Negative version of LP2951 LDO - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 09, 2019, 12:55:48 pm
content: I'm looking for a negative output LDO, 100ma.I found the positive version LP2951, however I'm unable to locate its negative version.Like the LM317 has a LM337 version, I was expecting the same for the LP2951.Am I missing something here or the negative version just does not exist for the LP2951?

author: Siwastaja
author_rank: Super Contributor
timestamp: September 10, 2019, 06:28:03 am
content: Most regulators on the market don't have a "closely related" "negative version" with a similar part number. Positive regulators are much more widely needed, thus a larger selection exists.Just do a parametric search on your supplier to find the negative voltage regulator you need, doesn't need to be closely related to any positive regulator you are using in the same circuit.

author: Andreas
author_rank: Super Contributor
timestamp: September 10, 2019, 06:50:44 am
content: LT1964 if you can live with -20V maximum input voltagewith best regardsAndreas

author: RES
author_rank: Regular Contributor
timestamp: September 10, 2019, 08:10:51 am
content: LT1175 ?-20V (-25V abs. max.), adj. Vout, 200-800mA, sense, current limit pins, shdn, SOIC8, 45uA op. curr., 10uA shdn curr.https://www.analog.com/media/en/technical-documentation/data-sheets/1175ff.pdf

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 10, 2019, 08:23:57 am
content: Thanks to all the suggestions.Price also is an issue.TheLP2951 => $0.5LT1964 => $3.6LT1175 => $5.8I was looking for something in sub-dollar range!

author: Siwastaja
author_rank: Super Contributor
timestamp: September 10, 2019, 08:30:05 am
content: Sub-dollar things are for mass market devices, which almost always run on single-ended supplies. A small positive regulator IC may be manufactured in tens of millions; a negative regulator in hundreds of thousands, maybe. Hence the price difference.A regulated negative rail is something a mass market device designerreallywants to get rid of, this is one of the first steps of product optimization.Hence we have some modern opamps, comparators, ADCs etc. which can read down to ground, or even slightly below (say -0.2V), with a single-ended supply.That's why negative regulators are a bit of niche, and as such, more expensive, with smaller selection.For price reduction, try to think how you'd get rid of the negative rail. If you really need it, does it really need to be regulated?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 10, 2019, 08:31:47 am
content: OK, so I did a recheck on Mouser,for negative voltage LDO's looks like there is nothing in the sub-dollar range.The best it shows is TPS72301DBVR which slots in at $2.7 but its max input is -10V.The next best in terms of voltage is LT1964 at -20V max input.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 10, 2019, 08:50:55 am
content: Quote from: Siwastaja on September 10, 2019, 08:30:05 amI am trying to build a audio mixer in the virtual earth configuration.So my thinking was virtual earth needs the non -inverting input to be at ground potential, so that pulls the inverting input to a virtual ground.Is my thinking flawed and if so is there some means to eliminate the negative power supply in a virtual earth configuration.

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 10, 2019, 02:54:22 pm
content: Quote from: ZeroResistance on September 10, 2019, 08:50:55 amWhat Siwastaja has said is correct, but audio - at least at better than low-end consumer grade - is one application where having bipolar supplies really helps simplify circuits and improve performance. That said, it is also an application where the regulation of the supplies doesn't need to be particularly good; just try to keep the ripple to a minimum.

author: edavid
author_rank: Super Contributor
timestamp: September 10, 2019, 03:28:24 pm
content: 1. For a simple design like this, a rail splitter would be as good as split supplies2. Unless you know exactly what will be connected to the inputs, it's better to AC couple them3. Don't you need a buffer after your volume control?4. For the negative supply, why were you thinking you needed an LDO? Would an extra volt drop really be significant?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 10, 2019, 05:31:51 pm
content: Quote from: edavid on September 10, 2019, 03:28:24 pmHow does one do that?QuoteThe inputs would be headphone outputs from other devices. What is the advantage of AC coupling, isn't it better to DC couple and get rid of the DC offset.QuoteSince the outputs are phone outputs, I didn't consider it necessary.QuoteI'm currently considering the good old 317 / 337 combo but was interested in the LP2951 due to its low dropout and thought it would be easy to find the negative version of it...Also i'm currently generating the negative supply with a charge pump, because I wanted to keep noise low and didn't go in for the switched inductor power supplies.

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 10, 2019, 07:43:17 pm
content: Quote from: edavid on September 10, 2019, 03:28:24 pmFunctionally, sure - it just turns the mid-point of a positive supply into the 0V reference - but this also elevates the "supposed to be grounded" shell of shielded interconnects to the mid-point voltage which can cause problems when connecting to other devices, especially if they share a chassis ground and/or are earthed.Quote from: edavid on September 10, 2019, 03:28:24 pmAgreed, though with proper bipolar supplies it is perfectly acceptable - and even preferred - to just use a coupling capacitor on the input, where even a small capacitance will still allow for a low frequency 3dB point in the single Hz range. The output coupling capacitor can - and should - be dispensed with.Always keep in mind that wiring 3 or 4 RC networks in series with overall gain present is a great way to make an oscillator...Quote from: ZeroResistance on September 10, 2019, 05:31:51 pmWith just a voltage divider, technically, but it's better to buffer the divider output with an op-amp follower so you don't have to use inconveniently small value resistors. Note that this circuit works best when the current drawn from each rail with respect to 0V is balanced, but that is pretty much always the case with audio.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 10, 2019, 09:24:32 pm
content: Quote from: MagicSmoker on September 10, 2019, 07:43:17 pmSo If I have 2 stages,1 Summing amplifier and1 Power amplifiersDC blocking caps at the Summing amplifier is sufficient?Is there No need to add it further down the line? Won't the summing amp stage add a dc offset at its output?

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 10, 2019, 10:04:09 pm
content: Quote from: ZeroResistance on September 10, 2019, 09:24:32 pmNo, I was saying that as long as bipolar supplies are used you don't need a coupling cap at the output of each stage. In other words, the output of one stage doesn't need a coupling cap if the input of the following stage has a coupling cap (which it should, for reasons already pointed out).For interstage coupling in an amplifier string with modest overall voltage gain* you can usually dispense with the DC blocking capacitor, but since you will only need a few uF to maintain a low-end frequency response down to a few Hz why not use one there, too? It's getting rid of the DC blocking capacitor between the power amplifier and speaker that is the real benefit** (for distortion, reducing "thump" at turn on/off, and cost).* - for example, you need 20Vrms to drive 8R speakers at 50W, and the standard line level signal is 1Vrms (0dBV), so that requires a voltage gain of 20; even relatively high offset op-amps like the NE5534 (5mV worst case) will only subject the speaker to a DC bias of 0.1V here. That's about 1.25mW of DC dissipation, which is basically negligible.** - as well as the DC biasing resistors needed for every op-amp if a single supply is used.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 11, 2019, 07:44:38 am
content: So I know impedance of at least one of the outputs that will connect to my Audio Mixer input.Its output impendance is 200ohms with max 4V rms.So would I need to match the impedance for this source. The circuit which I had posted earlier had just a 10K audio taper to attenuate the phone output followed by a 1uF dc blocking cap?

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 11, 2019, 10:25:20 am
content: You don't worry about matching impedance between outputs and inputs in audio, nor do you even want to. In fact, you want to present any signal source with a relatively constant (and relatively high) impedance so peaks aren't distorted from increased loading.You also want the DC blocking cap to work into a relatively high and constant impedance to minimize its size and maintain a consistent frequency response, so the better arrangement would be to put the cap first - that is, in between the source and the top of the pot - with the wiper from the pot going to the mixer input.In your example the volume control pot is 10k and the DC blocking capacitor is 1uF so the -3dB point (that is, when the reactance of the capacitor equals the resistance of the pot) will be 16Hz*. For high-fidelity applications you'll want the cutoff under 10Hz and this is best done by increasing the capacitance, rather than the resistance of the volume control pot, because the latter makes the circuit more noisy. I'd use at least 3.3uF for the DC blocking capacitor here, though there is no harm in going up to 10uF so long as you don't use ultra high-k ceramic types (like Y5V and Z5U) as they are piezoelectric and suffer a number of other pathologies that make them terrible for use in audio or other sensitive signal amplification applications.* - the equation to determine the reactance of a capacitor at a given frequency is Xc= 1 / (2 * Pi * f * C)

author: exe
author_rank: Unknown
timestamp: September 11, 2019, 11:29:41 am
content: Quote from: MagicSmoker on September 10, 2019, 07:43:17 pmCan shield be connected to "real" ground? In my understanding, any fixed potential is good for shielding.

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 11, 2019, 12:04:13 pm
content: Quote from: exe on September 11, 2019, 11:29:41 amShield is usually connected to earth ground in XLR (balanced) interconnects, but shouldn't be - yet still often is - in RCA and similar type interconnects. In the latter cases this results in the dreaded ground loop.

author: Zero999
author_rank: Super Contributor
timestamp: September 11, 2019, 12:56:51 pm
content: Quote from: ZeroResistance on September 10, 2019, 08:50:55 amJust AC couple all of the signal paths and bias the amplifiers to work at half the supply voltage.The volume control should be between the two amplifier stages. I've converted the second one to a unity gain buffer and put the volume control before it.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 11, 2019, 01:12:41 pm
content: Quote from: Zero999 on September 11, 2019, 12:56:51 pmThis is still great! but would it still count as an virtual earth mixer?The earth point has now shifted to 1/2 VCC.From what I understood about the concept of a virtual earth mixer, is that if all the input signals see a true virtual earth. There won't be any crosstalk between them.

author: Zero999
author_rank: Super Contributor
timestamp: September 11, 2019, 03:16:26 pm
content: Quote from: ZeroResistance on September 11, 2019, 01:12:41 pmI don't know what you mean by virtual earth mixer? Are you talking about using an op-amp as a rail splitter?Note that ideally the potentiometer should be AC coupled anyway, to cut down on noise when it's adjusted, but in this case the DC current should be no more than the op-amp bias current and the gain is unity, so it shouldn't be too bad.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 12, 2019, 12:18:41 pm
content: Quote from: Zero999 on September 11, 2019, 03:16:26 pmI read about it herehttps://sound-au.com/articles/audio-mixing.htm#s3

author: Zero999
author_rank: Super Contributor
timestamp: September 12, 2019, 02:40:21 pm
content: Quote from: ZeroResistance on September 12, 2019, 12:18:41 pmThat's got nothing to do with whether the circuit is operating from a single supply rail, a bipolar supply or a virtual ground. Both of the circuits I posted use the same virtual earth principle, as described in that article. The difference is that first one I posted was AC coupled and the second one DC coupled. In the AC coupled design, the DC is blocked, so only the signal matters, the fact that the negative input is fixed at half the supply voltage vs 0V doesn't matter.The circuit works by fixing the inverting op-amp input at a constant DC voltage. In most cases it's the same as 0V, because it's what the non-inverting input is connected to, hence virtual earth, but it doesn't really matter. Take the input resistors away and you'll see that the impedance into the inverting input is very low, due to negative feedback. If a current is injected into the input, the op-amp's output will move in the opposite direction, pulling current through the feedback resistor to counteract the change in voltage due to the input current.The circuit is a transimpedance amplifier. It converts a constant current into a constant voltage. In this instance, with a feedback resistor of 10k, 1mA of current in, will give -10V out, because the input will stay at 0V and 1mA of curenet will give a drop of 10V across the resistor.https://en.wikipedia.org/wiki/Transimpedance_amplifierKirchoff's current law states that the currents through all the resistors add together, at the node where they're all connected. Suppose you have 5V at two inputs, with 10k input resistors. The current through each resistor will be 0.5mA, giving 1mA in total, which also flows through the feedback resistor, giving an output of -10V.

author: Bassman59
author_rank: Super Contributor
timestamp: September 13, 2019, 01:19:00 am
content: Quote from: ZeroResistance on September 09, 2019, 12:55:48 pmEveryone else is pretty much on the money about negative LDOs.But I should ask: does your circuit need an LDO? I mean, I understand why you might want one, but unless you're really trying to minimize heat dissipation in your regulators, the 317/337 will do ya fine.That said, I just did a power supply that needed ±15 V, +5 V and +3.3 V, and I used LT1963A for the positive rails and LT1175 for the negative rail. And yes, the 1175 is frightfully expensive.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 14, 2019, 05:48:05 am
content: Quote from: Bassman59 on September 13, 2019, 01:19:00 amI'm limited by my input voltage that is going to be 9V dc.Then the LM317L / LM337L would drop around 2.5V.So that brings its down to max 6.5V at their outputs. Lets keep it 6 for giving it a little safety margin.So now I would be powering the op-amps with +6/-6V.The opamp is a non rail to rail type so its output would swing to +/- 2V of the supply rails.So 6 -2 gives me 4V as the max output swing.And I have to handle some line level inputs, which I though would be 1.2Vrms but I realised that one of the inputs would be around 4Vrms max.Having said that I'm still game for the 317 / 337 but I say that the LP2951 was at a similar price range so wondered it would be better to go in for that, if I get a suitable negative voltage alternative to the 337.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 14, 2019, 06:14:17 am
content: Quote from: Zero999 on September 12, 2019, 02:40:21 pmBy biasing the op-amp to mid supply level Vcc/2, won't the input signals see the mid Vcc supply rather that pure 0V.And if the input signals see that would there be a chance of crosstalk between them?

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 14, 2019, 12:00:20 pm
content: Quote from: ZeroResistance on September 14, 2019, 06:14:17 amYes, the (-) input will be sitting at Vcc/2 with no signal present if the (+) input is biased to Vcc/2 - ensuring both inputs are at the same voltage is the main function of an op-amp, after all - but why would that change the amount of crosstalk between mixer inputs compared to biasing the (+) input at 0V? The answer is it that it doesn't.When the input resistors are the same value as the feedback resistor in an inverting mixer biased for single-supply operation the output of the op-amp will be sitting at Vcc/2 with no signal present, right? If you apply a 2Vpp sine wave to one of the mixer inputs then the output will go up to a peak value of (Vcc/2)+1V and a trough value of (Vcc/2)-1V; still a 2Vpp signal, except centered around Vcc/2 instead of 0V, and inverted, of course. Remember, the (+) input in a single-supply inverting amplifier is biased at a fixed voltage, so the output of the op-amp has to generate a mirror image of whatever signal is applied to the (-) input to cancel out that signal, leaving just the fixed bias voltage; this cancellation of the input signal by the op-amp output is why this configuration is said to act like a virtual earth, and it is also why the input impedance to the (-) input in this configuration is that of the series input resistor.Since the junction of all the input resistors and the single feedback resistor is essentially earthed, every input is effectively isolated from the rest because no signal can make it past the (virtual) earth, regardless of whether that junction is at 0V or Vcc/2 or whatever.This is kind of a tricky concept to understand, but once you get it it will make perfect sense.

author: Mp3
author_rank: Frequent Contributor
timestamp: September 14, 2019, 10:48:09 pm
content: Hi, i know this might not be exactly what you want (i dont know exactly what you want) but i recently used this with a headphone amplifier which needed a bipolar power supply. It was cheap and does the job. I attached a female DC barrel plug extender to the Vin / GND and use a plain center positive 12v dc supply. Negative and positive voltage could be a little better matched (getting something like -14.75 and +14.9) but for the price it does the job.https://www.amazon.com/gp/product/B0752CK8DK/ref=ppx_yo_dt_b_asin_title_o03_s00?ie=UTF8&psc=1I did not feel comfortable making a rail divider on this amp so i got that while i await delivery of a pcb to build a bipolar psu.

author: dmills
author_rank: Super Contributor
timestamp: September 14, 2019, 11:09:27 pm
content: Quote from: ZeroResistance on September 14, 2019, 05:48:05 amSo why are you bothering with a voltage regulator at all?9V (Or even +-9V, so 18V is well within the ratings of most sane audio opamps (+-15 to 18V is what most pro gear goes for), so adding a regulator will just cost you precious supply voltage and headroom, run off the raw rail (well decoupled!).Regards, Dan.

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 15, 2019, 12:01:49 am
content: I have to ask,9 volts, from a battery?If so, a linear voltage regulator is very wasteful.And one of the primary reasons to use a regulator, ripple reduction, is not required with batteries.

author: exe
author_rank: Unknown
timestamp: September 15, 2019, 09:12:34 am
content: My friends, did we already established that regulation is actually required? A good opamp has quite good PSRR, so regulation might not be even needed.Probably, regulation itself is not even needed. What needed is something to remove ripple. So, consider a capacitor multiplier as a regulator:https://en.wikipedia.org/wiki/Capacitance_multiplier. Or may be just a simple RC-filter (just don't make R too big).

author: MagicSmoker
author_rank: Super Contributor
timestamp: September 15, 2019, 09:40:56 am
content: Quote from: exe on September 15, 2019, 09:12:34 amYep, see my post on the previous page...Quote from: MagicSmoker on September 10, 2019, 02:54:22 pm...saying the exact same thing as you.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 18, 2019, 02:54:41 pm
content: Quote from: Mp3 on September 14, 2019, 10:48:09 pmNice find! I see that its a switched inductor topology. How did you filter out the noise that thing is putting out?Currently I've opted for a switched capacitor charge pump for generating the -ve supply rail.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 19, 2019, 06:25:21 am
content: Quote from: dmills on September 14, 2019, 11:09:27 pmThe reason for me going for regulators, what the adaptor which I was getting the supply from is a switching type, and my thinking was the regulators would attenuate the switching noise, and provide a cleaner source of power for the audio analog circuitry.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 19, 2019, 06:26:36 am
content: Quote from: schmitt trigger on September 15, 2019, 12:01:49 amNo its not a battery, its a 9V adapter.

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 19, 2019, 09:08:24 am
content: If you would like high frequency ripple reduction, LDOs are not the best option. At powerline ripple frequencies they may have easily 45 or 60 db of ripple rejection.But at higher ripple frequencies, it drops to essentially zero. Some LDO datasheets clearly show this.In my humble opinion, it would be far better to have a transistor capacitance multiplier for HF ripple reduction.