should I ground the rectifier terminal ?

Discussion in 'General Electronics Chat' started by jerrytom, Aug 9, 2010.

  1. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    [​IMG]
    Hi guys,
    I want to get 5v dc from main supply(220v,50Hz), I'm using the circuit attached here, problem is bridge rectifier has four terminals(two for input and output) in which I'm suppose to use one output terminal and give the same to regulator, what about the other terminal of the rectifier?
    can I ground that, if so will I get 5v dc ?

    If the circuit is wrong then please post a valid circuit !
     
  2. tom66

    Senior Member

    May 9, 2009
    2,613
    213
    You wouldn't want to ground it to the case. The ground in the circuit here means anything connected to ground in other schematics/circuits gets routed to this point.
     
  3. kingdano

    Member

    Apr 14, 2010
    377
    19
    you could have a split ground set-up - one ground for power circuits which would be more tolerable of transients and ground loop current - and a second for other circuits, which you would want to have clean grounds.

    this is really only feasible on a PCB, but you could do it with grounding studs in a breadboard if you wanted.

    generally i tie everything to one ground/earth point, but i do not lay out my own rectifying circuits.
     
  4. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Unless a circuit has some sort of single ground reference point for the various signals involved all sorts of strange things can pop up. Your circuit looks fine.

    I remember having to re-engineer a system that had separate main power and audio grounds. There was a mute transistor on all the preamp cards that relied upon a control voltage which was referenced to the main current ground (drove all the cassette deck motors) and when put into record the decks had a habit of going into oscillation. Turned out the voltage drop in the cabling system was dropping about 0.6V when all the motors were running causing the audio ground to float about that much higher. Dirty solution, but I solved the problem by installing single rectifiers in series with all the motor supply grounds at the connectors to the decks. This lowered the voltage difference by enough such that the mute transistor did what it was supposed to instead of sitting there in a half-on state.
     
  5. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    then what should i do, leave the terminal floating, will that work?
     
  6. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    @kingdano, yes I'm doing it on general purpose pcb, you mean giving separate ground to rectifier and the other to regulator ?
    Is their any other solution as I want to do it on single board !
     
  7. kingdano

    Member

    Apr 14, 2010
    377
    19
    you can do it on a single board with multiple layers

    i mean that your power circuits (rectifier and regulator both fall under power circuits in this case) should use the 'analog' ground or 'power' ground and the digital circuits (logic etc) should use a separate 'digital' ground.

    have a poured ground plane for each digital and analog ground source and tie the two together at one common mounting hole/chassis point.

    if you only have a single or double sided PCB as an option, and dont have an experience PCB designer to work with, i do not recommend this approach.

    you should then use one common ground plane for all signals. you may run into problems with ground loop currents but likely you'll be ok.
     
  8. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Let's face it - no ground for the bridge and it's worthless. If you're wanting to build a single sided board just use jumpers or -0- ohm resistors across traces when and where needed.
     
  9. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    Your circuit is basically FINE, depending on how much AC ripple you can tolerate at the regulator's input.

    Some of the other messages are referring to grounding-practice details that are beyond the scope of your original question. You might have to deal with those problems later.

    Here is almost everything you need to use for linear power supply design:

    http://www.zen22142.zen.co.uk/Design/dcpsu.htm

    If you know the maximum current that you expect to be drawn by the load, then you can calculate the extrema of the top and bottom levels of the ripple-voltage waveform (after the rectifier, at the regulator's input), depending on the value of your first capacitor, which you can (if needed) change, in order to change the ripple voltage's max/min levels.

    Using "worst cases" of AC mains voltage as +/-10% (or whatever you expect), and of transformer regulation (if you know what that might be), you primarily only need to make sure that the minimum expected voltage level of the bottom of the ripple voltage waveform (i.e. the minimum expected regulator input voltage) will _never_ cause the regulator input voltage minus the regulator output voltage to be less than the regulator's "dropout voltage" specification. (Otherwise, the regulator's output voltage will not be smooth. It can get quite ugly.)

    But in the circuit that you have shown, with 12V AC RMS transformer output (about 17 Volts AC 0-to-peak) and 5 Volts DC regulator output, the dropout voltage constraint will probably NOT be the main problem (IF your 100 uF value is large-enough). However, you "might" need to worry about the power that the regulator will be forced to dissipate as HEAT, depending on the load current. (Usually, the greater the difference between a regulator's DC input and output voltages, the more you have to worry about dissipation of heat by the regulator. Caution: You "might" need to use a good heatsink on the regulator!)

    Here is a somewhat more-comprehensive (but still simple and practical) explanation of linear power supply design:

    http://sound.westhost.com/power-supplies.htm#rectifiers

    Cheers,

    Tom
     
    Last edited: Aug 10, 2010
  10. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    @kingdano, split ground analysis works !!!, I got 5.05 v that's cool


    1. but what's the difference after creating two separate grounds and then tieing together at common point ? what's going on their !
     
  11. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    @gootee,thank you for making me to decide proper capacitor value and the links provide good idea of different rectifications .
    cheers,
    gootee

    When I measured the output voltage I got these things =>

    1.dc =5.05v
    2.Ac=11v
    As regulator function is to provide a constant Dc voltage, why does it output Ac signals ?
    And seems that it's providing almost,Ac=2*Dc voltage.
     
  12. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    OK. I'll tell you. But remember that YOU asked <grin>!

    Actually, this is a _very_ important subject for anyone who designs physical circuits. I am not an expert. But I have learned a little bit about it, which I will try to share here:

    NO conductor is a perfect conductor. ALL conductors have distributed resistance and inductance, in some proportion to their length. So currents in ground-return conductors will always induce voltages back at the non-ground ends of the conductors!

    When could that be "a bad thing", you ask. OK, first here's one example:

    Imagine the two input pins of a typical single-ended voltage amplifier, which could be a circuit based on an opamp, discrete transistors, tubes, etc.

    The amplifier's input only "sees" the _difference_ between the voltage at the signal input pin and the voltage at the ground reference pin. The ground reference pin is connected to a ground conductor, which eventually connects back to the power supply ground.

    If the amplifier's input-reference-ground's ground conductor SHARES a length of conductor with ground-return currents from OTHER parts of the circuit, then the voltages that those OTHER currents induce, back at the non-ground end of the conductor, will also appear at the amplifier's ground reference pin. (This effect is sometimes called a "bouncing ground".)

    And since the amplifier input can only "see" the _difference_ between the voltages of its signal input pin and its ground reference pin, the actual result is that the induced "bouncing ground" voltage is ARITHMETICALLY SUMMED with the amplifier's input signal!

    That's "a bad thing", since the currents from the "other parts of the circuit", which induce the "bouncing ground" voltages, might be 120 Hz AC (hummm!), or just about anything!

    A LOT of types of circuits depend on having stable ground reference voltages at various points, in order to operate correctly and well, or maybe even to operate at all.

    With fast-changing dynamic ground-return currents, such as from the bypass capacitors of digital ICs for example, the ground-return conductor's unavoidable inductance can make things worse than you might (first) imagine. The resistance of a conductor merely induces voltages according to Ohm's Law, i.e. proportional to the current. But the voltage induced across an inductance is proportional to the rate-of-change of the current. So even a relatively low-amplitude current that has fast changes can induce a relatively large voltage. (For example, the fast edges of current pulses would induce SPIKE-like voltages!)

    The SOLUTION to the many problems that the sharing of a ground-return conductor can cause is simply to not allow "incompatible" ground-return currents to share any length of conductor.

    There are various ways to implement that, depending on what types of circuits are in a system, and what type of construction is used (e.g. printed circuit boards versus point-to-point wiring). You can find much more information about the finer points of good grounding implementation by searching for terms like "star grounding", "split grounds", or even "ground planes".

    -----

    Gee, WHILE we're talking about ways to avoid getting noise and other signals where you don't want them, I might as well also mention what I sometimes like to call "Faraday Loops".

    If you know about Faraday's Law (or Maxwell's Equations) [and now you WILL, at least a little bit], you know that a time-varying magnetic (or electromagnetic) field will induce a current in any closed loop of conductor that is in the field. (And note that the induced current will then induce voltages across any impedances that are in the loop.)

    Now imagine the input for, say, an (typical single-ended) audio amplifier circuit (opamps, transistors, or tubes, etc). Imagine that there are two wires that connect between an input jack and the input pins of the actual amplifier circuit. One wire is the input signal and the other wire is the input signal's ground reference.

    If those two wires are not very close to each other, all the way from the input jack to the amplifier input pins, then they will have some geometric "loop area" that they form between them. (Of course it won't actually be a closed loop unless something is plugged into the input jack.)

    If there are any AC or other time-varying electromagnetic fields in the air, where the two wires are, and there WILL be, then those fields will induce currents in the wires. And those currents will induce voltages across any impedance in the loop that they form.

    Unfortunately, in the case of an amplifier input, the voltage will be induced across the amplifier input pins (and across the source's impedance, and across the wires' own impedances). Can you say "HUMMMM"?

    The solution is to keep the wires (or PCB traces, etc) as close together as possible, over their entire length. A typical practical implementation is to at least twist the wires tightly together, and preferably to also have them be inside a shielded cable with the shield grounded only at the input jack end.

    And implied in the above is that the input jack should be isolated (insulated) from the chassis, so that the input ground doesn't connect anywhere except through the wire that is kept close to its corresponding signal wire.

    But there's MORE! A time-varying or AC _current_ in any closed loop of conductor will induce a time-varying magnetic field in the air.

    So loops of conductors can either be "receive" antennas OR "transmit" antennas!

    Going back to the audio amplifier example, that means that you should also twist tightly together the conductor pairs that have AC currents or other large time-varying currents in them, to stop them from making fields in the air that would induce currents in your input conductor pair, and other conductor pairs.

    So, in practice, you should at least tightly twist together ALL "natural pairs" of conductors, such as AC mains, transformer secondaries, rectifier/ground pair that goes to caps, DC/ground supply line pairs, signal/ground pairs, output pairs, etc. (For PCBs, keep them close together, or, use multiple layers so they can be one right over the other, which is sometimes also done using multiple signal and power ground planes, multiple power planes, various signal layers, etc. It's much more possible to do it well with more layers, if using PCBs.)

    One other thing: Always try to keep small-signal and noise-sensitive conductors as far away as possible from all large-signal and higher-power and highly-dynamic-signal conductors and devices. If two conductors with "incompatible" signals must get near each other, try to make sure that they are perpendicular to each other.

    Cheers,

    Tom
     
    Last edited: Aug 12, 2010
  13. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    Do you have a load connected to the regulator's output?

    What is the capacitance value of your smoothing capacitor (before the regulator)?

    Increase the smoothing capacitance value until the AC voltage measured across the capacitor is small-enough, and there is (almost) no AC measurable between the regulator's output and ground.

    Try 1000 uF, or 2200 uF, or even more, if needed.

    Basically, you can use as much capacitance as you want. More is better, as far as lowering the AC ripple amplitude.

    You should be using a low-ESR (equivalent series resistance) electrolytic capacitor that has a high-enough peak voltage rating and a high-enough ripple current rating.

    If you only have certain values on hand, you can put them in parallel, there, to sum their values.

    Cheers,

    Tom
     
  14. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Verify:

    1) AC from the transformer secondary winding going to the AC input pins of the bridge rectifier

    2) Negative pin of the bridge rectifier output going to the negative of the filter cap.

    3) Positive pin of the bridge rectifier output going to the positive of the filter cap.

    Unhook any loads and measure across the filter cap, you should have mainly DC with only a small amount of AC ripple.

    Next, put a load on it that will draw about half as much power as the transformer secondary is rated for.

    A light bulb is always good for this. If your circuit is built for 12V use a 12V auto lamp that's rated for about half the wattage the secondary is capable of, if it's made for 5V then use a 6V bulb or a resistor that will draw about half the current.

    For instance, if the transformer is rated for 6.3 VAC at 1A on the secondary use around a 10 ohm 5W or 10W resistor (Radio Shack has them) then measure across the filter cap again. You should see slightly less DC voltage and about the same or a bit more AC ripple on the output.

    I still don't quite understand why people are making this so hard, something sounded fishy from the start when the question arose about grounding the output of the bridge rectifier.
     
  15. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    Hi gootee,
    This was really very informative. Humm, I need to have an eye on this during my design and even I got to know about vcc sag !
     
  16. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    I had just 100 uf, yes when I increased capcitance value, AC voltage got reduced !!!
    Have a nice day :)
    gootee.
     
  17. jerrytom

    Thread Starter New Member

    Jan 20, 2010
    9
    0
    Hi marshallf,
    I checked as you said, but non worked as I have connected them properly.The real bug was capacitor value .
    Thank you,
    marshallf
     
  18. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    That brings up another related topic: Local "bypass" capacitors.

    If you have dynamic active devices to which your DC power supply is sending current on demand, you should remain aware of the fact that the inductance of the DC supply conductor/trace/wire might inhibit the timely supply of fast-changing currents.

    However else we think about local "bypass" capaitors, we should also think of them as "local power supplies". They can better and more-easily supply the fastest-changing current demands for the devices they serve, just because they are closer to the point of need.

    This also prevents disturbances in the supply voltage, since the demanded currents are not forced to flow through the inductance of the relatively long conductor from the power supply, where they would induce voltages proportional to their rate-of-change.

    This is one reason why a small-value capacitor should be connected abdolutely as closely as possible to the power supply pins of an active device. And typically you would also want a larger capacitor in parallel with it.

    Probably, enough "bypass" capacitance should be used to supply ALL of the DYNAMIC demands of the device, so that the main supply only has to steadily keep the capacitors charged. And the values of the smaller bypass capacitor(s) should be selected so that they can operate fast-enough for the device.

    The only common "gotcha" with simple bypass capacitor situations (that I can think of right now at least) is that they might form high-frequency resonances with the supply-conductor inductance. So, without sophisticated analysis or well-instrumented experimentation, it is usually better to use bypass capacitors that have some loss built in, so they won't "ring" (cause damped oscillations) as easily. That usually means not using low-ESR (Equivalent Series Resistance) electrolytics for the larger ones, and not using NPO or C0G ceramics, or film capacitors, for the smaller ones, where usually a lower-quality ceramic is safer. Sometimes a small resistance (10 to 50 Ohms, typically) should even be placed in series with the DC supply conductor.

    That's not to say that there aren't many cases where higher-quality bypass capacitors might give better performance. You just have to be more careful in using them there.

    Cheers,

    Tom
     
Loading...