one of the problems.

Why don't you more completely describe what you are trying to do. It's really hard for us to figure out what you are trying to accomplish, especially given the capacitor value that you mention.

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Your biggest problem is that you don't know what a virtual ground is, and you need one if you're going to power your op amp from a single 5 volt supply. Try

schematic

simulate this circuit – Schematic created using CircuitLab

with the proviso that, depending on the input impedance of the next stage, you may have to increase the value of C4.

As it stands, your circuit is attempting to produce a negative output from a negative input, and it cannot do that since its negative supply is ground. Technically, your op amp is attempting half-wave rectification, except that it gets more complicated than that due to feedback.

EDIT - Error in R1 connection - fixed.

Also, from comments in the OP - Even if you get everything right, you cannot drive a 23 ohm load with your op amps. They will not provide the current required, and will produce a low output voltage. Just like you're getting.

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  • \$\begingroup\$ But the reason I never did that this way was because I was under the assumption from some browsing on the subject that I needed a negative and a positive supply still for the initial op amp, which is a paradox of sorts \$\endgroup\$ – Dominic Luciano Apr 16 '15 at 20:51
  • \$\begingroup\$ No, you need dual supplies to allow the output to go both positive and negative. Op amps aren't magic. They can only produce an output within the limits of their power supplies (although driving inductors can produce subsequent higher voltages, but this does not apply to your circuit). And if you were under the assumption that you needed a both a negative and positive supply, why didn't you provide them? How can the need for two supplies (or a single supply modified as I've shown) be a paradox? The negative op amp input does not necessarily have to be tied to ground. \$\endgroup\$ – WhatRoughBeast Apr 16 '15 at 20:56
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    \$\begingroup\$ Right. And that's when you use a virtual ground such as I provided in my schematic. And the fact that you are unwilling to provide a negative supply for op amp A does not imply the existence of a paradox when you cannot provide a negative supply for op amp B. "I can't do A: therefor I can't do B" is not a paradox. \$\endgroup\$ – WhatRoughBeast Apr 16 '15 at 21:33
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    \$\begingroup\$ Please. Pay attention to my statement "Op amps aren't magic. They can only produce an output within the limits of their power supplies". Of course op amps can work without a negative supply. What they cannot do with a single supply is produce an output (at the op amp) above the power supply voltage or below ground. However, if you provide an intermediate virtual ground (as my schematic did), and capacitively couple the inputs and outputs, you can accept negative inputs and produce negative outputs. \$\endgroup\$ – WhatRoughBeast Apr 16 '15 at 22:45
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    \$\begingroup\$ Sorry. Since you don't understand capacitive coupling the discussion is at an end. \$\endgroup\$ – WhatRoughBeast Apr 16 '15 at 22:54

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