Best way to generate 10V from 12V? - Page 1
Forum section: electronics / beginners

author: legrady
author_rank: Contributor
timestamp: May 11, 2022, 04:38:14 pm
content: I am making a eurorack module, which involves some CMOS. The chips can handle up to 20V Vdd - Vss difference, so eurorack's +/- 12V are a problem. The best way to drop each supply a little bit? A string of 3 or 4 diodes to have a drop of 2 or 3 volts? 10V zener? 10V linear regulator?I want it to be simple, minimal, since the power shift isn't my project, just a pre-requisite.Tom

author: BrokenYugo
author_rank: Super Contributor
timestamp: May 11, 2022, 04:46:51 pm
content: How much current is needed?

author: bdunham7
author_rank: Super Contributor
timestamp: May 11, 2022, 04:53:50 pm
content: Quote from: legrady on May 11, 2022, 04:38:14 pmFor that you could use a positive and negative LDO, it won't get much simpler than that. How much current? If your needs are less than 150mA, TI has an all-in-one solution for you:https://www.ti.com/lit/ds/symlink/tps7a39.pdf?ts=1652207018154

author: magic
author_rank: Super Contributor
timestamp: May 11, 2022, 05:02:39 pm
content: Does it need to be symmetric ±10V?If not, drop the positive rail to 5V with LM7805 (LM78L05 if possible) for a total 17V supply to the chips and call it a day

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2022, 05:26:21 pm
content: The string of forward-biased diodes is the simplest and most efficient non-switching solution, just make sure you have a resistor to ground so that the minimum current through the diodes suffices to keep the "10 V" lines below the maximum voltages with zero load current. At 25oC junction temperature, a true 1N4148 diode has a typical forward voltage of 0.6 V at 1 mA, falling to 0.5 V at 90oC. At 100 mA, this increases to around 0.9 V at these temperatures (reading the Vishay graphs by eye).

author: james_s
author_rank: Super Contributor
timestamp: May 11, 2022, 05:27:56 pm
content: Is there a reason you want to power the CMOS chips between the + and - rails instead of between + and Gnd? The bipolar supply rails are normally only used for op-amps where the output needs to swing above and below ground.

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2022, 05:41:30 pm
content: I have used 4000-series CMOS devices in analog applications with appropriate split power rails.The original RCA ratings for Vdd are:CD4000A series: 15 V max, 3 to 12 V recommendedCD4000B series: 20 V max, 3 to 18 V recommendedBut, the JEDEC specs for the 4000B series are 18 V max, 3-15 V recommendedSeehttp://www.introni.it/pdf/RCA%20-%20CMOS%20Ic%20Databook%20SSD%20250C%201973.pdfpp 48-53.

author: DavidAlfa
author_rank: Super Contributor
timestamp: May 11, 2022, 06:56:24 pm
content: That would be at least 8 diodes, 4 for each polarity (2.4V drop), so you get ±9.6V.Check the LDO prices, a pair of 7809/7909/LM317/LM337 or anything similar might be cheaper or close to these 8 diodes, giving much better regulation.

author: eugene
author_rank: Frequent Contributor
timestamp: May 11, 2022, 07:12:17 pm
content: It's been asked, but no answer yet: how much current.What is the acceptable range for the CMOS (not just the absolute max)? How well regulated is the +/-12V? You might want to drop more than 2V from each rail to be safe.A bunch of series diodes is cheap and easy, but LDOs are cheap too and provide the benefit of removing noise from +/-12V supply. If I were trying to get something workingright nowand I had suitable diodes in hand, then maybe I'd do that. But if I need to order components anyway, I'd go with some kind of regulator. But that's just my opinion.As has been stated, the CMOS might be happy (maybe happier) running from the 0 and +12V outputs of the PS.

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2022, 07:25:05 pm
content: Quote from: eugene on May 11, 2022, 07:12:17 pmIn my reply above,The original RCA ratings for Vdd are:CD4000A series: 15 V max, 3 to 12 V recommendedCD4000B series: 20 V max, 3 to 18 V recommendedBut, the JEDEC specs for the 4000B series are 18 V max, 3-15 V recommended

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2022, 07:28:54 pm
content: Quote from: DavidAlfa on May 11, 2022, 06:56:24 pmDiode prices are up a bit compared to when I last bought some.Mouser has (TH) 1N4148 at $0.047 in quantity 100, $0.086 in quantity 10.They are rated for 300 mA max DC current.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: May 11, 2022, 08:19:45 pm
content: On volt regulators:There is a minimum voltage in to make the 10 volt output.There is a drop out voltage. If the drop out voltage is more than 2 volts then you cannot get ten volts out. I seem to recall that most volt regulators have a 3 volt drop out. This means the in volts have to be at least 3 volts more than the expected out volts. I think some regulators are better..So you probably will get about 9 volts out from a run of the mill ten volt regulator using 12 volts in.I do not think this will bother you though in your particular power regulator problem.You might find the spot in your power supply just before the final regulator and take the volts from there and put that through a ten volt regulator, If you require the ten volts exactly.

author: pigrew
author_rank: Frequent Contributor
timestamp: May 11, 2022, 08:31:11 pm
content: I'm going to assume that:have a dynamic load (load current changes)need reasonable regulation (voltage doesn't change much with input voltage or output current changes)have a dynamic load (load current changes)want symmetric suppliesthe load is <100 mAdon't want to waste much powerdon't care too much about costIn that case, a pair ofLDOslinear regulators is what I would use. You can buy a +9 and a -9VLDOregulator. Likely you'll need to use adjustable (since fixed -9 are not so available). LM317 and LM337 are an example. Make sure to add load resisters as needed since theseLDOregulators have significant minimum load current requirements. A LDO could be selected if you need the output to be closer to the supply rails.The proposed series diode would work, but provides bad regulation, and also you need to make sure you have some minimum load current (e.g., add a 10k resistor in parallel to your circuit, to make sure that diodes have an appreciable forward bias).Zeners may be difficult since the drop-out is low. Zener's (with a series resistor), for good regulation, need a current flowing through them similar to the current in your load. Also, the zener current will not be so well controlled since your drop out likely has a large range (1.5 to 2V?). It would be tricky to balance having enough current through the zener and too much that the zener is fried.A zener with a lower current, buffered by a common base BJT could work, as that will be more immune to variations in the drop-out voltage.If your load is more-or-less constant current, then having series resistors would work.The pair of LDO method is the simplest to design, should be reliable, and won't be too expensive (perhaps $1 or $2 total).On the other hand, if my assumptions are wrong, please share your correct requirements.(corrected based on james_s' reply)

author: srb1954
author_rank: Super Contributor
timestamp: May 11, 2022, 10:13:36 pm
content: Quote from: james_s on May 11, 2022, 05:27:56 pmIf you are using CMOS analog switches that have to handle signal swings below ground then you need to use bipolar supplies for those switches.

author: TimFox
author_rank: Super Contributor
timestamp: May 11, 2022, 10:16:51 pm
content: Also, to drive the switches, you may need to connect the digital drivers to the split rails.

author: eugene
author_rank: Frequent Contributor
timestamp: May 11, 2022, 11:50:17 pm
content: Quote from: TimFox on May 11, 2022, 07:25:05 pmThat's all digital logic with inputs and outputs that are referenced to GND. IOW, the chips have two power pins, one is positive supply voltage and the other is ground. Whatever other hardware the inputs and outputs are connected to needs to share a common GND with the CMOS logic.We don't know what else the +/-12V supply will power, but it might make sense to connect the 0V rail of that supply to common GND. If you need a positive supply, then you draw from the +12V connection. If you need a negative supply then you draw from the -12V connection. Power the CMOS with +12V (and GND/0V) and find something else to think about.

author: TimFox
author_rank: Super Contributor
timestamp: May 12, 2022, 02:58:37 am
content: The specs I quoted are for the voltage difference between Vdd and Vss. Except for some mixed-signal CMOS, there is no "ground" pin. When you see a ground pin on, for example, a switch, it really means the reference input for the low state of a logic driver.One interesting analog application for a simple CMOS inverter is to hang a large resistor from the output back to the input, and then drive it with a series combination of a resistor and capacitor.If you have split supplies, you get a cheap (nasty) amplifier that swings positive or negative.If you use positive feedback instead, you can square up a sine wave into a bipolar square wave.The usual use, however, for split supplies and CMOS is analog switching. In the 4000 series, the multiplexer chips could handle an analog range almost from Vdd to Vss, and had a built-in level shifter to drive the switches from a TTL or similar ground-referenced logic signal.

author: TimFox
author_rank: Super Contributor
timestamp: May 12, 2022, 03:10:44 am
content: Another simple and cheap way to drop your voltages are a pair of simple TL431s in series with each supply.The TL431 is essentially a programmable Zener.With no external resistors (only shorts), it will drop the voltage by a reasonably accurate 2.5 V, and can handle from 1 mA (minimum) to 100 mA (250 mW dissipation).This assumes that +/- 9.5 V is safe and effective for your load.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 12, 2022, 03:29:39 am
content: Quote from: srb1954 on May 11, 2022, 10:13:36 pmI found out the hard way that a 4066 isn't turned "off" with zero volts on the control input. They switched Ok, but the unused sections kept conducting, & the device got hot.I ended up with a weird split supply using 6.8v zeners, as that was all the Station store had.

author: james_s
author_rank: Super Contributor
timestamp: May 12, 2022, 06:36:32 am
content: Quote from: pigrew on May 11, 2022, 08:31:11 pmLM317 and LM337 are not LDOs, nor is a LDO needed here, 2V of overhead is enough for a standard linear regulator which the parts you mention are. Low dropout regulators are only needed when the minimum input voltage is within a volt or so of the desired output.

author: chinoy
author_rank: Regular Contributor
timestamp: May 13, 2022, 05:37:34 am
content: I would go with the diodes.Regulator ICs tend to fail. Diodes tend to last way longer.In my circuit I use both i.e. first the diodes and the regulator IC for just the last bit. The regulators are also chained. i.e from 12 v to 5v to 3v.

author: tepalia02
author_rank: Regular Contributor
timestamp: May 13, 2022, 11:07:35 am
content: I would have considered using a ready-made buck converter. Those are quite easy to use.

author: james_s
author_rank: Super Contributor
timestamp: May 13, 2022, 05:16:10 pm
content: I assumed this application is for audio, in which case a buck converter is not really ideal. If the current is just a few tens of mA there is not really any point anyway, a linear regulator is going to be about as efficient at such a low drop.

author: TimFox
author_rank: Super Contributor
timestamp: May 13, 2022, 05:23:48 pm
content: Non-switching voltage reducers, e.g. linear series regulators or series diodes/Zeners/LT431s, will have roughly the same inefficiency: series voltage multiplied by load current must be dissipated in the device(s), extra current comes from the control circuitry in a linear series regulator that further reduces the efficiency. This may or not be a big deal, depending on the application. Similarly, the noise from a switching circuit may or may not be a big deal.

author: David Hess
author_rank: Super Contributor
timestamp: May 14, 2022, 03:14:03 am
content: Quote from: pigrew on May 11, 2022, 08:31:11 pmIf the variation in load current is small, then the requirements for a zener shunt are easier. There are more complex ways to bias the zener shunt to handle a wider range of load current, but at that point an integrated series regulator is probably better these days.QuoteThat works well for higher supply currents, but lacks inherent current limiting so it can be pretty fragile.QuoteIntegrated LDOs have the advantage of current limiting and thermal protection.

author: legrady
author_rank: Contributor
timestamp: May 14, 2022, 10:39:48 pm
content: There's a 40106 inverter and a NAND gate and a 73193 up-down counter, which can run on a single supply. But the 4051 analog switches I want to have (near) maximum supplies, to get the lowest and most linear impedance on audio signals.Tom

author: pigrew
author_rank: Frequent Contributor
timestamp: May 15, 2022, 04:07:28 pm
content: Quote from: legrady on May 14, 2022, 10:39:48 pmI'm worried about supplying too high voltage to the 4000B logic... as TimFox mentioned, they range from +/-7.5 to +/-9V recommended limits. The 20V is the "absolute maximum" which guarantees that part will not be damaged, but does not mean that it will work well.73193 is a typo? Do you mean CD40193B? (I've never heard of 73 series logic, and it sounds like you want supply it with the +7.5 to +9 Vdd.)Since this is an analog application, perhaps someone can comment on the PSRR of the 4051B is? Would a noisy power supply capacitively couple into the signal? I'm not sure if a high PSRR regulator is important or not.These ICs are fairly low current (<1 uA). The LM317/LM337 may need up to 10 mA of load current to maintain regulation (check your datasheet), so a pair of 1k resistors may need to be fitted to ensure the minimum current limit is met. On the other hand, this low current means that a simple zener regulator with series resistor could work well (zener+series resistor+decoupling capacitor).So, what would I do? I'd plan to use +/- 8 V fixed regulators, but also provide a footprint on the PCB that allows the use of adjustable regulators in case of parts shortages.... But, also, I'd design the analog circuits such that the switch's series resistance is negligible (i.e., the load circuit has high input impedance).

author: TimFox
author_rank: Super Contributor
timestamp: May 15, 2022, 04:28:50 pm
content: Back when I used 4051As in an analog switching operation, I found that the most important analog parameter for my application was the change in channel ON resistance with the voltage at the channel with respect to the power terminals.Looking at the current TI datasheet for the 405xB series, I note that at 15 V total supply voltage, there is a broad flattish region in that variation of resistance for the channel voltage between -1 and +3 V (with respect to midpoint of the 15 V supply).