Calculating efficiency - voltage drops and AC current

[ACharnley]

Confused. How about a picture?

AC is 60Hz, AC voltage is read when no load is connected. DC voltage is read with load connected.

I think you're including the AC efficiency which I don't care about in this case. I only want to know how much wattage is going in and how much is going out, but when to measure the voltage - under load or with no load?

There isn't that much difference, 0.3V on the input side for the less efficient of the two.

 

[JimB]

Some interesting replies, but for my own question (before the hijacking!) can I verify that when comparing the efficiency of two devices connected to the same power supply (a transformer), I should be using the power supply voltage with no load on it, and the device voltage with the load connected.

OK, to try and give a direct answer to your question, you should measure
the input voltage and current to the device under test, while it is loaded
and
the output voltage and current from the device under test, while it is loaded.

If you measure the incoming supply volts with the device not connected, you will not get the true efficiency of the device under test. The results will be skewed by the load regulation of the supply.

JimB

 

[ACharnley]

Yet this seems odd, the more efficient device uses less amps so induces less of a voltage drop, which here means the input wattage goes up!

 

[alec_t]

But does the output wattage also go up at the same time?

 

[ACharnley]

Well no, as you can see it's less!

What I mean is, if the output wattage goes down, the voltage drop on the input goes down (so the voltage rises) which I'm using in the efficiency calculation.

In other words, output wattage goes down, input voltage goes up and input current goes down (but current will go down more drastically than the change in input voltage, which perhaps answers my thought).

 

[ChrisP58]

Efficiency is based on how much input power does it take to generate a particular output power. So, for an efficiency number to be valid, the input power, and the output power, need to be measured at the same time, and under the same conditions.

Otherwise, you have two, unrelated measurements.

Input voltage should be measured right at the input terminals of the device. And the output voltage should be measured right at the output terminals.

 

[Tony Stewart]

Confused. How about a picture?

AC is 60Hz, AC voltage is read when no load is connected. DC voltage is read with load connected.

I think you're including the AC efficiency which I don't care about in this case. I only want to know how much wattage is going in and how much is going out, but when to measure the voltage - under load or with no load?

There isn't that much difference, 0.3V on the input side for the less efficient of the two.

Your I(in) ac =o.35A has more error (high) due to measurement method errors. DMM rectifies the AC measures peak with a certain decay time and falsely assumes sine peak to RMS conversion, when it is a low duty cycle pulse current.

Either ask how, or measure it correctly. your error is at least 10% and AC current error is dependant on duty cycle and f.

Do you wish expert advice? Then state your goals (desired results) , specs and include DUT schematics.

or ask , how can I get say 12W easily at 5.0 V for $10 in parts or whatever, make or buy ..

 

[ACharnley]

So errata depends upon how accurate the multimeter is at calculating the AC current, and we're saying the devices (being switch modes) are not a resistive load so the current is voltatile?

 

[Tony Stewart]

Cheap DMM's are accurate to 1 or 2 % or better with quality instruments only when signal is sinusoid. Otherwise you must scope and use correction factors, or measure with peak and average to estimate crest factor (ratio of these two) to estimate RMS which is closest to true integral of digital samples of V*I or thermal method from hijacked discussions.

Again, your problem is not efficiency, as they are both equally inefficient for any LDO, but rather,,your problem may be marginal is input ripple Vdc min.- Vout, margin to dropout in LDO which contributes to input voltage regulation error and outpur load current regulation error in the two unknown designs. Old LDO's were 2-2.5V dropout. Modern Mosfet type LDO's are in xxx mV dropout depending on Rdson and Iout.

Without your details, on the design , which I requested,further discussion is unnecessary.

 

[ACharnley]

Unsure where LDO came from, these are both very efficient buck converters, one being 96%. Input signal is mains, so sin.

Assuming DMM is short for multimeter (about to check), then 1/2% is acceptable.

 

[Tony Stewart]

your spreadsheet values indicated poor efficiency of LDO , not 96%, sorry missing details , bad assumption.

Your digital multi meter (DMM) cannot measure pulse currents accurately , as explained previously.

If you used proper SMPS with PFC which converts AC input current to sine wave or use true RMS method.

 

[ACharnley]

The 60 & 70% are with diode rectification loss.

The SMPS I have is DC out, I was unable to find an AC one hence the use of a transformer.

I'm using a VC9808 multimeter, unsure how reputable the AC current sampling is but I'll try and find out.

 

[ACharnley]

Edited

 

[Tony Stewart]

I read schematic for your meter. I measures Vpp after gain using 0.01Ω shunt then scales sin to RMS so narrow pulse current reads higher than accurate depending on duty cycle.
Why not use old 250W PC PSU with jumper for Pon? (free surplus)

 

[ACharnley]

Unsure if the meter is good or bad then, but if I understand you right it's ok if the duty cycle is high, otherwise the pulsing has more of an effect (it will use the pulse as it's reading)?

Top effort anyhow!

I'm lost on the 250W PSU idea - I need 6V AC to mimic a bicycle dynamo. It also doesn't resolve the accuracy?

 

[Tony Stewart]

that's ok, I'm lost on your specs

What is your real requirement for output load?

 

[ACharnley]

So 6V AC in, device A and B converts to 5V DC. I'm using 5, 10 and 50 ohm loads to mimic USB 100mA, 500mA, 1000mA outputs.

The requirement being to calculate device efficiency to a reasonably degree of accuracy (ideally 0.5%).

 

[Tony Stewart]

There are standard methods;
Use a DSO with true RMS calculations using calc function Ch1 * Ch2 for V*I with RMS result.
Use a true RMS wattmeter that measues current and voltage, (not an RF 50 Ω true RMS Wattmeter)
Use a DSO to measure AC V and AC current with a 50 to 100mV shunt at peak A and compute data dump to PC in spreadsheet. Or observe waveforms and compute conversion factor based on Vpp and Vaverage or Rms based on power factor / phase shift / duty cycle.
Use an analog Wattmeter.
Design an analog watt meter using bridge and 2 quadrant multiplier with V*I
Use thermal method and load gen.with computed equivalent linear load that generates same temperature rise in load resistor as a DC variable source. like the obsolete IC thermal true RMS design, but using Discrete parts.

Since there are distrorted waveforms, linear approximations require a scope at very least to determine easiest approach, while observing V(t), I(t) with fixed RPM's , slow, med fast and dynamic Q transistor loads to determine efficiency profile.

I would use scope sweep ramp to control current on load (voltage controlled constant current) and measure average voltage with RC filter not peak to peak like DVM .
- input current using a 100mV AC shunt, precision OpAmp rectifier with gain then RC filter and DMM on Vdc.

Your AC to DC bridge diode and cap will reduce efficiency of generator by at least 30% due to crest factor , saturation. Try reducing input cap. SMPS choice buck-boost is preferred over buck.

cheap DMMs do not support this.

 

[Tony Stewart]

Is your object to power an actual USB device or LEDs? or battery pack? These are all nonlinear loads.