HQ audio amp with jellybean chips - Page 1
Forum section: electronics / projects

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 28, 2022, 07:06:57 pm
content: There are several 5 pin power opamps as LM1875 or TDA2050 that has been copied without protection circuits or current limiters and sold for less than $0.15/piece. Their characteristics is similar to TDA2050.Adding a low cost $0.1 opamp TL072 for two channels in composite and one gets a fabulous amplifier. To be HQ, the NFB must be 60db @20khz for each stage. Each stage has only 50db in OLG @20khz, so I came up with this universal topology in which the power chip can be any.About THD+N, the power amps have 0.05% with CLG 20, to go down to 0.005% with -1 and applying extra 20db NFB, It should not be higher than 0.0005%.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 28, 2022, 07:41:09 pm
content: Adding extra gain to the FB loop is not that easy. The power ampliers are usually not that simple in the response as the spice models. Also the load is usually not a perfect ohmic load and may include capacitive components. So the real world circuit has a good chance to oscillate or at least ring quite a bit at certain points. Things get tricky if the power amplifier part is not at least 6 times faster than the audio BW - many of the cheap ones don't and are just fast enough for the audio range.In the shown plan there is another small problem: the 120 kOhms resistors would a significant amount of noise. The TL072 is already not that great, but the resistors are worse. With such high resistors parasitic capacitance and the OPs input capacitance will add to the phase shifts.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 28, 2022, 08:04:42 pm
content: Thank you to have given some interest to this modest circuit.The circuit is a little bit more complicated than you can see at first sight. If you look closely you will see the driver to be Howland current pump with 240k output impedance. This makes the power amp run with -1 gain bellow 300khz and 20 above.The THD +N is implemented in the sim model of TL072 and the circuit has 0.00017% as stand alone, it is also with gain of -1.The loop gain curve shows over 80° phase margin for LM1875 and LM3886. I am using the later since 7 years and the sim model is perfectly match with the reality.You might wonder why I don't have Thiel LR on the output. This costly item I replace it with current feedback from 0.01 ohm PCB resistor, it transforms the inductive output impedance into capacitive after which capacitive loading simply decreases the frequency response.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 29, 2022, 02:28:06 am
content: It looks like the NE5532 has become a jellybean, but I read by some users, inside of these cheap 5532s are different circuits. I simmed it withNE5534 with 22p compensation to be equal, and I have still higher phase margin, the R6 and R9 resistors can be 10k each if you don't need volume control.

author: David Hess
author_rank: Super Contributor
timestamp: July 29, 2022, 03:12:04 am
content: That is a common way to get lower distortion out of a power operational amplifier. Frequency compensation needs to be carefully considered, but this is eased because the overall closed loop gain is relatively high.The "jellybean" design I would rely on uses discrete output transistors with current feedback to the output of the operational amplifier.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 29, 2022, 05:42:33 am
content: The common way of composite is this way,https://www.eevblog.com/forum/projects/composite-amplifier-lm3886-lme49720/This type, doesn't run after the dazzling 0's in distortion number, but to get 60db NFB for each stage, as you did remark, low THD+N is just a consequence.Why so high NFB? When an amp has 90° phase shift in OLG all the audio band, when a note is played, the NFB is the only parameter to bring the phase shift to 0° in transient. If only a single voice is present, there is no any problem, the story goes wrong when multiple voices, as a choral or a multiple instruments start about the same time with about the same frequency, the sound goes dirty similar to crossover distortion. The simplest example that I use since decades to qualify an amp to be audio is the song of Turtles, "Happy together", The refrain of 16 voices provokes in amplifiers with less than 40db NFB, distorted sound. The chip amps with 50db OLG@20khz once feedbacked to 26db CLG, only 24db NFB is left and they don't pass the audio qualifying test. To get clear Requiems of Verdi and Mozart, 50db is necessary to my ears when they were still hearing 14khz. Audio experts use commonly 60db, HQ audio opamps have 70db @20khz on OLG to provide a CLG of 10db only.In this topology, each stage gets 44db NFB in each stage as they are inverting unity gain with 50db OLG @20khz. The open loop gain of the amp is only 41.5db so than extra 20db NFB applied will result in 20db closed loope gain.

author: quadtech
author_rank: Regular Contributor
timestamp: July 29, 2022, 10:59:50 am
content: Another reference -https://www.diyaudio.com/community/threads/miniref-schematic-and-pcb-layout.184165/

author: mawyatt
author_rank: Super Contributor
timestamp: July 29, 2022, 12:27:25 pm
content: In the closed loop plots shown with ~11dB peaking at ~150KHz, this signifies quite a underdamped closed loop system. I would not expect a true 80 degree phase margin with that type of closed loop response. What is the gain margin you simulate? Could you show the Open Loop plots?Also, remember that the simulations are done with linearized AC equivalent device models that are characterized at small signal levels. With larger signal levels the models exhibit a much lower frequency response and lower effective gain (this is why op-amp specs show a Large Signal Loop Gain that is much lower than the small signal parameter). The effects of these should show up in significant ringing on large signal step responses, or other waveforms that exhibit a fast large signal crisp "edge".A technique we utilized long ago during simulations was to force a DC bias term into the system which would cause a large level DC output, and load the system with a DC compatible load for similar load current (doesn't account for the load reactance tho which can be included with a more complex load model). This creates an approximation for a large signal level, and then do an AC simulation at this high level bias to emulate a large signal AC level, then change the DC bias level to cover the system large signal full range (include + and - ranges) and repeat the AC simulations at each DC bias level. Then you can "see" the Gain and Phase margin deviate from the small signal levels and this explains the large signal dynamic responses "seen". Yes, this is a lot of work to do, but when you are designing chips like we did the cost of a design error and/or weak performance effect (significant ringing on large signal pulses for example) justifies the effort!! Much later Cadence addressed this AC linearized simulation limitation with specialized analysis routines for Large Signal AC analysis.Best,

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 29, 2022, 01:03:40 pm
content: The plots are loop gain, I post bellow the closed loop. With 22p lead capacitor, phase margin is 78° and gain margin of 6db.With 10pf it becomes 67° and 11db.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 29, 2022, 02:17:43 pm
content: With C4 22p instead of 6p I get perfect square wave, but I must review the saturation as LM1875 is sticky.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 29, 2022, 06:09:15 pm
content: I needed to add a pair of diodes to avoid oscillation at saturation.Bellow is the scope graph showing how my technique of current feedback replaces the Thiel output L//R. It needs 1uF upon the load to get it spiked, bellow 1u, capacitance loading keeps the square wave perfect.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 08:47:14 am
content: This amp is going one step further by doubling its output current by a current multiplier.This is my invention in opamp basic configurations.The dummy added power opamp is fully floating, the negative feedback tries to keep the input difference zeo by this the two resistors get equal voltage drops, then the current through them is proportional. Here they are equal 0.22 ohms then the dummy pours as much current as the main does. The loop gain is decreased by R2 and R5 and the delay by a 200nH inductor which can be implemented either on PCB, or by the leads of the 0.22 resistor.The plot shows how once compensated, the two current of the dummy follow the mains even in saturation.Now, The HQ jellybean chip amp can become 40W /4ohm amp.

author: David Hess
author_rank: Super Contributor
timestamp: July 30, 2022, 09:15:00 am
content: There are a couple of ways to parallel operational amplifiers. The one shown below seems more suitable, and goes back to at least the early 1970s.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 10:09:42 am
content: Thank you for suggestion.This configuration is valid for unity gain stable amps.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 11:39:47 am
content: I forgot that the model of LM1875 is stable down to 3. To more universal, I switched to LM3886 model and using high impedance values, as the chip is stable down to 2.5 with low impedance, And I looped the feedback from the output. This time the inductance is on the dummy side. With high gain, the bode plot can look good but very bad transient on pulse and with gain of 2, the pulse is perfect but dirty bode plot. What to decide?I replaced with 1875 keeping the same circuit and I get not so bad result.

author: mawyatt
author_rank: Super Contributor
timestamp: July 30, 2022, 04:21:11 pm
content: If the load can be floating (speaker) then why not use an "H" Bridge configuration?Best,

author: mawyatt
author_rank: Super Contributor
timestamp: July 30, 2022, 04:44:51 pm
content: Quote from: Kokoriantz on July 29, 2022, 01:03:40 pmC3 and R14 are not required since the impedance "seen" by the - input of the amp is dominated by R5. Also, your injection and sampling point seems incorrect for the Bode Plots, as some of the signal can bypass thru R48 and influence the readings. Maybe a better location would be into the + input of the op-amp, since all the various feedback branches congregate into the op-amp + input.Note you may be able to use a floating voltage source and not have to use the decoupling cap and inductor, and the Open Loop response T(f) is simply V- over V+ of the floating source when the source is pointed towards the + amp input. This works because the injected signal Vs at the source + terminal travels around the entire loop and shows up as Vs*T(f) at the - source terminal, where T(f) is the entire loop transfer function, and the source has zero impedance so it can't upset the DC bias or other parameters. So the voltage ratio is simply source V-/V+ or Vs*T(f)/Vs or T(f)!!Edit: Attached an image that shows why the floating voltage source works for showing Bode Plot of Open Loop Transfer Function.Best,

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 08:24:16 pm
content: Coming soon.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 08:41:53 pm
content: R14, which should be 120k is necessary for the offset and the bypass capacitor for the noise and HF feedback grounding.What is R48?Usual loop gain plot is indeed with LTspice with a floating generator but before the feedback H network. With my simulator, every time I must ask the postprocessor to perform the division. The error I get with this rudimentary way is that my source generator has zero ohm impedance but the feedback has that of the load in parallel the output of the amp. With 8 ohm load in series does make very little difference for that many prefer Tian probe. I rely on the square wave transient, it is not calculated as bode plot but simulated and as you indicated to measure with different dc offsets, the square wave or step impulse does the same.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 30, 2022, 08:46:31 pm
content: This is the 40W 4 ohms version with the same character as before.Next, to bridge for a gentleman's 80w single supply version.Edited: The offset is adjusted for dc blocking capacitor after the volume pot.

author: mawyatt
author_rank: Super Contributor
timestamp: July 30, 2022, 09:20:28 pm
content: Quote from: Kokoriantz on July 30, 2022, 08:41:53 pmThis was from your #8 post. R14 is 100K, but R5 is 5K from the first amp and thus dominates the impedance seen by the 2nd amp.Just noted the mention R48 is R4 which is 8 ohms, it looked liked R48 on the schematic! R14 is 100K, but R5 is 5KWe have no problem with LTspice utilizing the In-Line Floating source as shown in the attached hand analysis image shown above.Agree, a transient is a good way to evaluate a "system". Long ago we developed a "pinging" technique which introduced a small Dirac approximation doublet (has no DC value) as a test probe, used this for simulations and in actual lab measurements (we built a "pinger"!). This technique works very well for most systems since they can be "distilled" to an approximation of a simple 2nd Order System regardless of how complex the actual system is. With the classic 2nd Order response defined, the damping revels the effective stability based upon the response to the "pinging" waveform which approximates the impulse response without the average of unity so it doesn't "upset" the DC nature of the circuit. However, most prefer the Bode technique which shows the Gain and Phase Margin that most are familiar with.Best,

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 31, 2022, 12:17:20 am
content: The R5 is 2k but the positive feedback makes it infinity, Howland current pump. The power stage sees the impedance of the driver as R7+R8,120k +120k =240k which is equal to its feedback resistor, hence has a gain of -1 dc to 300khz. The bias current of the negative input sees two 240k in parallel that is 120k to be balanced at the positive input.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 31, 2022, 07:31:18 pm
content: Bridged 80W/8ohm single supply.For now this amp needs Thiel L/R on the output the time I figure out how to imply a current feedback. The feedback can be individual or differential with C8.I edited the post19 circuit.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: July 31, 2022, 11:15:57 pm
content: Power supply:To use low cost 48v SMPS with 150mv ripple, A voltage regulator is needed for TDA2050 for its very low PSRR.Any suggestions for a low cost high performance, 10mohm, regulator?

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: August 01, 2022, 01:44:21 am
content: I came up with this, the LM317 normally cannot function at 44v, what can possibly go wrong?The graph bellow is with 5A peak 50hz. I limited the reservoir capacitor to total about 10,000uF, as low cost SMPS don't start up with more.The 100uH is 0.1 ohm 6A is about 1.5cm in diameter, low price.Each channel needs its own SMPS to have the current limiter function as short output protection.Edit: Better power supply. L2 is PCB inductance reaching to each of four output chips ending with 100uF +100n.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: August 01, 2022, 04:41:58 am
content: It looks like I got lucky with the capacitive loading. With differential feedback, naturally the amp accepts capacitive load without altering its transfer function, why, I don't know . I am unable to see the square wave as in single supply mode the simulator doesn't end calculating but I do see with sine wave a flat response.

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: August 01, 2022, 08:11:53 am
content: I want to submit a wire/PCB cross section problem.American wire gauge.https://www.electricaltechnology.org/2022/04/american-wire-gauge-awg-chart-wire-size-ampacity-table.htmlAWG table shows the wire thickness and the maximum current that it should be used. For example, 1mm wire No18, No19 can pass 2A maximum for power transfer and 15A in chassis wiring. What does this mean. To my opinion, Thomas Edison is the author of this table as he ended up inventing the Fuse. Despite the Ohms law, whatever is the length hence the resistance of a wire, if the current exceeds the value in the table it goes higher resistivity, and gets cut. This phenomena is linked to the magnetic field that escapes from the wire. If the wire flows flat upon a steal plate, as the chassis, to constrain the field around it, the maximum current gets many times higher.Electrolytic, ceramic, low ESR capacitors, non inductive resistors have magnetic leads for this purpose as was once the metallic casing transistors.Coming to our TO220 chip, you can see that the leads are far be able to pass 5A as presented, but if you remark, the output pin is between the two supplies so that the magnetic field of the 3 leads cancel each other if they are maintained parallel.The leads are plied alternately for single layer PCB, nowadays, two layers become standard, which allows the first three leads remain parallel, and the output large track on one side flows in parallel to the supplies flowing on the other side. This is valid if the chips are mounted isolated from the heatsink.The problem I submit, how to do if they are not isolated as the single supply bridged, receiving the ground current from the heatsink?

author: Kokoriantz
author_rank: Frequent Contributor
timestamp: August 02, 2022, 03:40:13 am
content: Magnetic field canceled PCB.The chips are mounted isolated from the heat sink to receive the ground from the -V pin.The feet of the chips are aligned as Pins 2 and 4 to have the +v O -v perfectly parallel.The O runs from the chip from the top covering half of the +V and -V that run from the bottom. The +V passes to top to join the top rail of +V where the -v joins the ground rail in parallel with +V rail.The two outputs run to the edge connector in parallel on each side.I did not draw, the +V and the ground arrive to the rails also one on each side in parallel from the edge connector. If LC 100uH 10mF is used instead of external regulator, the 10mF is to install at the center between the two phases upon the rails.Bellow is a sketch not very clean, sorry for that, it shows one phase only as the other is identical except the -OUT rail is opposite side (left, right) on the bottom parallel to that of +OUT from the top.