Any way to use the AZ431L on the negative rail? - Page 1
Forum section: electronics / projects

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 05, 2022, 08:43:33 am
content: I'm on yet another font project. This time I want to use li-on batteries for a small symmetric source, with output regulated between 1.5v and 2.5v and maximum current of 100mA.As you can see in the schematic, for the positive rail, I used the AZ431L in the regulation and the MAX4080t as a current limiter. When looking at the negative rail, I noticed the impossibility of using the MAX4080t. As for the AZ431L, I would like to know if there is any trick to use it on the negative rail or will I have to elaborate something totally different?

author: ledtester
author_rank: Super Contributor
timestamp: November 05, 2022, 11:25:04 am
content: Have a look at this discussion:https://www.edaboard.com/threads/tl431-as-negative-reference.359992/The idea is to put the voltage reduction element on the anode of the 431 instead of the cathode, e.g.:Update: Or maybe these ideas using a PNP:(fromhttp://web.archive.org/web/20061013020027/http://hififorum.knaak.dk/Lovan/neg_tl431.GIF)(fromhttps://www.diyaudio.com/community/threads/tl431-and-pass-transistor-for-negative-voltages.370590/):

author: magic
author_rank: Super Contributor
timestamp: November 05, 2022, 04:50:31 pm
content: You are going to have problems with the positive side already. It's not exactly an LDO regulator and it won't produce 2.5V from a 3V input. The base must be 0.7V above the output voltage and then there is R4 which drops some more voltage to supply base current of the NPN and the minimum quiescent current of AZ431L.Real LDOs use a common emitter PNP, but this can't be driven by 431 type chips directly because of wrong polarity. Maybe a second common emitter inverter could be added to fix this. Similar kind of output stage is used inthis project, if you ignore the final NPN emitter followers, and it seems to be working for them. Compensation is achieved solely by capacitive feedback across the LM324, here it would be across the 431 like you already have with C1.Speaking of which, P1 presents a very variable resistance to the reference input and C1, so compensation will be all over the place depending on output voltage setting. Another problem with this arrangement is that output voltage increases to maximum when the pot's wiper loses contact. A safer connection can be seen on the schematichere. It's a fairly well known trick, not some crazy invention of mine that doesn't work.I think BD139/140 may be overkill for 100mA at such low voltage, perhaps 2N2222/2907 or BC337/327 could suffice.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 06, 2022, 12:03:09 pm
content: Ledtester,very interesting circuit that you presented to me, however, due to the narrow margin between the input and output voltages, the circuit did not work properly. I'm abandoning the idea of using AZ431L, starting to use R2R opamps.Quote from: magic on November 05, 2022, 04:50:31 pmBased on your guidelines revolvi change the project and use opamps. The next step will be to change the transistor to PNP and test the LDO configuration.Quote from: magic on November 05, 2022, 04:50:31 pmIn an output short circuit the current goes to about 170mA. In a worst case scenario, this gives about 650mW over the transistor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 06, 2022, 07:29:36 pm
content: Easy!Since the cells are shown separately, disconnect them from being in series. Make two units of the same design. Wire their +/- in common. Done!Tim

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 06, 2022, 11:57:53 pm
content: Quote from: T3sl4co1l on November 06, 2022, 07:29:36 pmWhy make it easy when I can make it difficult.It will not be possible to separate the batteries, as it will be part of a larger project.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 07, 2022, 12:20:04 pm
content: In the first tests with this LDO arrangement I am taking a beating. I made an oscillator instead of a regulator. There's a lot of gain in this arrangement and I'd like to know how to tame this beast. I spent R1 from 10R to 100R and it didn't help. I changed R2 up and down and with no effect. Should I add some more components and where, or just change some values like C1?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 07, 2022, 03:37:56 pm
content: Yeah, Q1 needs degeneration (and a B-E resistor). You have two stages with extremely high voltage gain, closing the loop on an op-amp that's only unity-gain stable. So it's a good, whatever, 100+ times higher loop gain than it's designed for, and that'll happen.How about a pair of adjustable LDOs? They already do it this way but are compensated for you. With a voltage divider for feedback, you can run an extra resistor from a DAC to the VFB pin, and that's control.Also to finish out the potential-non-answer category: I take it, switching regulators are out of scope here (e.g. for simplicity)? The efficiency gain is obvious, but maybe you aren't as concerned about that here.Tim

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 08, 2022, 10:38:05 am
content: Quote from: T3sl4co1l on November 07, 2022, 03:37:56 pmI got the desired stability with the increase of the value of R2 to 470R and a better regulation with the addition of C3. C1 and R3 help in the linearity of the output voltage, when the MAX4080 starts to act.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: November 08, 2022, 09:38:52 pm
content: I finished the tests and the results were satisfactory.Even on a breadboard, I decided to check the transient behavior and the noise level.1- Final Scheme.2- Transient (2v output with 80mA load).3- 640uV noise (2v output with 80mA load). A preamp with a gain of 100(40db) was used. Decreased 100uv of pre/dso noise.Thanks a lot for your help. It was a very learning project.