Tips on reducing ripple voltage on linear regulator - Page 1
Forum section: electronics / projects

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 01:32:10 am
content: I'm designing a headphone amp where ripple voltage on the power supply could really screw with the audio output because of the low voltages required by high efficiency headphones. this is the current design that iprototyped on a breadboard with semi good results. Good thing is nothing was set on fire but unfortunately at 600ma load at around 12 to 13 volts and has about 30mv ripple voltage. I want to know if there is anything i could do to reduce the ripple voltage. The ripple is in sync with the ripple voltage on the unregulated power supply. The transformer is 150va

author: bill_c
author_rank: Regular Contributor
timestamp: June 21, 2020, 01:49:16 am
content: Maybe this?https://www.eevblog.com/forum/blog/eevblog-1116-the-capacitance-multiplier/You using that -18V somewhere else? cause that doesn't look right.

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 01:51:16 am
content: the -18v is used at the emitter of the current source (bottom most bc547 unmarked because of my lack of attention) and i have a voltage multiplier at the vcc for the discrete opamp error amplifier

author: rdl
author_rank: Super Contributor
timestamp: June 21, 2020, 02:12:10 am
content: Why use an unregulated supply if you're worried about ripple?

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 02:17:59 am
content: the circuit above is just the power supply it's the actual regulator. The unregulated voltage gets regulated to around 12v by the circuit above

author: eblc1388
author_rank: Frequent Contributor
timestamp: June 21, 2020, 02:43:44 am
content: May I ask what the transformer secondary voltage is?Have you compared the circuit ripple performance against a simple 7812 voltage regulator under the same 600mA loading?

author: rdl
author_rank: Super Contributor
timestamp: June 21, 2020, 03:48:39 am
content: Yeah, sorry. I didn't look that far.30mV is a lot, I'd also look at 78xx or LM317 to see if they're any better. And be sure that the transformer has sufficient voltage headroom.Quote from: sahko123 on June 21, 2020, 02:17:59 am

author: Ian.M
author_rank: Super Contributor
timestamp: June 21, 2020, 07:16:50 am
content: Your current sink for the long-tailed pair isn't! It needs an emitter resistor. Also Zeners are remarkably poor voltage references when fed via a resistor from an unregulated supply - they only reduce the ripple by a factor of Rslope/(Rfeed+Rslope).I would suggest ditching the -18V supply to the tail sink, and re-biassing the long-tailed pair with a 4.8V reference by putting the two Zeners in series, then split the feed resistor to the Zeners and pre-regulate the feed with a 10V Zener. That + some decoupling caps across the Zeners, and 1uF across the upper resistor in the feedback divider should get you up to 100dB of PSRR from 100Hz to 10KHz.

author: Vovk_Z
author_rank: Super Contributor
timestamp: June 21, 2020, 08:49:12 am
content: Even simple Zener regulator (with bipolar transistor) will give about 30 mV riplle or less.Any integral IC regulator (7812/317) gives less then 1 mV ripple.If input AC voltage is low then there is an low-drop LD1084/LD1085/LD1086 etc.

author: 2N3055
author_rank: Super Contributor
timestamp: June 21, 2020, 09:49:22 am
content: Like Vovk_Z said, any integrated regulator will be order of magnitude better.If you really want something low noise LT3080/81 will have low noise in quite wideband, not only 50/100 Hz.If you really want to go down the rabbit hole, look on Internet for Walter Jung regulators...But all that should not be necessary if you have at least normal amplifier that will suppress PSU noise 50-60 dB..

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 04:33:35 pm
content: currently ive it down to around 15mv by adding 10uf to the zener and am going to add a capacitor multiplier to the negative side as well. ill get back on results

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 05:48:38 pm
content: as it turns out my probing was one of the issuesthe actual figures are 9.6mVpp and 2.6mVRms at roughly 1.2A

author: Ian.M
author_rank: Super Contributor
timestamp: June 21, 2020, 06:12:40 pm
content: That's getting over-complex. I've done a LTspice sim of a modified version of your circuit that has under 30uV ripple out for 1V pk-pk ripple in, costing you three diodes and some extra or bigger caps.Most of the improvement is due to increasing the AC loop gain by bypassing the upper arm of the feedback divider and powering the reference Zeners from the regulated output (apart from during startup). It also uses a diode and reservoir cap to reduce the ripple at the capacitance multiplier input (and boost the average voltage), and has various tweaks to RC time constants to get the corner frequency for LF ripple rejection well under 100Hz.

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 07:38:30 pm
content: How would you power the zener from the regulated output though my idea would be to have a 47k or so resistor from collector to emitter of pass element to get some voltage at the regulated output to start the zener initially but I think this'll defeat the point. Also what do you mean by bypassing the upper arm? Is it capacitoelr bypassing? (sorry replied from phone went on computer and noticed spice file)

author: David Hess
author_rank: Super Contributor
timestamp: June 21, 2020, 07:53:15 pm
content: Power the regulator circuits from the output rather than the input, or from another regulator. Watch out for ground loops. Fix the current sink although if the other changes are made, it can be replaced with a resistor. Just fixing the current sink might be enough to solve your problem.Is a completely discrete circuit required? Just replacing your discrete error amplifier with an operational amplifier and making the other changes would greatly improve performance.I prefer the configuration shown below for high performance regulation. Notice that the error amplifier and reference are powered from the regulated output. The integrated regulator used as a pass element provides fault protection but could be replaced with discrete transistors. If a completely discrete design is required, I would still base it off of this topology.

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 08:08:23 pm
content: Where you biased the Q2 would mean that the base of q3 is at 4.8v for the reference which would give me a dropout voltage of around the reference if i bias it from C3 through 5k6 resistor i could have a lower dropout voltage. What is the purpose of the D5? my guess was decoupling.

author: duak
author_rank: Super Contributor
timestamp: June 21, 2020, 08:50:15 pm
content: I found an interesting idea to reduce the noise of a power supply by adding a parallel instead of a series capacitance multiplier. It used an inverting power stage to apply an inverted version of the noise to cancel or reduce it. The article or paper was from either the US National Bureau of Standars or from NASA but I can't seem to find it now. It may have been in a collection of design solutions.I found and attached another article showing the concept and some sample circuits that were used to reduce broadband noise. Because these are all single ended, and cannot boost the output voltage, they will not help much if the regulator has no headroom. I think to eliminate ripple of more than a few mV would require a power stage that was AC coupled to the power supply output that could source enough current to the load to fill in the troughs of the ripple.

author: sahko123
author_rank: Frequent Contributor
timestamp: June 21, 2020, 09:10:09 pm
content: ah yeah i had a similar idea of having a unity gain phase splitter to invert the noise and ripple of thew unregulated and have it feedback to the power supply of the opamp after the capacitance multiplier to achieve a similar thing

author: SilverSolder
author_rank: Super Contributor
timestamp: June 22, 2020, 02:02:52 am
content: Quote from: duak on June 21, 2020, 08:50:15 pmJ. Linsley Hood has a circuit that does this in one of his books.

author: Ian.M
author_rank: Super Contributor
timestamp: June 22, 2020, 03:12:32 am
content: Quote from: sahko123 on June 21, 2020, 08:08:23 pmI think we have a rather different concept for the meaning of 'dropout voltage'. Dropout voltage is*usually*understood to refer to the minimum input voltage required by the regulator to maintain regulation, and is usually expressed as a voltage differential across the regulator, so adjustable regulators and regulators of different output voltages can be directly compared.However you appear to be concerned about the minimum output voltage,which obviously in this configuration cannot be less than the reference voltage. As your stated application was to power a microphone preamp, I didn't consider it to be important as long as it allows a suitable range of output voltages. If you are building a bench supply so need an output that can go right down to zero, you'd be foolish to start with this circuit. Its hard enough to get right with RRIO OPAMPs and a well stabilised negative bias supply.Regarding D5:Normally a capacitance multiplier follows the bottom of the ripple, as when there is insufficient collector voltage, the base capacitor discharges via the base-emitter junction to supply Iload, instead of the normal Ib of Iload/hFE, and only recovers slowly via its feed resistor. The Schottky diode D5 lets the supplemental reservoir capacitor C5 charge up to the ripple peaks of the Unreg18 rail, and as the load on the capacitance multiplier is small, the ripple at its input is significantly reduced, resulting in a lot less ripple on its output. It also lifts the capacitance multiplier output voltage, resulting in a lower dropout voltage for the regulator as a whole. It becomes increasingly important as you increase the load current, as that linearly increases the ripple on any unregulated supply consisting of a rectifier directly feeding a reservoir capacitor.

author: sahko123
author_rank: Frequent Contributor
timestamp: June 22, 2020, 06:47:39 am
content: This isn't a microphone pre-amp supply but a headphone power supply, still requires low noise and ripple. My definition for dropout voltage is the minimum difference between input and output so if under load the negative peak of the input ripple from the rectifier and reservoir capacitors drops to 14V and you have a drouput voltage of 4V then the regulator will go down to 10v and follow the input ripple on its way back up but if the dropout is lower the input voltage can go lower without affecting the output voltage as much

author: Ian.M
author_rank: Super Contributor
timestamp: June 22, 2020, 07:56:49 am
content: The dropout voltage is constrained by several things. The capacitance multiplier drops about 0.9V, the current mirror output side drops about 0.3V Vce + the drop across its emitter resistor, and finally the pass transistor Darlington pair contributes two Vbe drops so about 1.3V. Reducing the emitter resistors R6 and R7 to 100R will save you maybe half a volt. The action of D5 to lift the supply to the current mirror is much more important because it means at high load currents, if the ripple is over about two volts, the dropout voltage is reduced to the min. Vce drop of the Darlington of about 1V.It isn't in any way constrained by my choice of reference voltage - I could re-jig the circuit with 4.7V Zeners, and if I kept the tail current the same, the only side effects would be an increase of the min. output voltage to 9.4V and, I think, slightly poorer regulation with respect to load current due to the lower 'headroom' for the long-tailed pair.N.B. the exponential voltage source I used in the sim to approximate the ripple doesn't model the ripple increasing with load current, nor of course does it model transformer regulation. OTOH it does permit a DC operating point to be found, permitting AC analysis for the frequency response of the PSRR and load regulation. One issue that will crop up if you attempt to edit V1 in its Specialized Component Editor is it will loose the the extra (seventh) parameter 10m which sets the repetition rate of the ripple. To avoid this edit its value, currently:Code:EXP(17.4 18.6 0 0.2m 0.4m 3m 10m )and reinsert the 10m just inside the closing parentheses. Seehttps://www.analog.com/en/technical-articles/ltspice-using-time-dependent-exponential-sources-to-model-transients.htmlfor the extra EXP() parameters and general hints on pulse modelling using EXP() sources.