Linear power supply with multiple outputs. - Page 1
Forum section: electronics / projects

author: toto83
author_rank: Contributor
timestamp: March 22, 2018, 05:57:15 pm
content: Hello,I am creating an preamp circuit for a microphone. I wanted to do it with a linear PSU (I haven't got enough knowledge to do an SMPS.)I need two voltage rails : 48V for the phantom power and +- 15V DC for an op amp.I currently have a 30-0-30 transformer (yeah I know it will be overkill) I must use for this project.I firstly wanted to do an op-amp to create the +-15V with 30V DC The problem is, I need them all to have the same GND.How can I achieve this ?Thanks for your answersTom

author: Cerebus
author_rank: Super Contributor
timestamp: March 22, 2018, 06:12:35 pm
content: The current available for the 48V rail only needs to be small; 48V phantom power is supplied through a pair of 6k81 resistors, so you only need to supply just over 14mA. The raw supply for the 48V rail is easy to provide from a classicvoltage multiplier, followed by a bit of smoothing. Regulation on the 48V can be quite loose, but you do want it to be quiet.A 30-0-30 transformer isn't really the best starting place. If it's what you've already got lying around, so be it. But if you're starting from scratch then a very small (3VA, 5VA?) 18-0-18 transformer is the obvious choice (18 * sqrt(2) = 25V peak, doubled = 50V).

author: toto83
author_rank: Contributor
timestamp: March 22, 2018, 06:25:04 pm
content: Thanks @CerberusI already created the 48V rail, and i know my tranformer is a bad choice, but it is the only thing I have right now.Any Idead to create +-15V, with the other rail?

author: Cerebus
author_rank: Super Contributor
timestamp: March 22, 2018, 06:34:45 pm
content: Show us your schematicso faras it has come.

author: mariush
author_rank: Super Contributor
timestamp: March 22, 2018, 06:42:46 pm
content: Just use the 30-0-30 transformer and two linear regulators to get 15v and -15v, for example see :Replace if you want LM317 and LM337 with 7815 (or any fixed 15v linear regulator that accepts up to 30v+ at input) and 7915 for -15v (and since they're fixed linear regulators you don't need potentiometers or two resistors per regulator to set output voltage so R1..R4 can disappear from schematic)C1 and C2 ... 50v or higher rated electrolytics/tantalum/polymer , capacitance based on maximum voltage drop acceptable to you and current in circuit .. C (farads) = current / [ 2 x Mains Frequency (50 or 60hz) x ( Vdc peak - Vdc min)] and you'll want the minimum voltage to around 17v ... if i were to guess I'd say 220uF ..470uF would be common easy to find electrolytics and enough for an opamp.C3 and C4 some 0.1..0.47uF ceramic capacitors as close as possible to input pins of regulatorsC5, C7 something like 10..100uF electrolytic capacitors rated for minimum 25v (since the output is 15v and -15v and too close to 16v rating) ... C6 and C8 are not needed.Use a simple boost regulator (mc33063 or something similar) to boost whatever input voltage you have ( what's after C1 but before the linear regulator, some voltage between around 17v.. 30v-ish Vdc if your windings are 30v ac each)

author: C
author_rank: Super Contributor
timestamp: March 22, 2018, 06:55:19 pm
content: Might help you a lot to think of an op amp the correct way.Power to the op amp comes from the +-V pins.These pins also supply power to output.The + input is high impedance.The - input is a virtual 0 with the other probe connected to the + input.This is not ground, this is a voltage difference between the + input and - input.note that output changes based on this very small difference.Thinking virtual ground is bad news, thinking virtual 0 with respect to + input is good.Think of putting voltage dividers or resistor network around op amp. As op amp output changes based on input difference you want the resistor dividers/ resistor network are reducing the input difference.Would be a good idea to read/study some of the many how to's for op amps.If preamp current needs are low enough you could even use op amp's outputs as preamp power source.

author: toto83
author_rank: Contributor
timestamp: March 22, 2018, 07:40:08 pm
content: Quote from: C on March 22, 2018, 06:55:19 pmThank you for the advice.I think I have not been clear enoughMy problem is the following.I have these two windings.I want to create a +-15V with an opamp symmetrizer (with a voltage divider ofc) from 30V and a few bjts with the first winding and 48V with the second one.But, can I connect the +48V ground and +-15V <<ground>> (output of the opamp) ?

author: C
author_rank: Super Contributor
timestamp: March 22, 2018, 09:53:41 pm
content: Apply some common sense to problem.First the important equation.E = I x RWhat is a transformer.Simple version is just two coils of insulated wire. If you measure resistance between the two coils you should get a high resistance. With a high resistance very little current can flow.Now look at that equation.To have a current flow you have to have a voltage. If you have 0 voltage then no current can flow for any value of resistance.Said a different way, if you measure the voltage across a resistor and get 0 volts then no current is flowing.Think of a batteryYou can connect one contact to anything and no current flows because there is no path for current.Keep in mind the real world, every thing has some resistance, Even air & glass. A good insulator is just something that has a very high resistance so little current can flow.Now a good way to convert AC to DC is with a full wave bridge. You do not get DC but half sign wave AC but you have not added a connection to something else.You put a cap across the bridge. Now you have a DC voltage with ripple.You can build the equal of a battery with each winding of a transformer.You have option of where you connect batteries just like in above winding, bridge, cap circuit.Now think of a transformer winding with more taps. If you have equal number of turns on each side of tap then haveone AC signal between the two ends.One AC signal between tap and one end.one AC signal between tap and other end.If it's a center tap these last two are equal.Note that with a center tap voltage is increasing positive at one end and increasing negative at other end of winding.Now think a bit about you preamp.A ground connection has noise so connecting it to ground you could be adding noise.For current to flow you just need two connections.For stereo, you need four connections, unless you connect two connections together. Doing this means that you lose some separation of the two audio channels. You have one wire carrying the sum of the currents for the two channels. Wire has resistance, so you have different voltage drops on the three wires where with four wires you have two matching drops.Using just three wires is done just to save wire in one instance.A great preamp would use four op amp outputs to get the four connections needed.So think about what happens when you connect the two channels and if you connect a ground.

author: toto83
author_rank: Contributor
timestamp: March 23, 2018, 11:46:52 am
content: Thank you C, I think I understood.The preamp will be used for a XLR Mic, thus GND is needed.This is what i was thinking abouthttp://sound.whsites.net/project66.htm

author: C
author_rank: Super Contributor
timestamp: March 23, 2018, 03:04:57 pm
content: Ok low level adviceGround is bad news, When you see something labeled "GROUND" start thinking what kind of ground is it.Most times ground is actually 0 Volts, but 0 volts for "WHAT"?In stead of saying"0 volt audio reference""0 volt noise shield""0 volt signal reference""0 volt power reference""0 volt analog reference""0 volt digital reference"People shorten that to just ground. Then seeing ground, think ground is ground and start connecting ground signals and in the process connect noise of one to the other.SoXLR connectorhttps://en.wikipedia.org/wiki/XLR_connectorYou see Pin 1 is "Chassis ground (cable shield)"Not really a part of audio as audio is on 2 & 3. This is protection from surrounding world.Care should be taken on where and how this is connected to keep that protection high.Balanced audiohttps://en.wikipedia.org/wiki/Balanced_audioSo when you start connecting ground all over the place you get "Ground Loops" which are bad news.You might not really see a problem until you go to extreme example of use.Think of 1000 feet of cable connecting each of mike to preamp, preamp to down stream audio, preamp to power.That 1000 feef of wire has resistance, capacitance & inductance. You also have a time delay.If you shorten or lengthen the wire, all are still there, just the magnitude of vlaues changes and have a greater or lesser parentage effect on original.So in an ideal world when you start with balanced audio input, you have a balanced audio output to speaker and every thing in between is balanced.Fully balanced, you have one signal/wire going positive and other going negative. The current needed to make the change is on the two wires or signals. This costs more in electronics and sometimes wire so people try to be cheap and in process cause problems and you get less then you could have.A lesser choice would be to have one wire changing and other fixed. Here you are still keeping current needed to change on these wires but still have a change on other end of wire.So a smart person might first redraw that schematic such that you have different lines for each type of 0 volt use and put proper labels on them.Here you have three.Power supply 0 voltslAudio 0 voltsOther 0 volts.You might see that instead of connecting XLR mike pin 1 on input side of preamp. it should be connected differently to prevent other world currents from effecting preamp as much.Taking Care of return path could make a huge difference.An exampleIf you know any thing about a switching power supply, you know that it is high noise. You can reduce this noise effecting rest of circuit by using just three connection points & good examples from manufacturer of switcher will show this.So XLR mike pin 1 if connected to preamp circuit is connected in one spot only.A good balanced audio amp has equal currents on positive & negative rails. When you can not have this, keep currents small and get balanced again locally. A good audio amp with RCA jacks for input will treat each as two separate wires(signals) even with one wire not changing voltage(current does change with signal).The weaker the signals the more important this is.So a good balanced amp is a thing of beauty and this beauty counters many problems. Think of power supply noise. The noise is positive going on one rail and negative going on other rail, and circuit only has to adjust/compensate to supply changes.Now a balanced signal has a problem in that the signal is not always balanced and this can generate a DC offset. A PWM signal is one place this is used and the wanted effect. So a way to restore DC offset is needed.So if you put this all together for preamp,You want a separate floating power source of 30 volts with a local generated center tap. To reduce noise this power source only supplies the preampThe preamp should not be effected if power is balanced at -+15, +-16 or +-14 volts. Power supply voltage changes should have little to no effect.If you have to think ground then that poor audio RCA input jack shield is the ground. But notice that a good audio amp will have a ground stud to connect grounds so that RCA shield is back to being more audio 0 volts.

author: rstofer
author_rank: Super Contributor
timestamp: March 23, 2018, 03:49:53 pm
content: Your transformer secondary voltage is FAR too high for a standard 7815 regulator. At a guess, the peak voltage will be about 1.4 * Vrms which implies about 42Vpeak on each winding. A 7815 has a maximum input voltage of 35V and a recommend input of 17.5-30V. A 15-0-15 transformer would be a better choicehttp://hades.mech.northwestern.edu/images/6/6c/LM7805.pdfPage 2 has Max ratings, see 7815 variant further on in the datasheethttp://www.hep.upenn.edu/SNO/daq/parts/lm7915.pdfAll the other details given above are very interesting. The +-15V supply is conventional and the boost regulator has all the required information. But the +- 15V regulators aren't going to survive.

author: mariush
author_rank: Super Contributor
timestamp: March 23, 2018, 09:56:54 pm
content: Quote from: rstofer on March 23, 2018, 03:49:53 pmIt's a stupid microphone preamp circuit.If his opamp is gonna use something like 50-100 mA on each rail , then he could simply add a couple of resistors in front of the linear regulators to drop some voltage, for example let's say a 100 ohm resistor ... at 100mA it's gonna drop 10v and it's gonna dissipate 1w ... so just shove there a couple 100w 3w resistors , or maybe 2 x 4.7-5.6 ohm 1w resistors in series on each rail.Or play with some zener diodes... at low currents it would be fine. Or put a few 1n400x diodes in series to drop around 0.5-0.8v on each.

author: C
author_rank: Super Contributor
timestamp: March 23, 2018, 10:33:38 pm
content: And mariush's resistors adds isolation from other power use.

author: KD4PBS
author_rank: Contributor
timestamp: March 27, 2018, 04:09:33 pm
content: Keep It Simple, Silly.Resistors, zener diodes, and a half-wave bridge seem to be all that would be needed here. Get your +48V from the "+" phase of the transformer via a standard voltage doubler, then on through a current limiting resistor, a filter capacitor and then the cathode of a 1N5368 47V zener. Since the phantom power will only draw microamps at best, the current limiting resistor can be sized pretty large (resistance-wise), and give built-in short protection for this supply.Use a half-wave rectifier on each phase of the secondary to give you enough voltage to feed a resistor, capacitor, and a 1N5929 15V zener per phase arranged to deliver the + and - ends of your bi-polar supply. Diddle with the value of the current limiting resistors to allow for enough expected current for the op amp while keeping the zener within current limits.No need to resort to expen$ive linear ICs to do things we used to use discrete components for back in the day, and his transformer is sized perfectly to easily deliver all three voltages.KISS!

author: KD4PBS
author_rank: Contributor
timestamp: March 28, 2018, 03:57:32 pm
content: Like this...

author: Cerebus
author_rank: Super Contributor
timestamp: March 28, 2018, 05:25:47 pm
content: Quote from: KD4PBS on March 27, 2018, 04:09:33 pmA 48V phantom supply should be able to supply 14mA, as per IEC 61938.Quote from: KD4PBS on March 28, 2018, 03:57:32 pmTry moving those resistors toafterthe smoothing caps. Putting them before them, in combination with the zeners,createsadditional output ripple and raises the quiescent dissipation for the supply.Just looking at the 48V rail: Moving the resistor to after the smoothing cap almost halves the ripple in the loaded condition, which is still a horrible 3.2V, down from 5.9V. Clearly not an acceptable level of ripple for a microphone supply.Here's the LTSpice simulations demonstrating the above, and as a bonus, a variation with a simple low noise regulator with 4mV p-p ripple. It's thrown together, the values for all the partswork, but they are far fromoptimised.

author: KD4PBS
author_rank: Contributor
timestamp: March 28, 2018, 08:01:50 pm
content: Ripple depends greatly on the amount of current the circuit draws. We don't know what the current draw of the OP's circuit is. We do know that the typical mic will draw only a couple of milliamps for phantom power (sorry - I mistakenly wrote microamps). I've never seen a mic draw more, but I've only measured 4 of them in the past 30 years.Got a little ripple? Add capacitance. I was going with 50µF just as a WAG. No reason to not use 1000µF or more. When I did this on the +48 supply, even after adding a 15K load resistor, ripple drops to around 60 mV. I know it sounds like a bunch, but remember that the phantom supply will be applied common-mode on the mic output, so it won't be heard on the input.As for the +/- 15V supply, changing the resistors to 100?, the caps to 1000µF, and adding a second 1000µF cap across the load, I see only about 40 mV of ripple assuming a 100mA per leg load. As an added bonus, this ripple is in common-mode, so I think it should cancel itself out on the output of the OpAmp.Good catch on the position of the resistor!Still trying to KISS...

author: mariush
author_rank: Super Contributor
timestamp: March 28, 2018, 08:54:23 pm
content: Something that bugs me about these schematics and designs...You say that there's no need to buy regulators like 7815 or 7915 because you can use Zener diodes to do the job in a simple way.But if the guy wants to make ONE UNIT or a very low unit count of this product, chances are he'll find 7815 and 7915 or LM317/LM317HV and LM337 in local stores or at least within the country while some odd zener diodes like 1n5929 or 1n5368 may be impossible to find locally.Resistors to set the output voltage on such adjustable regulators are super common and he'll either have them in his local stock already or easy to find even locally (or harvest from some dead electronics he picks from trash)And then you have that third schematic and I'm starting to count the number of parts you involve just to output that 48v voltage ... you have 2 in4148, you have two transistors BD140 , BD139, you have a single 47uF capacitor, a single Zener diode and then you have some odd valued resistors (85 ohm , 60.4k ohm) and a 15ohm resistor and a couple of 6k81 resistors.Let's say he places an order for these parts.. how expensive would these be if he orders just 1 or 2 of each? It's just silly.Would make more sense to just buy 5 or 10 of those linear regulators and get a good volume discount and he'll most likely use those parts in the future. 1n400x diodes are everywhere and he could use whatever's cheap in volume or available at local stores, for example the same 100uF 100v electrolytic (or something like that) everywhere.He'll find a use for those extra parts within some reasonable amount of time, while extra zener diodes would probably just sit on shelves somewhere until he throws them away.And instead of 10 components just to get some stable 48v when to me it would just make more sense to me to use something like LR8N3 regulator for 0.61$ (even as 1pcs) :https://www.digikey.com/product-detail/en/microchip-technology/LR8N3-G/LR8N3-G-ND/4902368orLR8K4 in to252 packagefor 0.96$ if you want better heat dissipation.Has 12v dropout so I'd want the input voltage to be around 60-70v, has output current limit set above 10mA (between 10 and 30mA) and you only need 2 plain resistors to set the output and maybe a resistor as minimum load (you want min 0.5A load)And last but not least, you're also getting into pcb space... so many components would take place on pcb making it expensive (for a one off product) and reducing the options for cases... granted in this case the big transformer is gonna affect the design more than the small parts..

author: Cerebus
author_rank: Super Contributor
timestamp: March 28, 2018, 09:23:21 pm
content: Quote from: KD4PBS on March 28, 2018, 08:01:50 pmRather than design for a specific microphone, why not just design to the standard, which as I said expects a 14mA capability.QuoteWhy stress the CMRR with massive ripple when a handful of components gets you a quiet, regulated supply?Let's assume you opt for a 1000uF 100V smoothing capacitor, with that you've still got 156 mV p-p of ripple @ 50Hz. That's a 69 mVrms signal for the waveform that you end up with, which comes out at at suspiciously round -21dBm into 600R. Why put up with -21dBm common mode noise when you can design it out? With a typical pro-sumer microphone input that would appear as a 50 Hz hum at around -81dBm and any mismatch in the feed arrangements would make it worse (60dB CMRR assumes 0.4% or better matching in the phantom power feed resistors).

author: Cerebus
author_rank: Super Contributor
timestamp: March 28, 2018, 09:59:31 pm
content: Quote from: mariush on March 28, 2018, 08:54:23 pmNo, you can't use the LM317 and friends for a 48V supply because:So to use one, you're going to have to 'stand' the whole thing on a zener anyway. Moreover the OPs use of a 30V transformer means that the rail for the 48V supply is at 70V. That forces several design choices that one wouldn't normally make. My 'trivial' regulator design has around 2.5W quiescent power usage which I'd normally regard as horrendous for this kind of supply, but then I'd have chosen a transformer with an appropriate output voltage.Anyway the OP is in France. If we limit designs to what you can find in electronics scrap in the third world then let's start from that as a design goal, not add it in as a theoretical third party requirement half way through. There's nothing obscure about the 1n5929 or 1n5368 zener diodes they're just bog standard 5W zeners, any suitably rated zener would do.QuoteThe 6k81 resistors are, to all intents and purposes, part of the phantom power specification. You can use anything 'around' 6k8 but whatever you use needs to be matched to 0.4% or better for CMMR matching purposes. The other parts are, frankly, pretty arbitrary values and can be easily substituted if one has the faintest idea of what job they are doing. I did say "It's thrown together, the values for all the partswork, but they are far fromoptimised.".How many parts do you think are inside any off the shelf IC regulator. Alotmore than 2 transistors I can assure you.QuoteNobody asked for a BOM optimised, board area optimised, ready for production design here. We're just kicking around design ideas. If the former is what you want, buy a kit. If, on the other hand, you're interested in how things get designed, and how they work then continue listening. I suspect however that you're just bitching for the sake of it, your earlier contribution below seems to advocate "throw parts at it" and now you complain when a handful of jelly bean parts gets used.Quote from: mariush on March 23, 2018, 09:56:54 pm