Smoothing of rectified voltage with capacitor

  1. Dear forum

    I have s issue concerning rectification of a sinusoidal voltage with a diode. It is a experiment i am doing.
    The case is a 1N4001 diode in series with a lamp and a capacitor in parallel to the lamp. The circuit is supplied by a sinusoidal voltage with a rms value of 2.2 V.
    The set up can be seen on the attached file called "Half_wave_rectifier". (The switch is open so the diode does not get bypassed).
    On the attached file called "HalfWaveV_L" is 3 graphs showing the voltages of the lamp. The yellow is with zero capacitance and the orange and red is with the capacitances drscriped at the figure.

    My first question is why the wave has higher peak values with capacitors than without a capacitor? To that comes why the wavetops for the yellow graph is apruptly cut of, compared to the round tops of the orange and the red graph.

    I have also measured the rms current in the circuit for the various capacitances, and it gets bigger the bigger the capacitance is. I cant understand why that really is. The peak values of the voltage increases with the capacitance, so that kind of explains it, but then again, why does the peak values grow?

    The rms currents meassured was 0.26 A for the yellow graph, 0.4 A for the orange graph and 0.41A for the red graph.

    I understand that a capacitor in parallel works as a energystorage which can supply the system with voltage when the voltage of the power supply gets low.

    I hope you can help, that this is the right place for my post.

    Attached Files:

  2. jcsd
  3. mfb

    Staff: Mentor

    Interesting result. How did you measure voltage and current?

    The yellow graphs follows the power supply - probably similar to a sine wave (minus some offset for the diode).

    No, as the additional current is needed to charge the capacitor. It discharges itself via the lamp when the power supply is at a negative voltage.

    One guess: The capacitor increases current and therefore the average load of the lamp - it gets hotter, and its resistance might increase. If the power supply is not perfect and its voltage depends on the current, this might lead to a higher peak voltage. I would not expect this, but something has to explain your observations and I don't see anything better at the moment.
  4. Firstly I assume you voltage axis is measuring the voltage across the lamp (and capacitor).

    OK so the yellow curve is a poor quality half wave rectified waveform.

    When the capacitor is in circuit it charges to the peak value of the voltage (or nearly so depending upon the available current). this will be the true peak not properly shown by your yellow curve.

    The part of the orange and brown curves that you see in the part of the cycle when the diode is non conducting show the classic exponential discharge curve for the capacitor through a resistor (the lamp).
    The capacitor is releasing its stored charge through the lamp and the curve shows that it has the same polarity as the rectified supply - which is what you would expect.
    Note also that the smaller capacitor shows a deeper discharge - the curve coming coloser to zero.

    Nice experiment.
  5. Nugatory

    Staff: Mentor

    Why is does the yellow curve not reach to the true peak? asking, not arguing here.
  6. I don't have a good explanation.
    The inductance of a lamp is not likely to be sufficient.
  7. The 2.2 Voltage (rms value) which I imentioned was measured with a multimeter over the terminals of the power supply. The current was measured also with a multimeter but ofc. in connected in series.

    And thank you for your guess, but the differences in current are just very big.

    Yer, the graphs on the figure is the voltage across the lamp and capacitor. And it is measured with a osciloscope.

    This seems like a good explanation :). Now is only left to explain why the quality if the rectification is so bad.
    I do not udnerstand what you mean with that: "the inductance of the lamp is not liekly to be sufficient"? Sufficient for what?

    Attched is two graphs. The orange graph is of the voltage from the power supplied which was measured while the circuit was conencted. (It was measured with the osciloscope, and over the terminals of the power supply). And the red graph is the same as yellow before. (the voltage over the lamp, where no capacitor is connected).
    Maybe this can help with something.

    This graphs reminds me of another problem: the red graph is slightly shiftet compared to orange one. How can this be?

    Attached Files:

  8. The difference in height of the orange and red peaks on your new graphs corresponds to the forward voltage drop across the diode.

    Yes the capacitor introduces a phase shift.

    Yes the current will increase with capacitance, a capacitor represents an AC load in parallel with the lamp.

    Now was this a class experiment or what?
    You have made lots of (correct) observations but offered few explanations. Has your classroom worksheet drawn attention to these points or is this your own work?
  9. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    I think the height of the yellow peak (from its start value) is about the same as for the other curves. The finite 'on' resistance of the diode will be limiting the amount of charge that can flow during the conducting period so the peak output volts can only increase by so much in that time. The measured voltage 'sits on top of' the value which is stored in the reservoir capacitor. Would the 'diode voltage drop' not be the same when the diode starts to conduct - in all three curves?

    The rms current is higher with a reservoir capacitor because the average (height squared) of the voltage curves is greater (it's on for longer). If you had a 1F capacitor, the volts would be high all the time - the only factor to reduce the available volts would be the source resistance of the upstream circuit.
  10. There is a significant difference between the the lowest value achieved (=start value?) of the brown curve compared to that of the red curve, in fact the brown minimum is closer to the yellow minimum that the red minimum. Yet its maximum is pretty well the same as the red curve.
  11. Philip Wood

    Philip Wood 1,180
    Gold Member

    If you're keen to understand the behaviour of your particular circuit, this comment should be ignored.

    If, though, you're more interested in the effect of a reservoir capacitor on half-wave rectification, I'd swap the (severely non-ohmic) filament lamp for a resistor.
  12. mfb

    Staff: Mentor

    That would be a good test.

    In addition, you could disconnect the lamp and check the voltage with and without capacitors.

    That should happen with capacitor, not without.

    How does it get smaller with capacitor then?
  13. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    Yes, you're right about the brown curve. The brown curve drops faster than the red - as you'd expect and the brown curve starts rising very near to the yellow. The yellow curve has a flat top - perhaps indicating current limiting. But the diode handles 1A (spec sheet) so that shouldn't be a problem. Is the AC source limiting at a low value of current - but even that should give a slope at the top of the yellow curve. Does all this happen with different diodes and power supplies?
  14. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    A very good idea. At lower power, the dynamic resistance could be a lot lower. I think you may have got it.
  15. This is a experiment I am doing with a group for a project. Our supervisore told us that maybe there can be a problem with the measureent of the current as a rms value. This is a mode on the multimeter designet for ordinary alternating current, which is not the cast here.
    And that maybe the graphs are shifted because the readings was not taken at simultaniusly on the oscilloscope. (meaning that some shift can have occured in the pricessing of measuring the graphs.)

    I will think about all the comments and write back later. I dont have much time right now.
  16. There is an alternative way to measure current, that is to place a very small resistance value in series with the power supply, say 0.01 ohm or maybe even 0.1 ohm.

    Then you can hook up an O'scope across the resistor and measure the voltage, your current will have to be figured out with a calculator, I=E/R so if you have 1 volt across a 1 ohm resistor you have 1 amp of current and so forth.

    If you measure 100 mv across a 0.1 ohm resistor, that is also 1 amp of current assuming DC voltage.

    and 1 mv across that 0.1 ohm resistor, 1/100th amp, 10 milliamps. That would be showing peak to peak on the oscope so you would need to convert that first to RMS which I believe is (2^1/2)/4 which would be 0.707/2 which is 0.3535 times the peak so in that case if you saw 100 mv p/p you would have an RMS of 35.35 millivolts RMS. That is the DC equivalent of the ac voltage, so 100 mv p/p would be the same as if it were DC at 35.35 mv.

    So using those figures, .035 (35mv) /0.1, the current in RMS terms would be 0.3535 amp, 353.5 milliamps. You then have a current reading as accurate as the Oscope and the accuracy of the resistance. Getting a 0.1 ohm resistor would mean you would have to pretty much take their word for it that it is a 1/10th ohm since you won't see that on most DVM's.

    you would need a milliohm meter, not something floating around much.
  17. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    Any ordinary meter will only give you rms values by assuming a sinusoidal waveform. These waveforms are definitely not. One good way to measure the 'proper' rms current would be to use the data output from your oscilloscope. What form is its output? It seems to have ported to an output / printing device so the numbers are there. If you look at the voltage across a 0.1 Ohm resistor, the data you get will tell you the current (scaled by 1/10) at each instant in time so you can square and sum those values and divide by the number of values. I appreciate that may be hard to organise in software but you could always get instantaneous values of current off a printed version of the graph and do a pencil and paper calculation. Draw and measure vertical bars every few mm over one whole cycle.
  18. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    If you compare the two graphs you have produced, the 'raw waveforms' are not the same but the input sinusoid in the second graph has a strange 'flat top' on it and an even flatter bottom. What sort of source are you using? Is it a pukkah transformer or a signal generator of some sort? If it is then you may be running into source problems which will confuse things.
  19. Okay guys

    We are turning in the project today, so we cant make changes to the set up. We are asuming that the graphs was shifted compared to each other when it was meassured with the oscilloscope. The graphs was not emassured simultainously so the oscilloscope might have taken different zero points.

    We have dropped the observations of the currents as the meassured currents was not sinusoidal, and the ammeter we used was set to meassure rms value of a sinusoidal current.

    I just need to get one thing clear:
    When the capacitor is charged by the rectyfied voltage (V_rec), it has some voltage betweeen its plates (V_c).
    When the diode is reverse biased it is clear that only V_c contributes to the voltage over the lamp (V_L), but it is unclear to me what the situation is in the second half of the forward bias where v_rec is falling. From the attached graphes it is evident that the exponential behaviour if the red and orange graphs begins before V_rec reaches zero.
    Is it true that that V_c+V_rec cannot exceed the maximum value of V_rec at any time?
  20. sophiecentaur

    sophiecentaur 14,689
    Science Advisor
    Gold Member

    As soon as the voltage of the source is below the volts on the capacitor, the capacitor will be discharging exponentially because there will be no charging current from the source (diode is reverse biased)
    It is instructive to look at the time variation of the current delivered to the capacitor. Current only flows in short bursts of high value at the top of the AC cycle. The rest of the time it will be zero.
  21. This is a fundamental and crucial statement that you would be well advised to take note of.

    I do not see that you have measured this effect from the data offered and you should have done. Perhaps a recommendation for the report?

    I think we have done enough towards writing your report for you.

    Go well in your studies.
    Last edited: Dec 21, 2012
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?