Did you look at the output with a scope? I think the superreg needs 5-6V headroom. 2.5V sounds way too less.
I am going to set my voltage around 18V-20V for the Waynes Preamp. I will no longer be trying for +-24V.
I don't have a scope to check but it seems the parts I will try to use should work with the Waynes.
I don't have a scope to check but it seems the parts I will try to use should work with the Waynes.
Enough threshold is necessary for any voltage. If you do not have a scope, you should at least use a mV meter to check the output is clean.
20 VAC ends up to approx. 30 V DC when rectified. => it works. OPA1611 will probably work fine.
The documentation for this regulator says that, when building one for low output voltages like 5V or 3.3V, the zener on the output can be replaced by a LED (in opposite bias) which, in that position, serves the same purpose. Like in the early Walt Jung schematics in the first Audio Amateur articles (except it wasn't a LED but a 1N4148 if memory serves me well). The documentation says that the zener voltage should be around half the regulator's output voltage. Apparently this has to do with taking the power supply for the opamp from the regulated output rather than the unregulated input voltage.
What if the regulator output is not that low - can I use any number of LEDs in series to jointly amount up around half the output voltage? Like for example 3 LEDs for 12V output or 6 LEDs for 24V output? Like suggested by Walt Jung on page 5 of this here article on his site (paragraph entitled "A Series-Stack LED-Zener Equivalent").
The reason for asking is the apparent noise performance of such a "series-stack LED-Zener equivalent" as pointed out by Walt Jung in that source. Considering that Zeners are not exacty low-noise devices. If this works in principle, are there any downsides to using the series LED stack compared to a zener (other than PCB space and the cost of multiple parts compared to just one).
What if the regulator output is not that low - can I use any number of LEDs in series to jointly amount up around half the output voltage? Like for example 3 LEDs for 12V output or 6 LEDs for 24V output? Like suggested by Walt Jung on page 5 of this here article on his site (paragraph entitled "A Series-Stack LED-Zener Equivalent").
The reason for asking is the apparent noise performance of such a "series-stack LED-Zener equivalent" as pointed out by Walt Jung in that source. Considering that Zeners are not exacty low-noise devices. If this works in principle, are there any downsides to using the series LED stack compared to a zener (other than PCB space and the cost of multiple parts compared to just one).
kunlun121, the only thing that matters when it comes to output noise of the regulator is the input noise of the opamp.
https://www.linearaudio.net/sites/linearaudio.net/files/V4 JW F7_0.pdf
AD797 Jung, look at the noise coming from the AD797. Compare then Jung/Sjöström AD825 which on paper has slightly higher noise. Both my and Jung/Didden's design have the curves up top of each other. OPA1611 would probably be a very good candidate when it comes to noise.
https://www.linearaudio.net/sites/linearaudio.net/files/V4 JW F7_0.pdf
AD797 Jung, look at the noise coming from the AD797. Compare then Jung/Sjöström AD825 which on paper has slightly higher noise. Both my and Jung/Didden's design have the curves up top of each other. OPA1611 would probably be a very good candidate when it comes to noise.
It has to do with letting the opamp operate with the output at around mid supply. Without a diode or LED or zener in series with the output, the opamp output would be close to the output voltage, with a danger of running out of manoeuvering room.
So it really isn't important whether you use a LED, a zener, a string of stuff or even a voltage reference, as long as it eats about half the supply.
If you are worried about noise, put a cap across it, although that could in some cases give start-up troubles, but you'll see that soon enough!
Jan
Oops. Sorry about that. Let's see if this works. List to Walt Jung's publications here. It's the article near the middle of the list with file name LEDs_as_Audio_Voltage_References_P1.pdf
Hi all!
The LM329 appears to be unavailable in my country. What would be a suitable replacement?
The LM329 appears to be unavailable in my country. What would be a suitable replacement?
I'm located in Hungary. There are many electronics stores here that I usually order from, but none of them have it available. I need it for a 12V application. If there's no other option, I can order the entire component set from abroad, but the shipping cost would be a bit higher in that case. I have another question: according to the description, the Zener voltage should be roughly half of the output voltage. In the case of a 12V output, would a 6.2V Zener be suitable, R6 620R R7 1k?
An alternative for the LM329 should be the LT1029. That is a 5V reference, but with adjusting the voltage divider, it can be used to have the regulator produce the 12V.
Also, I see that a letter from Germany to Hungary is very cheap. If you cannot manage getting LM329 or LT1029, I can get some and send to you for my own cost.
In LTSpice using a zener does work, but I think is inferior because of worse temp coefficient and noise. But that is only a somewhat educated guess. The experts here will be able to answer that better.
Also, I see that a letter from Germany to Hungary is very cheap. If you cannot manage getting LM329 or LT1029, I can get some and send to you for my own cost.
In LTSpice using a zener does work, but I think is inferior because of worse temp coefficient and noise. But that is only a somewhat educated guess. The experts here will be able to answer that better.
The LM4040C-2.5ID is available. Based on the table, in this case, R6 should be 3.6k, which would give an output voltage of 11.8V.
edit: Or LM4040A-10ID
edit: Or LM4040A-10ID
10V reference does not work I think, it is too near to output voltage. 2.5V ref works in Spice.
I’ve ordered the components, and in the meantime, I’m also able to get the LM329 locally. Thanks.