Virtual ground and stability driving a capacitive load - Page 1
Forum section: electronics / projects

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 03:55:26 am
content: I've been working on restoring an old precision current source. One of the portions of the circuit that used components for which direct replacements were not available, is a DC supply that generates +/- 12V. The supply works by rectifying a secondary output of a transformer into ~31VDC, and then running that through a linear regulator to get 24VDC. This 24V is then "split" using an op-amp "virtual ground" circuit (see the portion of the schematic from the manual below).So because I needed to replace this circuit, I just built something myself out of an LM317 regulator to supply the 24V, and an op-amp and a simple push-pull circuit with discrete transistors to supply the virtual ground. The discrete transistors are there because the op-amps I had available couldn't supply enough current. This all seemed to work great, however, I noticed a little bit of oscillation on the output of the equipment. It didn't occur to me that my virtual ground circuit might be the source of this, but now I am starting to think that it is. I had already ripped up the circuit to rebuild the thing on a separate, small PCB, before I thought of this, so I can't just check on the scope to see. However, I did mock the thing up in LTSpice, and I definitely see oscillation in the simulation.I think the root cause of that is the capacitive load on the output (C215 and C216 in the original schematic). That's an obvious source of stability problems when working with op-amps.So my question is: Does this seem plausible? Should I try and compensate the circuit to remove this oscillation, or just remove those two capacitors? My inclination would be to just remove the capacitors, as I don't see that they are adding any value.Relevant portion of original schematicReplacement circuit (from LTSpice model)

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 04:14:14 am
content: 1. Firstly, it is semi-good practice to have a 10k resistor (rather than the dead short you have) between the VGND line and the inverting input of the opamp, to equalise error due to input bias current. Maybe offset on the virtual ground isn't such a big deal in this application, but it also allows you to do step 2 below:2. Random, uneducated though: add a capacitor from opamp output to inverting input? (Without the 10k resistor, the cap would just be fighting against VGND). The capacitor will settle down the speed of the opamp, and hopefully bring it to a stable state. Having said that, I'm fairly bad with control loops and the cap might just make things worse.

author: coppice
author_rank: Super Contributor
timestamp: June 15, 2015, 04:46:34 am
content: Op-amp instability while driving capacitive loads is nicely addresses herehttp://www.ti.com/ww/en/analog/Amplifiers-eBook/#chapter-1-2

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 04:47:46 am
content: Quote from: rs20 on June 15, 2015, 04:14:14 amI've actually tried both those things in the simulation, and some others as well - with a range of op-amps it oscillates. It's a fairly low level oscillation, but if it's happening in the real circuit, it would definitely couple through to the output.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 04:53:13 am
content: Quote from: coppice on June 15, 2015, 04:46:34 amThat basically says stick a resistor between the output of the op-amp/transistor stage and the capacitors/load. Ok, these are also things I have tried in the simulator (I was going through this reference:http://www.analog.com/library/analogDialogue/archives/31-2/appleng.html), but unless I make the resistor prohibitively high, it doesn't help. And if I do that, I can guarantee it's going to cause issues with the circuit this is going into. The whole idea of using the transistor output stage is to allow it to source and sink enough current. Sticking a 500 ohm resistor on the output will guarantee I can't meet that goal.

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 04:59:10 am
content: If you're just leaving the two 100 ohms resistors in the circuit, and adding your 500 ohm resistor afterwards, then clearly that creates a voltage divider, which is why you need such an unacceptably high resistance there.I'd be curious to see what happens if you take your negative feedback off the emitter of Q1, rather than from VGND. That will prevent the voltage divider effect. Obviously this will make the VGND line the wrong DC value, but see if it fixes the oscillation? Then again, everything we're doing here is more or less equivalent to my original suggestion :-/

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 05:07:38 am
content: Quote from: rs20 on June 15, 2015, 04:59:10 amI think you've made an error there. The feedback is coming off the junction of R5 and R6, which means the op-amp will act to make that node match the voltage on it's non-inverting input (12V, the "virtual ground" midway between 0 and 24V supply). Moving the feedback to the emitter of Q1 will mean that it is trying to keep the emitter of Q1 at 12V, and then the voltage at the load will actually be somewhat less due to the voltage divider between R5, R6, the series resistor you've suggested I add, and the load resistor.

author: coppice
author_rank: Super Contributor
timestamp: June 15, 2015, 05:12:34 am
content: Quote from: motocoder on June 15, 2015, 04:53:13 amIt doesn't just say stick a resistor in. It gives you a proper explanation of the root cause of the problem. Adding a resistor alters the relationship between the closed and open loop response curve in a beneficial way. If you can't tolerate the rise in output impedance you need to find an alternative way to achieve the same effect. Its not easy, without adding considerable complexity. These kinds of virtual ground circuits with a single op-amp are common, and are typically poorly done. You often see huge amounts of capacitance, clearly trying to slug the output of the op-amp so hard it can't fight back.:-) . Its not a proper solution, and the amps usually fight harder than you might expect.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 05:27:30 am
content: Quote from: coppice on June 15, 2015, 05:12:34 amThanks for reviewing and commenting. I do appreciate the help reasoning through the options here.I actually understand why it oscillates, and I read several articles on driving large capacitive loads with an op-amp before posting. I can't tolerate a high output impedance because this piece of test equipment places the equivalent of a 200 ohm load on this virtual ground, and it switches from sourcing to sinking current when you adjust one of the controls on the front panel (the output polarity). If that wasn't the case, I could just adjust the 24V supply to set the VGND at exactly 12V above the actual ground because the load when it's set on a particular polarity is relatively constant. However, can't do that for the reasons stated, and I need to keep the output impedance low.So I am not hearing any actual solutions so far that will work for my scenario. It seems like the options are:- Remove the filter caps, and hope any remaining capacitance in the load is low enough not to cause stability issues.- Use an op-amp without built in compensation, and then compensate the crap out of it using its compensation pins- Use some other form of frequency compensation. I did try some ideas there based on the article I posted, but in simulation, I was still seeing oscillations.

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 05:43:05 am
content: Quote from: motocoder on June 15, 2015, 05:07:38 amYeah, I did say the dc value of vgnd would be wrong, and that I was just curious as to whether it fixed the oscillation. If it did, you could generate a feedback signal couple to vgnd in dc and q1e in ac, but like I said that's moving back towards my original suggestion. Don't mind me :-)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 15, 2015, 06:13:10 am
content: What was the original op-amp? LM309 or something?Part of the reason most voltage op-amps are unstable with capacitive load is they're dominant pole compensated. The original appears to have pole-zero compensation (R+C across two pins) which might be enough to keep it happy.Putting a buffer after a generic amp is probably not going to help. The current capacity of that circuit doesn't look much more than the amp itself, and it needs compensation.If higher-than-average current capacity is required, you should shop for an op-amp with suitable ratings, and capacitive load drive capability. A common case is up to 18V supply rating and 60mA output, for something to do with LCD driving. Or if you need more, there are driver/buffer kinds, or you can go to the trouble of building (and compensating) your own (or using a proper split supply to begin with).How much load is actually on that rail? A few mA?Tim

author: dom0
author_rank: Super Contributor
timestamp: June 15, 2015, 12:20:16 pm
content: Op-amp based virtual ground circuits are almost all total crap, since they will always become instable with some capacitive loads.This one is much better. Unconditionally stable.https://www.mikrocontroller.net/topic/258266#2675651Stability comes at a price here (lower precision, higher output resistance), but neither matters in usual applications.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 02:47:29 pm
content: Quote from: rs20 on June 15, 2015, 05:43:05 amWell, I don't have the real circuit to test with, but at least in sim putting a large enough series resistor in before the caps does fix the oscillation - just as the analysis says it will. Could you sketch out a circuit showing the feedback approach you are proposing? I've been tossing that idea around, but it's challenging because the op-amp is configured as a voltage follower, so no matter what you do, the minimum gain is 1.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 02:54:05 pm
content: Quote from: T3sl4co1l on June 15, 2015, 06:13:10 amWrong. Using a variety of op-amps, the voltage virtual ground was getting pulled several volts to one side or the other by the load. My circuit, using a 741 op-amp and the transistors as configured, was able to source/sink up to 80mA while keeping the virtual ground within a millivolt.Quote from: T3sl4co1l on June 15, 2015, 06:13:10 amRequirements here are 24V. Read original post for details.Quote from: T3sl4co1l on June 15, 2015, 06:13:10 amYes, I'll just rip out the transformer and replace it with one that can supply the other 3 rails, plus the split supply for the +/-12V. Let me know where to find one, and make sure it fits the original PCB.Quote from: T3sl4co1l on June 15, 2015, 06:13:10 am40mA or so, both source and sink. Re-read original post.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 02:55:56 pm
content: Quote from: dom0 on June 15, 2015, 12:20:16 pmBut they do matter here. Thanks.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 02:56:39 pm
content: There's a good article at the link below. I am going to try the compensation technique they suggest.http://www.maximintegrated.com/en/app-notes/index.mvp/id/5597

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 15, 2015, 03:11:14 pm
content: Quote from: motocoder on June 15, 2015, 02:54:05 pmIf you merely didn't want help, just say so...

author: c4757p
author_rank: Super Contributor
timestamp: June 15, 2015, 03:11:49 pm
content: Quote from: dom0 on June 15, 2015, 12:20:16 pmThat won't be unconditionally stable with some parasitic layout inductance thrown in the mix. Chances are the ESR of the capacitors will dampen it, but that might not be the case anymore when you get some nice ceramic decoupling caps and whatnot in there too.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 05:11:17 pm
content: Quote from: T3sl4co1l on June 15, 2015, 03:11:14 pmI do want help, but your suggested solutions didn't meet the project requirements. No offense intended, and I appreciate the suggestions anyway. I shouldn't post before my morning coffee

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 05:13:45 pm
content: Quote from: rs20 on June 15, 2015, 05:43:05 amSo, I thought about your suggestion for separate AC and DC feedback paths, as well as similar comments in the Maxim app note I posted earlier in the thread. This lead to the modified circuit below, which seems to be stable in the simulator. I'll breadboard it up and try it out and see if that is the case in the real world too. Thanks for the idea!

author: dom0
author_rank: Super Contributor
timestamp: June 15, 2015, 05:33:49 pm
content: Quote from: motocoder on June 15, 2015, 02:55:56 pmThat it needs to be precision wasn't stated by you, but good to knowbtw. with "higher output resistance" I mean something like a few Ohms.c4757p: I couldn't see any instability even with long leads and ceramic chip caps on the output.

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 05:59:29 pm
content: Quote from: dom0 on June 15, 2015, 05:33:49 pmI know simulation in this scenario can be misleading, but it had to be a lot more than a few ohms to stabilize it in LTSpice. However, with the separate DC/AC feedback paths, this isn't an issue.For those that are wondering why this is a requirement, the Keithley 225 reference circuit generates a 1V reference above this VGND or below this VGND depending on which way the output polarity selector switch is set. This VGND, and hence the + and - 12V off of it, "float" on a bus that sits between the gross current limit sense resistors and the final output current sense resistors on the high side. So it is important that VGND is clean and doesn't change much when the output polarity is changed as any mismatch there leads to mismatched output current when you toggle the polarity.

author: dom0
author_rank: Super Contributor
timestamp: June 15, 2015, 07:53:46 pm
content: Quote from: motocoder on June 15, 2015, 05:59:29 pmI referred to the circuit I linked above ; I only realized later that saying "higher output resistance" in this thread might lead people to believe the linked circuit has 100+ ohms or something like that output resistance.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: June 15, 2015, 08:00:33 pm
content: The first spice circuit you posted has some issues. 1/ The two monsters R3,R4 in theemitters cause the output impedance of the followers to be raised from ~ 1ohm upto 100 ohms.(simply work out what that does to the time constant of any capacitive loads on the output)2/ in your spice sim did you right click and give those Electrolytic's some esr ?.this has to be included because the inherent ESR of large electrolytics makes them stable in these type of circuits and some circuits rely on this to create stable conditions (the original you posted relies both on the opamp comp cap and also the esr of the output caps to provide it's loop stabilty ).Also when using largish caps on the output it's not really needed to bias the push pull as class AB since this is a Voltage regulator and not an Hi fidelty amplifier, so crossover distortion doesn't matter (it takes a few micro seconds for the opamp to swing and bring the complementary transistor into conduction and when there is also a largish capacitor on the output this will dominate the output slewing rate during any load changes as it will hold the output V steady by sinking/sourcing current during fast load changes, (some spice load stepping experiments here and you might notice the transient load response is barely effected by class AB, or B biasing ).I have included an alternate simple circuit below for you compare with your version, the loop response plot is included if your interested, it's got good GBW with < 10 uS recovery . (the components I chose are just example g.p and most general purpose opamp in lm324 class will do the job, and BJT again general purpose medium power but will need to handle your ~ 2 Watt dissipation so to220 devices or somthing. Notethe preload resistor which I included to always insure a small UNBALANCED load is present (if by a long shot you happen to have a perfectly balanced loads on both rails (within 1 mA) then both output transitors switch off which might cause some very minor ringing during fast transients (says spice but I doubt in a real circuit it would).On the output caps if you choose > 10uF low esr aluminiums with esr in 1-6 Ohm esr range then you should not need any Feedback cap at all but if you wanted to include one for peace of mind or perhaps sub zero temp workingthen use a ~ 330p from OP out to - OP in( esr of alco's goes up ALOT when very cold ).Regards

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 08:14:38 pm
content: Quote from: Kevin.D on June 15, 2015, 08:00:33 pmI think you are forgetting that the high gain op-amp with negative feedback lowers the effective output impedance (look where the negative feedback is coming from - it's after the 100 ohm resistors).Quote from: Kevin.D on June 15, 2015, 08:00:33 pmIdeal caps with zero ESR isn't the issue here. It's a widely understood fact that op-amps have stability issues when driving large capacitive loads.Quote from: Kevin.D on June 15, 2015, 08:00:33 pmI couldn't disagree more. Since the circuit is designed to sit exactly midway between the two input rails, without the biasing scheme the transistors would continually be sitting right in the dead zone. This definitely has a detrimental effect on stability and output ripple of the circuit.Quote from: Kevin.D on June 15, 2015, 08:00:33 pmThanks, Kevin. I will take a look at this when I get home this evening.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: June 15, 2015, 09:34:12 pm
content: It took you all of 3 minutes to answer you must'nt have thought about your reply muchQuote from: motocoder on June 15, 2015, 08:14:38 pmIt's irrelevent where the opamp feedback is coming from .The transistor has to source current via it's emitter (it's the emitters resistance that the source current has to come through that is Re+RE where re is intrinsic emiter resitance(small) and RE is any external emitter resistor ) . Think of it like this,Re is small but even if your bjt turns hard on then max current it can source is now limited by RE . thus any load cap can only be charged at at time constant of RE*CloadQuote from: motocoder on June 15, 2015, 08:14:38 pmthen widely misunderstood it seems:) .The problem isn't 'large caps' it's actually 'small caps' like ceramics with low esr's (Electrolytics with there large esr's are nearly always stable ,) the cap forms a pole you see with the output impedance Rout (in your case the 100 ohm resitor you put in the emitter) . this cap creates a pole at 1/2 pi C Rout+ESR so you see the bigger Rout or C then the lower frequency (bad) this pole goes . BUT any resistor in series with the capacitor forms a zero or 'anti pole' which cancels the first pole .This anti pole is at 1/2 pi C esr , so you see if ESR OR C is bigger this zero also moves down in frequency moving down with the pole and this is why you can put massive electrolytics with aparantly little esr on your power supply and it will be stable .Since if we increase capacitance but esr stays more or less the same the anti pole still goes down in frequency with the pole.Quote from: motocoder on June 15, 2015, 08:14:38 pmDid you even read anything what i typed before you answered ? .The time it takes for the opamp to turn on the complement bjt is small compared to how long a 10 uF capacitor can source current to the load for ,thus you may hardly wont even notice it .THATS WHY I suggested you do some load steps with lt spice and posted a bode lot so you could see a real plot of the loop response speed and a 10uS recovery'you strongly disagree with that' ?Let me ask you what you think the GBW of your is ? (with some real capcitive loads on it).but I think you dont need evidence you just to know yeh

author: motocoder
author_rank: Unknown
timestamp: June 15, 2015, 10:24:47 pm
content: Quote from: Kevin.D on June 15, 2015, 09:34:12 pmTurning the other cheek on that comment...Quote from: Kevin.D on June 15, 2015, 09:34:12 pmNo, it's not irrelevant. Let's just look at one transistor, the NPN. It's connected to 24V. Assuming 1V drop across the transistor at saturation, it can source (24 - 1 - 12)/100 = 110mA and still keep the output rail at a stable 12V. That said, I just checked my original circuit, and it IS a typo (should be 10 ohms). Doesn't change my point.Quote from: Kevin.D on June 15, 2015, 09:34:12 pmOk, I guess the EE community at large is all wrong on this issue, and Kevin is correct. Duly noted.Quote from: Kevin.D on June 15, 2015, 09:34:12 pmIt is indeed a pole in the overall transfer function created by the cap.Quote from: Kevin.D on June 15, 2015, 09:34:12 pmI did read what you said. I just disagree. By not biasing those two transistors, the op-amp has to slew rapidly from above to below the midpoint (12V) as the load changes and pulls the signal above or below the 12V mark. So instead of reaching a steady state value with minor output changes as the load varies, the op-amp is swinging wildly over an approximate 1.4V (2x one Vbe drop) range. That has all kinds of potential to inject noise and instability into the system.

author: rs20
author_rank: Super Contributor
timestamp: June 15, 2015, 11:49:16 pm
content: I think there's confusion here around what "source impedance" means in the context of a circuit like this. You're both right insomesense.A. For small changes in output current, the op-amp enforces that the Vgnd voltage stay very constant, so for small currents (and sufficient low frequencies) the output appears to be an extremely low-impedance source, 1 ohm or better.B. If you're only interested in the ability of the circuit to provide a bulk amount of current, then yes, the 100 (or whatever) ohm resistors make the source reach a limit at roughly 11.xV / 100 ohms. Still performs better than 100 or even 50 ohms tacked on an ideal voltage source, but nowhere near 1 ohm in this sense.So just saying that "the circuit has an output impedance of x" seems like a vague oversimplification.

author: motocoder
author_rank: Unknown
timestamp: June 16, 2015, 12:10:59 am
content: Quote from: rs20 on June 15, 2015, 11:49:16 pmThere's no question that the 100 ohms was a mistake. If the circuit is sourcing 100mA, that's 1 watt being dissipated in that resistor. I've since corrected the model to use 2.2ohm emitter resistors (which I think is just enough to help with temperature stability), to add some realistic ESR to those outpuit caps, and to move the series isolation resistor that is used to grab the AC feedback to the output of the op-amp before the transistors. At least in simulation, it's looking very good.

author: dom0
author_rank: Super Contributor
timestamp: June 16, 2015, 07:05:43 am
content: Actually the open-loop output resistance of the circuit is that one thing (okay, together with the o.l. gain/phase response of the op amp, i.e. where it's dominant pole is) that determines stability when driving capacitive loads, since that's whats creating the phase shift of the feedback in the first place.

author: motocoder
author_rank: Unknown
timestamp: June 16, 2015, 01:16:21 pm
content: Quote from: dom0 on June 16, 2015, 07:05:43 amThat is true, and I definitely saw that in the sim when I lowered the value on those two emitter resistors.

author: motocoder
author_rank: Unknown
timestamp: June 22, 2015, 06:49:06 pm
content: Have some updates on this based on building some of the circuits discussed in this thread. Yes, they do oscillate as shown in the simulation

author: motocoder
author_rank: Unknown
timestamp: July 02, 2015, 12:30:49 am
content: I ended up building the circuit using the push-pull discrete transistors and putting it in the Keithley Current Source for which it was intended. The Keithley is well within specs, but it bothered me that I could see a few millivolts of noise on the output. This seemed to be originating from my +/-12V (+24 V with virtual ground) supply.So I kept tinkering with things, and end up with a different circuit. Along the way, I discovered that the primary source of the noise was actually the LM317 I was using to generate the 24VDC from the 31V rectified AC. There's a simple fix for that shown in the TI data sheet. It involves adding a 10uF capacitor from the adjust pin to the negative supply, and a diode to discharge this cap if the output gets shorted to ground.I also got rid of the discrete transistors, and substituted an LT1010 buffer. This unity-gain buffer has no trouble driving capacitive loads, and has a pretty good amount of output drive. I am still using an op-amp with the feedback connected to the output of the LT1010. This lowers the effective output resistance and is necessary to keep the output within specs as the load changes. I have no issues with noise or stability with this circuit.I can post a schematic if anyone is interested.

author: helius
author_rank: Super Contributor
timestamp: July 02, 2015, 01:22:55 am
content: Count me as interested. One of my projects is building a buffered virtual ground for a modular analog synthesizer that is made up of NJM45580 op amps and NJM13600 OTAs. Each module has ~30 uF of capacitance on each rail and there are 5 modules.

author: motocoder
author_rank: Unknown
timestamp: July 02, 2015, 03:33:39 am
content: Here's a rough schematic. I am using one half of a TL082, not an LT1056 as the schematic shows, as that is what I had in my parts bin. Also, the feedback resistor is a 4.7K, not 10K. I tested it driving about 50uF of capacitance. It may do better - didn't check.I am using an LM317 to turn the unregulated 31V into regulated 24V. The calibration procedure for this current source has a fine adjustment of the -12V supply, which is the voltage from the virtual ground to the negative rail of the input to the LM317. That's why I'm using an LM317 versus a fixed 24V regulator. The LM317 is configured as shown in figure 13 in this data sheet:http://www.ti.com/lit/ds/symlink/lm317.pdf

author: motocoder
author_rank: Unknown
timestamp: July 02, 2015, 10:20:09 pm
content: If you looked at that circuit and thought, "wow, that's just trivially simple", you're right. BasicallyI just added an LT1010 to the output of an op-amp buffering a voltage divider. I just installed this circuit, plus the LM317 24V regulator with the improved noise supression cap, in my Keithley 225, and it's working much better than the previous circuit. All the issues I had with noise when trying to calibrate the thing are gone. Performs better than the original now.

author: jimon
author_rank: Regular Contributor
timestamp: July 03, 2015, 05:05:08 pm
content: Beginner question, What does capacitor to ground on voltage divider do ? (C3 on last diagram, or C1 in first post in this thread)If it's just power rail filtering, then why not from rail-to-ground, but voltage-divider-to-ground ? Or it's just there are more 12v caps than 24v caps ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 03, 2015, 06:50:39 pm
content: If noise could be induced to the voltage divider (say from nearby electrostatic sources), the capacitor will tend to keep the AC voltage ground-referenced instead.Note that this is undesirable for a virtual grounding system, because the virtual ground itself is intended to be the reference for the system! All inputs, outputs and bypasses should be referenced to the virtual ground instead, so that it is treated as ground (as odd as that might sound).If symmetrical PSRR is desirable, then a cap divider should be used in parallel with the resistor divider. Using large enough caps, this can suffice for many purposes, including power amplifiers.It is possible to use virtual grounds in a common-negative (or "single supply") system, but those always suffer from startup or shutdown transients, because coupling capacitors are necessary to isolate the -V referenced connections against the amplifier bias voltage (which is around virtual ground potential).Tim

author: motocoder
author_rank: Unknown
timestamp: July 04, 2015, 05:15:49 am
content: Quote from: T3sl4co1l on July 03, 2015, 06:50:39 pmWhat Tim says is correct, but let me add one data point that may not be obvious. For this supply, which is for a Keithley 225, virtual ground noise relative to the "-12V" rail is what is important. For example, you adjust the LM317 supply until the virtual ground is exactly 12V above this rail. I won't go into detail, but read the Keithley 225 manual for details. That's why there's a cap to the lower rail but not to the upper rail.