For example, a customer charges his/her Electric Vehicle (EV) overnight (from home) and is charged by the energy company for the relevant number of kWh consumed.

The formula for kWh can be broken down as follows:

Power (kW) = Volts (V) x Amps (A)

Energy (kWh) = Power (kW) x Time

So how does this translate to EV battery charging and PicoScope?

The capture to follow is the first experiment looking at EV charging over a 6-hour period with PicoScope

The instrument panel indicated approx. ¼ charge remaining with a range of 11 miles prior to charging and 55 miles ¾ charge thereafter

The scope used was the PicoScope 4823 and the connections are as follows:

Channel A 12 V battery voltage

Channel B 12 V battery current

Channel C HV battery voltage (using x200 differential probe)

Channel D HV battery current

Channel E Type 2 Mode 2 (Main’s charger) AC voltage (using x100 differential probe)

Channel F Type 2 Mode 2 (Main’s charger) AC current

Channel G Proximity Pilot (PP)

Channel H Control Pilot (CP)

The test above captures the connection of the Mains Type 2 (Mode2) charger into a domestic 3 pin power outlet (Max 10 A) at 2 m 53s and disconnection at 6 h 11 m (Total charge time approx. 6 h 8 m)

I have included the desired measurements (between the time rulers) for each channel and will summarize below:

Channel A 12 V average battery voltage 13.55 V during charging

Channel B 12 V average battery current 1.974 A during charging

Channel C HV maximum battery voltage 342.5 V during charging

Channel D HV average battery current 4.762 A during charging

Channel E Type 2 Mode 2 (Main’s charger) AC RMS voltage 239.3 V during charging

Channel F Type 2 Mode 2 (Main’s charger) AC RMS current 7.882 A during charging

Channel G Proximity Pilot (PP) average voltage 1.522 V during charging

Channel H Control Pilot (CP) Positive duty 28.57% during charging (measurement error)

The support question was; exactly how much energy is consumed by the vehicle against how much is charged by the energy company?

**Let’s take a look at current**

As we can see from the results above, the mains charging current (RMS 7.882 A) is split between the 12 V and HV batteries. If we add these together 1.974 + 4.762 = 6.736 A total DC charge current

Looking at these values as a percentage of DC current against AC RMS current we have:

6.736 A (DC) / 7.882 A (RMS AC) = 0.85 x 100 = 85 %

Approx. 85% of the

**RMS current**applied by the mains charger is converted to DC and then split between the 12 V and HV batteries (Which equates to a 15% loss of total mains current)

For further information regarding the above, PicoScope 7 Automotive looks at HV charging current split using 3 x BNC+ current clamps within the EV Guided Tests along with a supporting video.

Click on Guided Tests > Electric Vehicles > Charger vehicle tests > Charging current split

**Looking now at power loss**

Main’s power utilized = 239.3 RMS V x 7.882 A RMS = 1,886.16 W (1.886 kW)

Power utilized by 12 V battery = 13.55 V x 1.974 A = 26.75 W (divide by 1000 for kW) = 0.027 kW

Power utilized by HV battery = 327.7 V (Average) x 4.762 A = 1,560.51 W (1.561 kW)

Total power utilized by these batteries = 1.561 kW + 0.027 kW = 1.588 kW

Power loss = 1.886 – 1.588 = 0.298 kW

Efficiency = Total power utilized by batteries / Total power delivered from mains

1.588 kW / 1.886 kW = 0.842 x 100 = 84.2% efficient

Power loss = 100 – 84.2 = 15.8%

Of course, 100 % of the RMS current, power and energy applied by the mains cannot be used for DC charging as numerous components will “consume” during the AC to DC conversion process along with coolant pumps, fans, AC compressors, relays, networks and ECU’s (Not to mention natural losses such as heat, cables and connections)

The thermal image below reveals the varying heat losses about the charging system

If we are to now look at the energy consumed from the mains supply (energy company) using RMS values the customer charge should equate to:

Power (kW) = Volts (V) x Amps (A)

239.3 RMS V x 7.882 RMS A = 1.886 kW

Energy (kWh) = Power (kW) x Time

1.886 kW x 6-hour charge time = 11.316 kWh consumed from the mains supply

Estimated cost at 14 p per kWh is £1.58 which is not bad for the estimated range of 55 miles

Let’s take a look at Energy loss (which should be the same as power loss)

Let’s take a look at Energy loss (which should be the same as power loss)

Main’s power consumed = 239.3 RMS V x 7.882 A RMS x 6-hours = 11.316 kWh

Power consumed by 12 V battery = 13.55 V x 1.974 A x 6-hours = 0.162 kWh

Power consumed by HV battery = 327.7 V (Average) x 4.762 A x 6-hours = 9.366 kWh

Total power consumed by these batteries = 9.366 kWh + 0.162 kWh = 9.528 kWh

Energy loss = 11.316 – 9.528 = 1.788 kWh

Efficiency = Total energy consumed by batteries / Total power consumed from mains

9.528 kWh / 11.316 kWh = 0.842 x 100 = 84.2% efficient

Energy loss = 100 – 84.2 = 15.8%

Here is where I need to repeat the above tests whilst also monitoring workshop mains consumption as I don’t have this data at present. Theoretically the energy charge should be £1.58 and I do wonder if this will be the case!

I will repeat this test over a longer time period using both PicoScope & PicoLog looking at the pros and cons of both. I hope to capture a charge cycle beyond 12 hours to a point where charging is halted by the vehicle rather than disconnection by the owner

I hope the above gives some insight into mains charging of EV’s and I will report back with more data ASAP

Take care…..Steve