How do I create a -0.3V rail from -20V - Page 1
Forum section: electronics / projects

author: peter-h
author_rank: Super Contributor
timestamp: December 04, 2020, 02:24:31 pm
content: I've been in electronics since the 1960s and obviously I know how to do this by throwing a load of bits at itBut I need to do it with very few components.The application is a -0.3V rail for a micropower op-amp. The reason is a bit complicated but basically it is to enable it to handle voltages a little below 0V. It must be at least -0.22V and no more than -0.32V. The temp coefficient doesn't matter so long as it is between these two from -25C to +60C.I started with a schottky diode (ZHCS400TA) and a 4k7 resistor. The resulting current of about 4mA produces -0.19V, so I need a little more. I could put more current through it, but don't have much more available. A silicon diode e.g. LS4148 produces about -0.4V. Even a big silicon diode (say BYG10M) will do 0.4V. So I looked for a tiny schottky diode and found an obscure Toshiba one on mouser.com, rated at something like 20mA max, but the spec sheet suggests that will be below 0.2V also.Is there some cunning way to do this? Is there some way to use Vce(sat) of a silicon transistor like a BC847C?The current which the rail has to source or sink is about 100uA.I thought about doing a bodge and assuming the -20V is fairly predictable and just put a ~20R resistor in series with the ZHCS400...Thank you for any tips

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 04, 2020, 04:56:51 pm
content: Plenty of ways to do this, depending on the details of what you need. What other rails are available? Would you like ratiometric tracking? How much power can you burn? How robust does it need to be under single faults?Two methods are particular favorites of mine for their straightforwardness.The first is just to use a plain old resistor divider buffered with an op-amp. This usually requires you to have higher voltage positive and negative rails available, though not always. You can get ratiometric operation for free if you use an inverting configuration on the amp and drive the center of the divider to ground.The second is to take a voltage reference and play about the same trick. For voltages near to ground like your -0.3V, this can require using an adjustable reference and putting an extra tap in the feedback network (i.e., one extra resistor). Since the output is coming straight out of the divider, it won't be particularly low impedance, but for 100uA that may be okay. If you have to buffer it, this option is less appealing. But it does end up being absolutely referenced to ground, which is nice. (Also be aware that there are references available with different regulation polarities. This can be handy for bipolar work. As an example, for an adjustable shunt regulator having anode (A), cathode (K), and feedback (F) terminals, you can get ones that regulate the voltage K-to-F or you can get ones that regulate F-to-A.)

author: peter-h
author_rank: Super Contributor
timestamp: December 04, 2020, 05:02:44 pm
content: Sure; any sort of voltage ref (1.2V, 2.5V, etc) plus a resistor divider plus an op-amp buffer will deliver a "perfect" result, but that's a lot of parts.I was wondering if there is some clever way to do this using perhaps a transistor and a resistor. Connecting base to collector of a transistor just gives me a 0.6V "zener", or with two resistors you can create a "zener" anywhere from 0.6V upwards (with a -0.3% per degC tempco). But no obvious way to get 0.3V or so.One idea I had was subtracting the schottky diode drop from the LS4148 diode drop, which would give me 0.6-0.2 = 0.4V. Their tempcos will partly cancel, too.Re other rails, I have +3.3V and +5V.

author: exmadscientist
author_rank: Frequent Contributor
timestamp: December 04, 2020, 05:19:56 pm
content: Quote from: peter-h on December 04, 2020, 05:02:44 pmYou shouldn't need both a reference and a buffer (at 100uA anyway). One or the other will be fine.Quote from: peter-h on December 04, 2020, 05:02:44 pmYep, that's the problem. The reason I prefer the other solutions is that they're dead simple: easy to understand, easy to design, easy to verify. And when you look up the price of a crap-grade TL431 or bottom-of-the-barrel LM324 and find that they're basically equal to a diode when you include assembly costs, they start to look pretty good.But I've no doubt it can be done the old-school way

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: December 04, 2020, 06:06:31 pm
content: The late Bob Pease showed a crazy clever way to do this. I can't find the original article, but here's some text-https://www.planetanalog.com/my-encounter-with-the-world-of-bob-pease/edit- sorry, I didn't see that you had -20, not +20.

author: SiliconWizard
author_rank: Super Contributor
timestamp: December 04, 2020, 08:02:14 pm
content: Sorry if I miss something, of course, but if you need a slightly negative power rail for an opamp, I am not sure I understand why the exact voltage would matter, as long as it has a minimum guaranteed value. So, in particular, I don't understand why a simple solution with a silicon diode wouldn't work. Of course a diode-based solution will give you a voltage that depends on temperature anyway, but with a typical silicon diode, you should get something like 0.3V to 0.7V for a wide range of temperatures. Which to me looks adequate. But again, I may miss the reason why you exactly need a voltage as close to -0.3V as possible.

author: moffy
author_rank: Super Contributor
timestamp: December 04, 2020, 11:54:04 pm
content: Just use two resistors in series connected to the -20v rail, I assume it is stable enough. A 10k and 150 ohm should do it. The 100ua will only shift the voltage by 0.015v.

author: Vovk_Z
author_rank: Super Contributor
timestamp: December 05, 2020, 04:57:52 am
content: I don't understand stability requirements. So I would use 0.2 V Shottky plus 100 R diode in series with a 1mA current across them (20 k series resistor).

author: Whales
author_rank: Super Contributor
timestamp: December 05, 2020, 07:04:25 am
content: If you need more current than a resistor divider can provide: try a resistor divider + BJT acting as an emitter follower. If the -20V isn't regulated enough in the first place then create a regulated supply with eg a 3.3V zener (but this starts to increase partcount).

author: peter-h
author_rank: Super Contributor
timestamp: December 05, 2020, 08:00:10 am
content: The reason for the range of voltage is that the op-amp in question (a voltage follower) needs to pass through -0.23V, but at the same time I don't want to start to turn on some protection diodes downstream (BAS16). 0.4V starts to turn them on. Also the other half of the op-amp feeds the input of a delta-sigma ADC (ADS1118) and I don't want to go anywhere near to turning on the protection diodes in that, either.I wasn't able to find the article on thehttps://www.planetanalog.com/site. There is one about a transistor with light shining on it going negative when there is no negative rail. Well, that's true. I found a photodiode acts as a generator when enough light is shining on it, too. I used one as an ambient light sensor and wondered where the 3V was coming from when the supply was only 2.5

author: Ian.M
author_rank: Super Contributor
timestamp: December 05, 2020, 08:39:33 am
content: You want to save power and component count. If Moffy's suggestion of a passive divider isn't good enough, e.g. because you don't want to 'burn' 2mA from the -20V rail or the rail isn't stable enough, then I strongly recommend the use of an inverting OPAMP, powered between either your +3.3V rail or your +5V rail and your -20V rail, with its input resistor fed from any stable positive reference you have available or your most stable positive power rail.With an appropriate choice of resistor ratios, that gets you a high quality low impedance -0.3V bias rail with only three parts, that only loads the existing rails by a little more than the +/-100uA load on the bias rail. If you need lower noise on the rail, add a fourth part - capacitance across the feedback resistor.Another option if you need a higher current negative rail, but don't mind that it can only sink current would be any 'jellybean' negative adjustable regulator rated for -25V in, with its feedback potential divider returned to your +3.3V rail rather than ground, and of a ratio to give -3.6V out, leaving you with an actual output of the difference, -0.3V.Trying to get 'clever' without significantly increasing your part count, at best will only save you a few cents per unit, probably with severely degraded load and thermal stability and much worse power consumption. Design with what's cheap and effective in 2020, not 1970!

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 05, 2020, 02:04:40 pm
content: What about use a regular diode to get about -0.6V, but then have a clamp circuit to prevent the outputs going below ground?

author: peter-h
author_rank: Super Contributor
timestamp: December 06, 2020, 09:33:21 am
content: How would you do the clamp circuit?There seems to be nothing in the "silicon world" which gives anything between 0.6V and the schottky diode drop of about 0.2V.The idea of inverting the +3.3V rail or using a -3.3 or -5V regulator and a resistor divider is a good one

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 06, 2020, 01:55:03 pm
content: Quote from: peter-h on December 06, 2020, 09:33:21 amNPN transistor clamp with a reference transistor connected as a diode (E to ground, B and C connected to a resistor to positive supply) and clamp transistors connected with B to the reference voltage, E to the signal, and C to the positive supply. If you connect the collector of the clamp transistor through a resistor, you can use that as an indication of clipping.

author: poorchava
author_rank: Super Contributor
timestamp: December 06, 2020, 02:35:24 pm
content: There is a TI IC that does exactly that: LM7705, although it generates -0.23V

author: exe
author_rank: Unknown
timestamp: December 06, 2020, 04:11:33 pm
content: Quote from: peter-h on December 06, 2020, 09:33:21 amPutting several schottky in series? "Exact" voltage can be adjusted by bias current and diode power rating. Power diodes with larger silicon give lower voltage drop. Still, voltage drop is temperature-depended, so not very suitable for precision.

author: m k
author_rank: Super Contributor
timestamp: December 06, 2020, 05:54:03 pm
content: Quote from: peter-h on December 06, 2020, 09:33:21 amChange the world.https://www.nteinc.com/diodes/general-purpose.php

author: peter-h
author_rank: Super Contributor
timestamp: December 06, 2020, 07:47:21 pm
content: LM7705 is a brilliant solution! I now recall I have used it before, years ago... Actually, no, that was a TL7705I did consider two of the schottky diodes but that gave 0.4V, but experimentation found that a 33R in series with the schottky gave me 0.3V with not too much variation over temperature, supply voltage, etc. I tried various size silicon diodes and none of them got near 0.3V Not even a big 5A one, at 5mA (that gave 0.4V).Another way would be a fairly stable 3.6V "zener" circuit hanging off the +3.3V rail and with a pulldown resistor to the -20V. A TL431 with 2 resistors would do that.