Too much ripple

Discussion in 'The Projects Forum' started by svdsinner, May 21, 2014.

  1. svdsinner

    Thread Starter Member

    May 17, 2011
    39
    2
    I just built a basic power board for a project that needs a +5V, GND, -5V outputs. Current requirements are tiny (100ma max, and <<10ma normal)

    Here is my schematic: (Pretty much straight from the 7805 datasheet)
    [​IMG]
    The problem is that both +5 and -5 lines ripple by roughly .25V. Oscilliscope reading:
    [​IMG]

    I tried adding some electrolytic caps (47uf) to smooth things, but it had almost no effect.

    What do I need to do to smooth these lines? I'm using them for an ADC that needs to take accurate readings between 0 and 3.3V, and this is destroying accuracy. I'd like to get this ripple down by an order of magnitude or 2.
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,148
    3,058
    Who built your ICs? Lots of knock-off parts out there.

    Any idea if oscilloscope measurement noise might be an issue?

    Just throwing out ideas.

    [edit] Wait a minute - where did you get that circuit? It's not a standard split supply at all. r1?
    Here's the one from the Fairchild data sheet.
    [​IMG]
     
  3. alfacliff

    Well-Known Member

    Dec 13, 2013
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    that diagram is ok for after a filtered supply, but 47 mfd is way too small to filter hum. the caps shown are to prevent oscilaition of the regulator chips, not filter the dc.
     
  4. #12

    Expert

    Nov 30, 2010
    16,330
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    Problems!

    The capacitors in the first post are not what the datasheet calls for, the 1k resistor limits the lower chip to about 5 ma, and you can label this as positive and negative all you want, but you can't pull electrons out of a 7805 chip. I think you should try a different circuit.
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,544
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    Yeah, you sorta can. Before the uA7805 there was the original 3-terminal regulator, the LM309. The first circuit is an incorrect variation of an early LM309 trick that showed up before true negative voltage regulators appeared.

    ***With a center-tapped transformer***, bridge, and two big caps typical in a bipolar supply, you can put a lower 7805 in series with the transformer/capacitor center tap and connect the upper 7805 ground pin to its output. The lower 7805 ground pin is the -5 "output". With lotsa decoupling, this works surprisingly well. This "regulating the ground" technique was fairly common back when.

    ak
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,033
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    But the op does not have a center tapped transformer. :confused:
     
  7. crutschow

    Expert

    Mar 14, 2008
    13,033
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    As noted the ops circuit can't work. Note the path for the current for a +5V load. It has to go out of the top regulator but into the output of the bottom regulator, and most regulators can only source current, not sink it. Also the 7805 has a 2V dropout, so even if the circuit worked you would need at least 15Vdc on the input.

    Below is the simulation of a circuit that uses two low-dropout positive voltage regulators that can both source and sink current.

    Note that if you draw 100mA for any length of time you may need a small heat sink for the devices.

    Edit: I modified the schematic to avoid overvoltage on the U2 EN pin.

    Edit 2: Just noticed that the polarity shown on C4 is backwards.

    Split Supply.gif
     
    Last edited: May 21, 2014
  8. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    That's why I called it an incorrect variation.

    ak
     
  9. #12

    Expert

    Nov 30, 2010
    16,330
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    Nice history lesson, but not relevant to the O.P. Let's try to keep a focus on the problem at hand.
     
  10. prashanth58

    New Member

    Jan 30, 2010
    8
    1
    Mr Sinnah1 ur original diagram is n,t going 2 solve ur prob. The Subsequent diagram drawn by Mr Wyneh is much ,much better excep 4 a few imp observations. U had asked 4 + as well as -5v Rt? Going by the same diagram all that u should do is limit the input voltage to 9 or 12V Max, an instead of .337 A 2.2 Uf Capacitor used@the input, use at least 330uF (Min)& about 1000uf or so @the opt of the reg Ic instead of .1uf&1uF used in his diagram. Do observe the proper polarities of the capacitors as well as the terminals of the Negative regulator.(U Could get confused rather easily here. REMEMBER,the center terminal on a 7905 isnt grounded )...Happy building...:)-)
     
  11. prashanth58

    New Member

    Jan 30, 2010
    8
    1
    Pz pardon a few errors in writing as I tend2 write2fast...( besides there is a character limit that limits the amount of info that we can post )(;-)
     
  12. svdsinner

    Thread Starter Member

    May 17, 2011
    39
    2
    I think you guys are missing the obvious facts that:
    1) The circuit is working (except for the noise that needs smoothing) It is just a basic combination of the +5v and -5V regulator applications from the datasheet, and R1 is just lowering the input voltage to the second 7805, since it It is putting out correct voltages (+/-0.01V) on a multimeter. I need an Oscilloscope to spot the noise in the rails.
    2) There isn't any functional difference between a center tapped transformer and a switched mode wall-wart powered by a 2-pronged (non-grounded) plug. They both simply output two relative voltages, and either can be grounded to make any particular value GND.
     
  13. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,810
    834
    I agree with this part of your statement. Between text speak and run on sentences in one massive paragraph, it's hard not to get confused.

    If you think and write fast, try writing your response offline in a text editor. Then edit it removing text speech, breaking it up into reasonable paragraphs and correcting punctuation. Then copy and paste the polished response here.
     
    DickCappels likes this.
  14. #12

    Expert

    Nov 30, 2010
    16,330
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    Let me spell this out for you. This is an international forum. You already know that. The part you are not considering is that a fair amount of our participants do not speak English in the first place. When they try to use a translation program to understand you, it does not work. Your contributions are useless to some of the people here (even if your answer is correct). You simply must find a way to participate with (mostly) proper English.
     
  15. svdsinner

    Thread Starter Member

    May 17, 2011
    39
    2
    I did some experimenting in Spice, and forgive me, the schemat

    it appears a few minor changes -should- fix things.
    1) I typo'd in the OP schematic. I was using 100nf caps, not 100pf caps.
    2) Bump C4 to >= 2.2uf
    3) Add a 10R resistor between the positive terminal of the wall wart and the rest of the circuit.

    The biggest change was #3. I know enough about resonance and noise to have guessed it might work, but I'd love to hear from someone smarter than me as to why that should work.

    Note: My 7805 spice model is not necessarily 100% accurate. I'll report back after I make the actual changes and put it back on the oscilloscope
     
  16. crutschow

    Expert

    Mar 14, 2008
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    It's certainly not obvious to me. :rolleyes:

    1) I don't see how. Show me the current path for the current through the positive load if there's no negative load equal to or greater than the positive load.

    2) A center tapped transformer can generate a distinct and separate positive and negative output voltage relative to the center tap common. A two output terminal supply cannot. There is most definitely a functional difference between the two. And whether the wall-wart plug is grounded or not makes no difference to what the output is doing.
     
  17. svdsinner

    Thread Starter Member

    May 17, 2011
    39
    2
    The load isn't shown in the schematic. My existing project is the load.

    The schematic I posted uses a virtual ground circuit made with the two 7805s in the place of a center tap to create the exact same effect. And, yes, it is critical that the +12v power isn't grounded, otherwise the negative terminal would be GND, and a virtual ground could not change that since my project also has a USB connection which wouldn't be very happy if the usb power was +10V and +5v instead of +5V and GND.
     
  18. crutschow

    Expert

    Mar 14, 2008
    13,033
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    Precisely the point. You need to add the load to see the whether the circuit operates (or not).
     
  19. svdsinner

    Thread Starter Member

    May 17, 2011
    39
    2
    So I made the changes I mentioned, and I still have too much ripple. However, I took some additional measurements with the oscilloscope, and the +5V and the -5V lines are staying precisely 10V apart. It is only my virtual GND that is unsteady.

    What can I do to encourage the GND to stay centered?

    And, FWIW, crutschow, the load is primarily between the +5V and the -5V rails, and ranges from 6.32K to 10K. (And yes, I know to hook it up before testing. :rolleyes:) However, I need the GND set where it is to enable some ADCs to make readings inside their 0-3.3v range. And, if the GND is unsteady, the ADC readings are also unsteady, which defeats the point of 12-bit ADCs.
     
  20. crutschow

    Expert

    Mar 14, 2008
    13,033
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    So I don't think you will achieve what you want with your circuit. I believe a 10V regulator with a virtual ground at 1/2 the voltage is likely the best solution, as I previously suggested.

    Below is the simulation of a virtual ground circuit which uses an op amp along with a push-pull transistor current booster circuit for good stability. V2 is 10V that could be generated by a 7810 regulator connected to your 12V wall wart.

    The circuit on the right in the dotted box is just to generate a current pulse in the virtual ground circuit to check the virtual ground stability and is not used in the actual circuit. As shown, a 10mA pulse cause a peak change in the virtual ground voltage of about 3mV and it recovers to the steady value in about 30ms.

    Virtual Gnd.gif
     
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