Guru
Joined: 02/02/2017
Location: AustraliaPosts: 1432
| Posted: 06:00am 05 May 2017 |
Part 7 Primary current waveforms with different inductors plus efficiency calculations
It is time to look at this inverter prototype under significant loads.
My test gear can provide a DC supply up to 10 Amps only.
So this means I have to use the 2 x 100A 12 Fiamm batteries to supply the required grunt.
Under load they run at about 25V so I removed turns from the primary (was 25 T at 0.7Vrms/turn) to now 19T.
Test conditions now are:
AC load 480Wrms and it’s a quartz halogen light so the power factor = 1.0
AC output 230Vrms
DC supply around 25.5V
I test 3 inductors.
2 turns in the ferrite E core (50uH)
7 turns in the laminated steel toroid (16uH)
13 turns in laminated steel toroid (50uh)
Yellow trace is DC bus current, 6.1 Amp/div.
Blue is AC output
Pink is primary winding current, 24.4 Amp/div.
The oscilloscope calculates the RMS voltage of this which I use later.
first is the E core at 50uH at idle
then under 480W load. Notice the large primary current swings from PWM.
This is still eliminated at the secondary winding though.
next is the 16uH steel toroid at idle. MORE PWM current swings than the 50uH E core
then under the same load. LESS PWM current swings than the 50uH E core.
hmm.
last is the 50uH steel toroid at idle. same low PWM current swings as the 50uH E core
and under load, we get the lowest PWM current swings of all.
Now I want to show the efficiency of this setup.
I measured the primary winding resistance at 15.4 mOhm, and secondary at 45 mOhm
The primary is wound with 2 x 6mm2 copper. This is nowhere large enough for a real inverter. The inductor is wound with a single 6mm2 copper wire.
I measured the inductor plus current sensor at 5.4mOhm
(easy to measure these, just put a fixed current into the winding and get the millivolts, R = V/I. done)
We have the primary winding current as calculated by the DSO.
Test conditions:
AC Voltage 219Vrms (measured by a Power Mate Lite wattmeter. very good specs as a matter of fact)
AC power 434Wrms
Primary current 37.1Arms
DC Bus current 19.4A
DC supply 25.3V
Power in = 25.3 x 19.4 = 491 W
loss = 57 W
efficiency = 434/491 = 88.4 %
This is not so good. Let’s see where some losses may reside.
Secondary winding sees 434/219 = 1.98 Amps and I2R loss = 1.98 x 1.98 x 45mOhms = 0.18W
Primary winding I2R loss = 37.1 x 37.1 x 15.4e-3 = 21.2 W
Inductor copper loss = 7.4W
So the primary winding circuit accounts for over 1/2 of the loss
The mosfets are IRF3808. I have 3 per H bridge leg so there are 2 x (3 in parallel) also in the primary circuit. Rds(on) is 7mOhm. I figure series resistance is 4.6 mOhm
So maybe the loss when the fets are ON is 6.3 W
So far there is 6.3 + 7.4 + 21.2 + 0.18 = 35W loss in the primary circuit.
If I use 4 x the copper area (50mm2) then the losses will drop by 3/4 of the 28W primary winding+inductor
Then I should get a loss of 36W total at 434W output which is 92.3% efficiency.
Now this is looking like a real inverter.
When running these tests the H bridge remains at ambient temperature with no fans. My calibrated finger can’t feel any extra warmth from the 20 second test runs at load.
wronger than a phone book full of wrong phone numbers