power
I have a Uni-T UT139C multimeter. I want to measure the power usage of my Blitzwolf USB charger. Can I just connect my multimeter in series with the charger at the wall and measure how much power the charger uses ?
And how can I calculate watts in AC circuits? it's same as DC (VxI)? Or it is different than that. I searched about it in the internet and it says we need to know the power factor. Is it true? if so, what is power factor..?
In order to overcome these sort of sampling measurement errors you can make the problem "easier" by adding a large electrolytic capacitor on the power feed to the development board after the series measurement resistor. This is an "attempt" to measuring average current taken by the dev board. OK, I say you are measuring "average" but it depends how big your capacitor is in uF terms and how long your "violent spikes" are.
If you have a 1 ohm series resistor and a 100uF capacitor, the time constant of this is 0.1ms so you'll need more capacitance or more resistance. To get reasonably decent measurements go for a time constant of about 10ms - this should suit your multimeter.
As an aside, your multimeter may be perfectly capable of averaging over a 1ms period and giving you the results you think you may not be getting.
USB-C will use the Power Delivery specification, a first connexion is done at 5V, then "negotiate" whether it can use a higher profile to charge. There are 5 profiles available :
There are 4 connection point for the power (2 on each side - see pinout below), as far I know, they are all equal and may be connected by a single cable (I think it will be at the cable manufacturer discretion). These additional connection allow to go for higher current without having massive voltage drop at the connection. Coupled with higher voltage, that gives a lot higher charging power.
All in all, I guess that the laptops will as well charge with 5V (on USB A charger), just far slower. And based on what I saw from Apple for their new Macbook, the charger is 29W, so most likely a Profile 3 (a bit under spec), it seems then to be only 12V.
It seems that additional profiles have been added by some manufacturers, for instance, Qualcomm Quick Charge 2.0 seems to be an implementation of the Power Delivery but using 9V too. This technology though does not use the Power Delivery specification as it uses the D+/D- lines of the USB 2.0 port to negotiate the voltage.
Qualcomm Quick Charge 3.0 brings it one step further and now allow to "negotiate" any voltage from 3.7V to 20V by increment of 200mV. No data found so far about the current at each voltage.
Best Answer
Strictly speaking yes it absolutely is. The instantaneous product of voltage and current is power and will have an average value that is true power. Here are a few examples of different types of loads: -
When the magneta and blue curves multiply you get the red curve (power). The mean of the power waveform is the watts you are billed on.
Proper watt meters instantanously multiply voltage and current waveforms to get true average power.
If the true watts are 1000 W and the product of RMS voltage and RMS current is 1000 VA then power factor is unity. If the VA is only 500 then power factor is 0.5.
See my top right diagram - it shows the average power at 50 % of what it is when voltage and current are in phase (top left). If you take the arc cos of 0.5 you get 60 degrees; in other words power factor can tell you how displaced the voltage and current waveforms are.
From this follows the much abused formula that...
Power = V.I.cos (phi)
This implies that if you know the RMS values for voltage and current and, using (say) an oscilloscope, you estimate the phase angle displacement, you can "calculate" watts. Where this gets abused is that more often than not one or both voltage and current waveform shapes are NOT pure sine waves and this can lead to significant errors in power calculation.
No such error is introduced when voltage and current instantaneous waveforms are multiplied together.