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This question doesn't relate to a specific meter, I am using AMPROBE AMP-220 manufactured by Beha-Amprobe. I believe am an experienced engineer but only in the digital world, not really in power and AC field.

I am apparently confused by understanding the current measurement, specifically the difference in the three measurement functions:

  • AC A
  • DC A
  • DC+AC A

The circuit under test is a 1200VA/60Hz transformer with two 37Vrms secondary coils sharing one wire (center tap), also called 37V/0/37V.

schematic

simulate this circuit – Schematic created using CircuitLab

My calculations of expected current is I = Urms/R; that is I = 37Vrms/1.6 Ohms = 23.125 Amps

I do see this current... but on 'unexpected' meter setting. I was planning to use the AC setting, but the current is on DC+AC setting.

Clamp meter is properly aligned with the wire. Can someone please in layman terms explain me why I measured these values:

  • AC setting shows 9.7A
  • DC setting shows 20.5A
  • DC+AC setting shows 22.95A (closest to my calculation)

Thank you for clarifying this. While my circuit currently has high power resistor to test the newly purchased clamp meter, the goal is to have lead-acid battery instead and to measure the Irms charging current. Which setting of the clamp meter is correct to measure this charging current?

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    \$\begingroup\$ Check: electronics.stackexchange.com/questions/442195/… \$\endgroup\$ – Huisman Jun 17 at 13:46
  • \$\begingroup\$ The answers below help in understanding what these different measurements mean and how they relate. Please note that the model of a lead-acid battery is very different from a resistor. \$\endgroup\$ – vangelo Jun 17 at 16:42
  • \$\begingroup\$ @vangelo, yes thank you. This post question is only seeking to understand the high end clamp meter functionality, battery charging is another story. \$\endgroup\$ – EmbeddedGuy Jun 17 at 16:47
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This answer was given before the OP revealed he was using output rectification

It's possible that there is some form of rectification in your secondary circuit and that the two components of your current are actually 20.5 volts DC and 9.7 volts AC. If you do an RMS mathematical combination of the DC and AC currents you get 22.68 amps RMS.

$$RMS = \sqrt{DC^2 + AC^2}$$

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  • \$\begingroup\$ Thank you @Andy, you are starting to bring light to my problem. The description of my test circuit is not fully correct, I thought the simplifying my circuit description to (what I understand) it is helps, but perhaps it only hihglights my ignorance. Let me rewrite my question above. \$\endgroup\$ – EmbeddedGuy Jun 17 at 14:17
  • \$\begingroup\$ So @Andy, thank you I accepted your answer, but please be so kind and now with redrawn schematics please explain why I even have DC component, when the common wire 'shouldn't know' I have rectification? I thought both diodes assure I have both half-waves going through? \$\endgroup\$ – EmbeddedGuy Jun 17 at 14:40
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    \$\begingroup\$ Both diodes feed only forward current to the load so, you have full-wave rectified current into the 1.6 ohm resistor. Your secondary coil arrangement and diodes are a classical method for obtaining a full-wave rectified voltage on a load. Hence you have an average current of 20.5 amps and this is \$\dfrac{2}{\pi}\$ smaller than the peak current. So reverse the formula to get 32.2 amps peak. Compare this with the peak current calculated from 37 volts across a 1.6 ohms with a diode drop of 0.7 volts = 32.08 amps! \$\endgroup\$ – Andy aka Jun 17 at 14:46
  • \$\begingroup\$ I see where you going with your comments to John. To stay in practical world, if the load will be replaced with a lead-acid battery, and the goal is to measure the Irms charging current... which meter setting should be used? AC or DC or DC+AC? I have a small 1mOhm resistor in series I measure the voltage on it, and I wanted to verify my calculation of RMS current on purchased clamp meter, and am seeking the correct setting to use. \$\endgroup\$ – EmbeddedGuy Jun 17 at 15:41
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    \$\begingroup\$ To measure RMS current with that meter use the AC + DC measurement range. \$\endgroup\$ – Andy aka Jun 17 at 15:58
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Andy's answer was already very clear regarding how the measurements relate, but I'd like to add something crucial without much detailing:

  • AC measurement removes the DC component before calculating the True RMS value

  • DC measurement removes the AC component, hence it is equal to the average, since it's DC

  • AC+DC measurement considers both components, giving the equivalent measurement to the True RMS meaning: what is the DC value that would equivalently heat the resistor

In your circuit the resistor is connected to a full-wave rectifier, consequently, the current never circulates in the reverse direction of the (ideal) diodes. Don't think of the DC component as a fixed current always flowing in the same direction. In fact, due to diode voltage drops, the current is zero for very short times, twice for each cycle.

The average current (DC) that charges the battery you mentioned would relate to its charge process.

The AC+DC True RMS measurement relates to how much heat is dissipated by your resistor.

The True RMS AC reading would not help in this example, except if your measurement device doesn't have a AC+DC mode. In this case, you can use the formula Andy mentioned.

Instead of a full wave rectifier, check the several readings obtained from a 1 V peak sine wave added to a 1 V DC offset:

enter image description here

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  • \$\begingroup\$ thank you for the answer. Do I read you correctly, if the load will be replaced with a lead-acid battery, and the goal is to measure the charging current... I should use the DC+AC setting on the clamp meter to get the true RMS of the current? \$\endgroup\$ – EmbeddedGuy Jun 17 at 15:57
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    \$\begingroup\$ I've edited my answer to try to explain this point. \$\endgroup\$ – vangelo Jun 17 at 15:59
  • \$\begingroup\$ Hmm.. I would think, that "how much heat is dissipated" is what is needed to know how much energy was sent to the battery? The battery manufacturers specify complex charging steps, some of which require fixed-current stage. They assume true DC current. Since we have here pulsating DC, to follow their specs, I would think we should use DC+AC true RMS current measurement, not DC? \$\endgroup\$ – EmbeddedGuy Jun 17 at 16:26
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    \$\begingroup\$ A recommend you don't extrapolate your measurements with a resistor to try to understand how the two phase charging circuit would work. I was just trying to give you physical interpretation to what you are measuring. Are you trying to know how fast your battery will charge? \$\endgroup\$ – vangelo Jun 17 at 16:32
  • \$\begingroup\$ No, I've built a relatively complex battery charger with CPU controlled thyristors, not diodes. (The resistor we use to test the transformer heating to compare various manufacturers). My charger software measures current going to battery, I wrote the function to do RMS calculation (maybe I shouldn't do RMS?). I wanted to compare that to purchased clamp meter and this post question was seeking to understand which meter function does that. Now you are asking what is another question: am I correct to measure RMS current if the battery specification mentions true DC charging current requirement. \$\endgroup\$ – EmbeddedGuy Jun 17 at 16:43
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I have never seen a transformer output marked as Xx Vrms. However since primary is supposedly marked 120v, let's assume that's correct.

Next - any meter showing rms reading is ASSUMING that the wave shape of the parameter is a perfect sinewave. In your case it is not. In fact we don't even know if the voltage is actually 37Vrms or something else.

Hence because of the above you will get wrong readings. Correcting the readings based on actual waveshape Vs assumed waveshape, As Well As actual voltage will require extensive calculations.

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    \$\begingroup\$ "Any meter showing rms reading is ASSUMING that the wave shape of the parameter is a perfect sinewave." This is not correct. See True RMS measurement on Fluke's website. \$\endgroup\$ – Transistor Jul 21 at 19:11
  • \$\begingroup\$ You are right! I was thinking of cheapo meters of long ago. \$\endgroup\$ – Kripacharya Jul 22 at 6:07

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