Audio Amplifier Question

Hi Daniel,

First of all, sorry for the delayed reply. Thanks for providing your circuit, this has been helpful in understanding your present implementation.

While I presume this is not your highest priority, I have evaluated the THD+N of various portions of the circuit. Depending upon the cleanliness of your oscillator, I was able to obtain about 0.2% THD+N from a lab oscillator at 60kHz into your filter with a 50k ohm load (the digipot). So, based on this I don't presume the THD+N to be any better than this. The LMV321 has about 0.1% THD+N at 60kHz from the datasheet and I tested the LM4951 at 60kHz under various loads and don't see the THD+N being above 0.1%. If your circuit needs a lower THD value then a little more effort might be required to reduce harmonics with the filter.

One question I have is: how do you control the signal amplitude (digipot) with respect to the changing load impedance? Is this done based on some panel/GUI adjustment or is there a feedback mechanism that is not shown in the schematic? I'm just trying to get an idea of the control mechanism.

I will need a little more time to investigate whether there is a more elegant solution to simplify your circuit, but so far I don't have a solution. We have analog subsystems with volume controls, and power amplifiers, so there could be some simplification there, but these guys have many more features than what you might require. Is PCB area savings your number one goal?

Best Regards,

JD

John De Celles said:

First of all, sorry for the delayed reply. Thanks for providing your circuit, this has been helpful in understanding your present implementation.

No problem, I appreciate the help.

John De Celles said:

I tested the LM4951 at 60kHz under various loads and don't see the THD+N being above 0.1%.

Can you share your circuit with me, I recently orderd some samples, and when I get them I'm going to bread board it and measure the THD, I want to see if I can drive it directly from the digipot and bypass the transformer.

John De Celles said:

If your circuit needs a lower THD value then a little more effort might be required to reduce harmonics with the filter.

Actually I've done a little bit of digging, and have been testing some of our existing products, and have found out that a THD of about 5% would be acceptable.

John De Celles said:

how do you control the signal amplitude (digipot) with respect to the changing load impedance? Is this done based on some panel/GUI adjustment or is there a feedback mechanism that is not shown in the schematic? I'm just trying to get an idea of the control mechanism.

The signal amplitude is changed by a micro that gets feedback from two peak detectors that monitor voltage and current, that ironically was another circuit that i was looking at to improve.

John De Celles said:

Is PCB area savings your number one goal?

PCB space is my number one priority then low power requirments.

John thanks a lot I appreciate all the help. Let me know if you have any more questions.

Hi Daniel,

Wrt LM4951 THD tests, I used the circuit you provided previously; ie feedback resistor = 18.7k so that the gain was the same. I drove load impedances of 8 ohms and 30 ohms.

Thanks for supplying the info on your control mechanism. Based on this I presume that the optimum thing we can do is replace the digipot, the LMV321 buffer and the LM4951 power amp with one IC; assuming we can find the right part. This way, your sensing circuitry can remain the same with your micro controlling an integrated volume control. The code would have to be changed, but I'm assuming that since you are trying to reduce ICs and PCB area that this is on the table.

In order to converge onto the right IC can I confirm the supply voltage range? You had mentioned 3.1V min, but was is the nominal and maximum supply voltage?

I am looking at the LM48100Q or the LM4865 or LM4875. Can you take a look at these and let me know what you think. I will continue looking at options on my side and if you have any other details that you feel are pertinent that would be much appreciated.

Best Regards,

JD

John De Celles said:

Based on this I presume that the optimum thing we can do is replace the digipot, the LMV321 buffer and the LM4951 power amp with one IC; assuming we can find the right part. This way, your sensing circuitry can remain the same with your micro controlling an integrated volume control.

Thats an interesting proposal. I was primarily looking to remove the transformer since it takes up a pretty big chunk or real-estate. I was hoping I could push the LM4951 to a higher gain to compensate for the absence of the transformer, but as far as alternatives go thats not bad, it would remove a few components and overall might be a little more efficient.

John De Celles said:

You had mentioned 3.1V min, but was is the nominal and maximum supply voltage?

To answer your question the battery is about 3.7V fully charged, and about 3.4V nominal, once it falls below 3.1V the device is designed to automatically turn itself off.

John De Celles said:

I am looking at the LM48100Q or the LM4865 or LM4875. Can you take a look at these and let me know what you think. I will continue looking at options on my side and if you have any other details that you feel are pertinent that would be much appreciated.

I'll take a look at the specs, hopefully I'll have a response by Monday for you. I was hoping I could find something similar to the audio amp but something that possibly had an internal up-converter so I could get the output I needed inorder to remove the transformer but if I can't find something or the LM4951 is to noisy then I'll have to continue to work to see how small I can make the transformer, or optimize it all together.

Let me know if you see any issues with me pushing the LM4951 to a higher gain.

Thanks for all your help, I appreciate it.

Hi Daniel,

Can we clarify the end-load requirements?

I presently have 9mA from 30 ohms to 280 ohms and 2.6Vmax from 281 ohms to 10k ohms. Is this on the output of the transformer?

Some possible options I see are:

1. Use the LM4951 in its true bridge-tied load configuration, this will give you more voltage swing; the IC has plenty of current drive. I presume that this may not be possible with your detection circuitry for your feedback mechanism; please confirm. Would a bridged output work for your solution?

2. Try to use a simple analog subsystem like the LM48100Q to integrate volume control buffer and power amp into one IC; different programming would be required for attenuation control. I presume transformer would still be required. This is probably the most work on your end.

3. Keep all circuitry the same, but replace the transformer with a DC-DC boost to power the power amplifier. Once again, please confirm end load requirements. Please review attached quick sim from web-bench. This would probably be the easiest, smallest and probably cheapest solution.

Let me know what you think of these options. I'm sorry that I'm still not 100% clear on the end load requirements; I wasn't able to find the transformer turns on the web.

Best Regards,

JD

4857.Boost_for_LM4951.pdf

Hi Daniel,

Thank you for clearing up the loading requirements. Based on that information the worst-case loading requirement occurs at 280 ohms with 9mA of current for 23mW of output power. The 2.6Vpk output voltage also becomes an issue for your minimum 3.1V supply voltage. Hence the need for the transformer.

What I was suggested previously was possibly using the LM4951 in its intended operation as a bridged output; ie the load is connected across Vo1 and Vo2 of the IC. These outputs are 180 degrees out of phase, creating twice the peak voltage across the load each half cycle. This is a great way to get more output voltage using small power supply voltages. So, by using the 3.1V min supply voltage we should be able to obtain 3Vpk into the load without using the transformer.

I connected up the LM4951 using a 3.4V nominal supply with 18.7k ohm feedback resistor as you have in your schematic. The closed-loop gain of the amplifier is 1.87 x 2 or 3.74 due to the bridged output operation. So, with a 700mVpk 60kHz input signal I obtained a 2.6Vpk output across the 280 ohm load. This is shown in the attached scope photo.

I then proceeded to check the THD of the amplifier at this frequency, as this was also one of your concerns. The harmonic analyzer stated 1.8mV/1.85V x 100% = 0.097% THD at 60kHz into 280 ohms at the 2.6Vpk. I also quickly ran an FFT showing the 2nd harmonic at -60dB or again 0.1% THD. The analyzer is limited in how high the harmonics can be stated, but this gives you the idea that it is not completely terrible.

If you can utilize the bridged output topology (example as in Figure 1 of the datasheet) then I would recommend at this point that it is the simplest way of eliminating the transformer, thus reducing some of your PCB area requirements.

I hope this helps a little.

Best Regards,

JD

2161.LM4951_BTL_Output_061312.TIF

0601.LM4951_THD_061312.TIF

John,

I'm going to try what you're suggesting, my only other concern related to this project is the feedback that this signal creates. At the following link is the circuit we use to measure current and voltage feedback, these circuits are half peak detectors, however their output is not linear, I was hoping i could find an amplifier, preferably in a small dual package that could detect those low current peaks and voltages and report them linearly any thoughts?

http://www.freeimagehosting.net/b3i1a

Hi Daniel,

I've had to reread your request a few times. I'm not sure I understand what you are looking for when you say "report them linearly". Since you are using half-peak detectors, I think you are referring to the output essentially not being continuous.?. Would this be a correct assumption? If so, then maybe converting to a fullwave rectifier would solve that problem?

My initial understanding regarding the output not being linear led me to analyze your circuits with respect your previous loading requirements. They look like they would work just fine.

Best Regards,

JD

Sorry for the confusion I should have been a little clearer on what I was thinking and what I needed, sorry about that.

John De Celles said:

Since you are using half-peak detectors, I think you are referring to the output essentially not being continuous.?. Would this be a correct assumption?

The output is continuous however the amount of amplification I am getting is very small with respect to the current, I'd like to increase the max output voltage, so I was hoping someone might be able to suggest an amplifier that has a higher max gain than the one I'm currently using that can give me a signal with a peak voltage for my application of around 2V, for both current and voltage.

John De Celles said:

My initial understanding regarding the output not being linear led me to analyze your circuits with respect your previous loading requirements. They look like they would work just fine.

The voltage works pretty well with the divider, however the current feedback with such a small current has a few issues.

However my other concern and I think it got lost in my last post, is if I use the circuit which you suggested by bridging the output, I'd need to change the analog feedback to compensate for the fact that the new output signal is technically floating and I need feedback to my micro with respect to ground. So I would need a hopefully simple and easy way to reference it to ground for the feedback, any suggestions on that? I’ve seen some Instrumentation Switched Capacitors that might work, but was curious of what else might be out there. Since I’m trying to keep it simple and small using an instrumentation amp is not the best idea unless there is a really small one I could utilize.

Hi Daniel,

If you need more gain from the current sense circuit you could increase the value of R13 to what you need the output to be. Right now the amplifier is set for a gain of 11 and with 9mA into 10 ohms, you are amplifying 90mV, resulting in an output of about 990mV.

Each circuit should work in a differential mode and be able to be tied directly to the bridged output of the LM4951. For the voltage sense circuit you will need to add another resistor from the R3, R5 GND node to the other output of the LM4951. That GND node will no longer attach to GND. The same is true for U2-B; don't attach the node of R7, R6 to GND. I am concerned about how R7 attaches to the load, however, because this will load the output load down.

JD

John De Celles said:

If you need more gain from the current sense circuit you could increase the value of R13 to what you need the output to be. Right now the amplifier is set for a gain of 11 and with 9mA into 10 ohms, you are amplifying 90mV, resulting in an output of about 990mV.

Each circuit should work in a differential mode and be able to be tied directly to the bridged output of the LM4951. For the voltage sense circuit you will need to add another resistor from the R3, R5 GND node to the other output of the LM4951. That GND node will no longer attach to GND. The same is true for U2-B; don't attach the node of R7, R6 to GND. I am concerned about how R7 attaches to the load, however, because this will load the output load down.

Yea it will add 10 ohms to the load when its connected and nothing when the load is open, which is fine, since the minimum threshold of the load is 30 ohms we will always know when something is not connected or when something is connected; is that what you were concerned about?

I'm going to try out your suggestions next week since it will likely be very slow here I will have a great opportunity to spend a lot of time in the lab, I'll let you know how it turns out.

John De Celles said:

Each circuit should work in a differential mode and be able to be tied directly to the bridged output of the LM4951. For the voltage sense circuit you will need to add another resistor from the R3, R5 GND node to the other output of the LM4951. That GND node will no longer attach to GND. The same is true for U2-B; don't attach the node of R7, R6 to GND. I am concerned about how R7 attaches to the load, however, because this will load the output load down.

I finally had a chance to try the circuit you suggested aside from the gain for the amplifier a little crazy it works great, adjusting the gain should be simple I just need to recalculate some resistor values throughout my circuit. I did however have a problem with the feedback circuitry there is a constant DC level on the output of the amplifiers now that corresponds to the DC offset between the ground and the floating output signal from the amplifier. Any thoughts on what I might be doing wrong or what I can do to return the output from the feedback amplifiers closer to 0 when I first turn on the input signal at its minimum?

Hi Daniel,

It's been a while since I've looked at this design, so I'll take a look and re-familiarize myself with this tomorrow.

thanks.

JD

Here is a redline defining the modifications I made based on your recomendations, this morning while reviewing it I knoticed tgat SP1 and SP2 should also be removed. So I'm going to get that done and run my tests again.

I am using an Agilent 33220A to inject a small sine wave (60kHz @ 10mVp-p to 300mVp-p) into the buffer designated U12 pin 1. I am using a standard tektronix scope to measure the signal at D2 and D3. I'm also powering the VCC externally to 3.3V using an old lambda style supply.

Hi Daniel,

If I'm not mistaken, all you really need is another coupling cap at the output of VO+ as you have for VO-. You have C7 at the output of VO-, so using an appropriately sized capacitor for the load off of VO+ should do it. Is there a reason that I'm missing such that this doesn't solve the issue? Let me know.

Best Regards,

JD

John De Celles said:

all you really need is another coupling cap at the output of VO+ as you have for VO-. You have C7 at the output of VO-, so using an appropriately sized capacitor for the load off of VO+ should do it.

I added the capcaitor per your suggestion.

But I'm still having a problem with the feedback, below are scope shots of both.

Voltage feedback, at 2.6Vrms at the output, with the output resistance being 100 ohms.

Current feedback, at 2.6Vrms at the output, with the output resistance being 100 ohms.

With the output voltage set to 2.6Vrms the output of both feedbacks should be between 1 and 2 volts DC, however I have the above. I checked all the rework to make sure it was done properly. Any ideas what might be the issue?

Hi Daniel,

Have you verified that without any load attached to the LM4951 that the output is as expected; ie a clean 60kHz sinewave? I presume that the transformer is no longer in the circuit.

You can verify proper operation of the detectors with TINA, and/or you could short out the diodes and verify that the op amps are working as expected; ie proper DC  bias and clean 60kHz output signal without clipping.

Best Regards,

JD

John De Celles said:

Have you verified that without any load attached to the LM4951 that the output is as expected; ie a clean 60kHz sinewave? I presume that the transformer is no longer in the circuit.

Yea I did that again today, the sinewave is a little destorted when I go to 6.8Vpk-pk. The circuit is just as I posted earlier, the transformer has been removed.

John De Celles said:

You can verify proper operation of the detectors with TINA, and/or you could short out the diodes and verify that the op amps are working as expected; ie proper DC  bias and clean 60kHz output signal without clipping.

I shorted out the diodes as you suggested and I found I have a DC bias of about 1.6V even when the input signal is turned off. This could make things complicated when I try to read the output signal from the peak detectors, I need to find a way to remove this and only leave the peak detector as the only component of the output. This is going to be the more difficult part.

Hi Daniel,

A 1.6V DC bias on the outputs of the op amps is as expected; ie approx 3.2V supply divided by two. If that bias is not desired for circuitry after the op amps then you should be able to add coupling caps at their output, so that only the AC component will get through.

I don't see anything fundamentally wrong with the circuit.

The sinwave from the LM4951 should be clean. if it is not, I would check the sinewave at the input and check on the loading of the LM4951. What is shown in the schematic looks fine, but loading is a common cause of signal distortion.

Then check the signal at the output of the detectors; the sinewave should be clean, undistorted and have the appropriate gain. If not, then again, connections and or loading could be the culprit. Once you have verified that the detectors have the appropriate bias and gain, you can add back the diodes to see the rectification effects.

This system is a single 3.3V supply, so each IC should have approx a 1.6V DC bias. Capacitive coupling is generally used to prevent the bias from being amplified. It is important to know the load after each coupling cap to ensure that the 60kHz signal will be properly passed without distortion.  If unsure of the loading, you can try bigger capacitor values and verify the quality/amplitude of the signal.

I think at this point, you have essentially accomplished your original goal of reducing the circuit size by removing the transformer. The detection circuits are essentially the same as before except the inputs are now differential, so the solution really should work as before.

Best Regards,

JD

John De Celles said:

A 1.6V DC bias on the outputs of the op amps is as expected; ie approx 3.2V supply divided by two. If that bias is not desired for circuitry after the op amps then you should be able to add coupling caps at their output, so that only the AC component will get through.

This is where I'm getting a little confused the feedback circuit is made up of two peak detectors, so if I block the DC bias by using a decoupling cap wont I effectivly block the peak detectors signal, even though there is a pulse from the wave itself the peak detector holds that state and expresses that as a constant voltage?

John De Celles said:

The sinwave from the LM4951 should be clean. if it is not, I would check the sinewave at the input and check on the loading of the LM4951. What is shown in the schematic looks fine, but loading is a common cause of signal distortion.

I have still have to take the LM4951 through its paces but hopfully the load wont distort the wave to much otherwise this solution wont work, unfortunutly so I'm keeping my fingers crossed.

Hi Daniel,

I'm confused myself. Didn't your detector circuit always have a DC bias? I presume that the circuits following the detector take that into account.

If the DC bias is not as expected: Since the input is now configured differentially with no tie to either ground or 1/2 VCC, we will probably need to setup a proper DC bias, say on the non-inverting input. You can use a resistor divider between VCC and ground, using high-valued resistors (say 100k) and a large valued cap (say 10uF) on the non-inverting node to ground. You may need to do this as it appears that the detector circuit may be floating.

You may want to go back to the original setup and examine the original operation of the circuit, then make these changes and ensure that the results are the same with the higher input signal amplitude from the LM4951.

JD

John De Celles said:

I'm confused myself. Didn't your detector circuit always have a DC bias? I presume that the circuits following the detector take that into account.

No there was no DC bias, when the detectors come up untill their is a signal output from the device they rest at 0VDC then when the signal output comes up they act accordingly.

John De Celles said:

If the DC bias is not as expected: Since the input is now configured differentially with no tie to either ground or 1/2 VCC, we will probably need to setup a proper DC bias, say on the non-inverting input. You can use a resistor divider between VCC and ground, using high-valued resistors (say 100k) and a large valued cap (say 10uF) on the non-inverting node to ground. You may need to do this as it appears that the detector circuit may be floating.

Yea I had that idea I was hoping to put it to the test, I'm glad we agree on our implamentation.

John De Celles said:

You may want to go back to the original setup and examine the original operation of the circuit, then make these changes and ensure that the results are the same with the higher input signal amplitude from the LM4951.

I have been doing that all along, I always like to have a control sample.

I'll take another look at the device and I'll check to see if my sinewave is accurate at my different loading conditions.