Hey, Ive just designed my own LM1875 based amplifier thats fed by a SMPS through a filter PCB. On this filter / regulator board i have 2x lm317, one set up at 32v and another at 12 with both of these having a voltage divider to provide a bias point for the TL071 and LM1875 on the amplifier.
I've noticed that the 16v (VD 32v) output are only putting out between 5 and 11v depending on the position of the 10k pot on the amplifier board. I assume the Pot is changing the load on this? Is there a way or a change i can make to the PSU filter board to stop this loading occuring? I'm getting reduced output as i get asymmetric clipping due to a low bias point on the LM1875. Sorry for the handwritten schematic 🙂
I've noticed that the 16v (VD 32v) output are only putting out between 5 and 11v depending on the position of the 10k pot on the amplifier board. I assume the Pot is changing the load on this? Is there a way or a change i can make to the PSU filter board to stop this loading occuring? I'm getting reduced output as i get asymmetric clipping due to a low bias point on the LM1875. Sorry for the handwritten schematic 🙂
Correct the polarity of the coupling capacitor after the potmeter, that should solve it.
Another issue could be the current handling of the LM317 for the +32 V.
Another issue could be the current handling of the LM317 for the +32 V.
Cheers, I was thinking the same re the 317, but even with no signal and as low as possible it's happening. As for the reversal of the electrolytic cap, can you explain why it would possibly have an effect? Not disputing, just trying to learn 😁
Hi. Electrolytic capacitors like yours 10uf are polarised , they works as capacitor only in one voltage polarity. If voltage polarity reversed ,like +16v on negative lead, and positive to ground, capacitors have big leakage current, and without current limiting resistor explodes. Imagine something like diode in parallel to capacitor ,what hapens if diode begins to conduct ?
According to your schematic, the negative pole of the electrolytic capacitor is biased at 16 V, while the positive pole is tied to ground via the potmeter. That means there is a large reverse voltage across it.
An electrolytic capacitor is also an electrolytic diode, when you reverse bias the capacitor more than about 1.5 V to 2 V (for aluminium electrolytic capacitors, 0.5 V for tantalum), it goes into conduction. Hence, you have a path that can conduct DC from the voltage divider via the electrolytic capacitor and the wiper of the potmeter to ground.
Connect the positive pole to the +16 V side and the capacitor will work as a capacitor and block DC current, except for some small leakage.
An electrolytic capacitor is also an electrolytic diode, when you reverse bias the capacitor more than about 1.5 V to 2 V (for aluminium electrolytic capacitors, 0.5 V for tantalum), it goes into conduction. Hence, you have a path that can conduct DC from the voltage divider via the electrolytic capacitor and the wiper of the potmeter to ground.
Connect the positive pole to the +16 V side and the capacitor will work as a capacitor and block DC current, except for some small leakage.
Or use bipolar capacitors?
https://www.digikey.ca/en/products/...4VEx8YnJALQI0ApQAnGqTuT6ElIp6q5xqG5guQBuLBJAA
https://www.digikey.ca/en/products/...4VEx8YnJALQI0ApQAnGqTuT6ElIp6q5xqG5guQBuLBJAA
Maybe try KiCad for producing nice schematics. The latest version is 6.0.5 which is easy to use.
Sorry, I've used it as ive done the pcbs in it, but I was away from home so I only had the sketch from my surface.
Thank you everyone. Good to see that I can continue learning and improving 😊 I'll reverse that cap and see what I can do. I'll report back on the result later
You can greatly dimish the load-dependent voltages on the VD's by dividing the divider resistors by ten.
These are 22k if I read it correctly, and the loading-dependent current causes excessive drop on the top resistor.
Use 2.2k instead of 22k, that should fix it.
Jan
These are 22k if I read it correctly, and the loading-dependent current causes excessive drop on the top resistor.
Use 2.2k instead of 22k, that should fix it.
Jan
Yes, I see the reversed cap. But even a correctly connected cap will/may leak. 22k is just too high a value for a voltage divider if you want it to be stable.
A cap from the voltage divider mid point to gnd also would improve signal-dependent stability.
Jan
A cap from the voltage divider mid point to gnd also would improve signal-dependent stability.
Jan
Actually there should be 3 resistors. Two for divider, and capacitor in parallel to bottom one ,which is going to ground , and third resistor to lm input.
Yes, good point, definitely required to avoid shorting the non-inverting input signal. In fact, the way it is now, with the 2 x 22k and no cap, the divider has an equivalent Rout of 11k and that saves the day, signal wise.
So it's a balance between a 'hard' 16V which would short the signal, and lower resistor values for better stability.
You'd probably not even need the 16V bias when you use the 6V bias (adapted) at the opamp, and delete a couple of elcaps.
Jan
So it's a balance between a 'hard' 16V which would short the signal, and lower resistor values for better stability.
You'd probably not even need the 16V bias when you use the 6V bias (adapted) at the opamp, and delete a couple of elcaps.
Jan
Lm1875 has higher supply voltage, than preamp ,so bias voltage is different (6v vs 16v), doing so (without capacitors ) will cause unsymmetrical clipping and noise when changing volume.
You can use the opamp to amplify the DC only. It's only a factor of 2 or so.
But all this shows why a bipolar supply is often a better solution.
Jan
But all this shows why a bipolar supply is often a better solution.
Jan
What I was hinting at, is the fact that the voltage divider is loading the signal path, and that an 11 kohm output impedance is therefore quite reasonable. You could obtain the same with a 2.2 kohm-2.2 kohm voltage divider and an extra 9.9 kohm resistor, but when you apply a star to triangle transformation, you see that that is a terminal equivalent of a 22 kohm-22 kohm voltage divider and an extra resistor that just wastes some power without doing anything useful.
It's a different matter when you want to improve the supply rejection by decoupling the voltage divider output, of course.
Without using third resistor , from lm input to divider , you are supplying power supply ripple and fluctuations to the lm input ,thats bad practice .Regardless the fact you use regulated voltage for amplifier itself ,even small voltage drop on it (lets say 1mv drop when you play music) will get into input, and will not improve sounding ,may even cause amp to oscillate.In your place ,i would not use low resistance resistors ,all three can be 22k , and add cap to grnd like 10 uf from divider midpoint ,to remove ripple from input .
Yes, understood. I feel for the OP, with us taking his design apart. Then again, that's the way to get smarter ;-)
I still think a bipolar supply is much more elegant and saves us from these conflicts, even if, as noted, it costs a bit more.
Jan
No! Don't! I'm loving it. Onky way to learn is from mistakes.
As is it works, but im going to be able to squeeze more out of it with these tweaks (the ones I won't need a new pcb design for anyway 😉)