measuring phase angle with dual trace scope

Discussion in 'General Electronics Chat' started by PeteHL, Apr 18, 2015.

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  1. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Last year in Dec. I built a LED lamp that is 12 of the star-type LEDs in series powered by house current (mains to you Europeans). Now I want to measure efficiency of the lamp, that is, Lumens per watt of the lamp.

    Unfortunately I don't have a watt-meter and don't want to buy one as I would probably only use it rarely. But I do have a dual trace CRO (old style cathode ray oscilloscope).

    In order to obtain the actual power dissipated by the lamp, I measure the apparent power at input to the lamp, V times A, and then multiply that by the cosine of the phase angle.

    See the attached figure of my lamp. To get the phase angle, I connected the CH1 probe across the secondary winding of isolation transformer T1. The Ch2 probes I connected across the leads of resistor R2. The voltage drop across R2 is in phase with input current. Thus the extent to which the CH2 voltage leads the CH1 voltage tells me (in milliseconds) the extent to which current is leading voltage.

    The CH2 voltage waveform crosses the 0 V horizontal axis 3 ms. before the CH 1 voltage. Actually the CH2 wave goes to 0V and is flat at 0V for about 2 ms. before going above/ below 0V. So I'm taking the 0V crossing time of the CH2 voltage to be when it starts to rise or fall after staying flat at 0V.

    The frequency of the house current is 60 Hz which is a period of 16.7 mS. Therefore the phase angle of my lamp I take as being 3 ms divided by 16.7 ms times 360 degrees equal to 64.7 degrees.

    Using a scope, would you say that this is a correct way to determine the phase angle?

    Thanks,
    Pete 12LEDS~1.png
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Why not use a 120 to 35 Volts transformer?
    That way you can leave out the huge capacitor (C1) and the two large resistors (R2 and R3).
    The current will be now 1.25 / 2.4 = 0.52 Amps.

    Bertus
     
  3. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,221
    If you want the power out of the transformer then, because of the non-linear current waveform generated by the rectifier capacitor supply, you can't get an accurate power measurement by looking at the waveform and simply measuring the phase-shift between voltage and current.
    The correct (but tedious) way is to instantaneously take a sample of the voltage value and a sample of the current value at the same instant and do this for many samples over one half cycle.
    You then multiply each voltage sample by its corresponding current sample to get the instantaneous power and take the average of those samples to get the total (average) power.
    An Excel worksheet can help with the calculations.
     
  4. ian field

    Distinguished Member

    Oct 27, 2012
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    Scopes are rarely well enough calibrated to give much more than a general visual impression.

    Somewhere in my travels I've seen a relatively simple logic gate phase comparator - AFAICR: it can be as simple as an AND gate to detect coincidence of the 2 inputs as a percentage. use that to gate a constant current generator and drive a moving coil meter.

    There are more complex logic arrangements that give a more direct indication of relative phase position - well worth a google search along those lines.
     
  5. ScottWang

    Moderator

    Aug 23, 2012
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    Because we are not allowed to discussing transforemerless PS, and your circuit still in danger as the C1 is too big, normally, it should be less than 3.3uf.

    I think if you could replacing by a AC120V/12V/0.2A transformer is better, you could using current source or I prefer to using a voltage and in series with a resistor to limited the current less than 16mA, assuming that the LED spec is 3V/20mA/5mm.

    For a AC120V/12V transformer.
    DC Vout = 12Vac * 1.414 = 16.968 Vdc ≅ 17 Vdc.
    Vdrop = 17V - 12V = 5v, 5V is more than 3V for the Vdrop of LM317, so it can be keep in a stabilize situation.

    Assuming that LED=3V/20mA/5mm, it just using 80% of 20mA, I_led = 20mA*80%=16mA.
    You need 3 LEDs in series +R and 4 strings of LEDs in parallel.

    Calculating current limiting resistor:
    R = (12Vcc - (3V_led *3))/I_led
    R = 3V/16mA
    R = 187Ω, choosing 180Ω.
    I = 3V/180Ω = 16.7mA
    I_total = 16.7mA * 4 = 16.7 = 66.8mA, we can count it as 70mA.

    Calculating the AC current for transformer.
    The rate of Iac transfer to Idc is (1/0.6)
    Iac = 70 mAdc *(1/0.6)
    Iac = 70mAdc *1.67
    Iac = 116.9mA ≅ 117mA

    So, if you choosing a transformer as 120Vac to 12Vac/150mA or 12Vac/0.2A or some more then it will be enough for your led application.

    Or maybe you can buy a adopter as AC120V/12Vdc 0.5A to use.
     
  6. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Sorry, but I can't respond to this without revealing what I am really doing, which is forbidden to discuss! :)
     
  7. PeteHL

    Thread Starter Member

    Dec 17, 2014
    59
    0
    My guess is that I can't do this with a DMM, so I would be out of luck. What equipment would one need to do the type of sampling that you propose?

    best regards,
    Pete
     
  8. PeteHL

    Thread Starter Member

    Dec 17, 2014
    59
    0
    Thanks, I will look to see what I can dig up along those lines.

    -Pete
     
  9. PeteHL

    Thread Starter Member

    Dec 17, 2014
    59
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    Thanks for all of this, but you apparently missed on the schematic that the LEDs are of the higher current-rating "star"type, in my case, rated 700 mA. This is why the capacitor value is five times greater than what you propose.

    Thanks for the response,
    Pete
     
  10. ScottWang

    Moderator

    Aug 23, 2012
    4,850
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    Oh, sorry.
    700 mA is a little high then the voltage should be rising up to 32Vac when Vdc = 32Vac *1.414 = 45.2V, and Iac = 700 mAdc *1.67 = 1.2Aac, adding some more then it will be about 1.5A, so 32V/1.5A transformer should be enough, and it just using one string of led.

    Using 24Vdc 3A adopter is another option for 6 leds in series with 2 strings, and quite cheap.
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,638
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    Hello,

    I assume you want to use the leds directly on the mains when you say this in a post:

    I am closing this thread as it violates AAC policy and/or safety issues.

    Quote:
    6. Restricted topics.

    The following topics are regularly raised however are considered “off-topic” at all times and will results in Your thread being closed without question:

    • Any kind of over-unity devices and systems
    • Automotive modifications
    • Devices designed to electrocute or shock another person
    • LEDs to mains
    • Phone jammers
    • Rail guns and high-energy projectile devices
    • Transformer-less power supplies
    This comes from our Tos:
    Terms of Service
    There will be enough sites where automotive questions can be discussed :
    Member selected automotive forums

    I am closing this thread as it violates AAC policy and/or safety issues.

    Quote:
    6. Restricted topics.

    The following topics are regularly raised however are considered “off-topic” at all times and will results in Your thread being closed without question:

    • Any kind of over-unity devices and systems
    • Automotive modifications
    • Devices designed to electrocute or shock another person
    • LEDs to mains
    • Phone jammers
    • Rail guns and high-energy projectile devices
    • Transformer-less power supplies
    This comes from our Tos:
    Terms of Service
    There will be enough sites where automotive questions can be discussed :
    Member selected automotive forums

    Bertus
     
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