Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

[gyulasun]

The MOSFET data sheet says the ON resistance is max 6 milliOhm (at 10 Vgs and at 75A drain current), so in case you have two such FETs in series, the 12 mOhm at the cca 8.5 A peak current would cause only 102 mV drop across the FETs and we still have at least a cca 2.3V drop.

As the current increases from zero to about 7-8 Amper, the voltage drops from 10 V to cca 8 V in about 2 millisecond, this what the scopeshot says.
One test would be to further reduce the supply voltage to say 4-5 V DC and see how the curve on the top changes if any. 

Gyula

[gotoluc]

The MOSFET data sheet says the ON resistance is max 6 milliOhm (at 10 Vgs and at 75A drain current), so in case you have two such FETs in series, the 12 mOhm at the cca 8.5 A peak current would cause only 102 mV drop across the FETs and we still have at least a cca 2.3V drop.

As the current increases from zero to about 7-8 Amper, the voltage drops from 10 V to cca 8 V in about 4 millisecond, this what the scopeshot says.
One test would be to further reduce the supply voltage to say 4-5 V DC and see how the curve on the top changes if any. 

Gyula

Okay Gyula,

the below scope shots are 2 different circuit scenarios using the minimum voltage the motor needs to rotate. Keep in mind the air gap between E's and I cores is large and not idealized since the rotor flexes.  This was built as a proof of concept.

I've eliminated potential resistance in the circuit and obviously the Amp meter is not connect at all in the below scope shots.
First scope shot is, removing the cheap Watt meter I have in top of my dual capacitor bank meters and next scope shot is removing my dual cap bank meters and connecting directly in the large 100,000uf source capacitor.

The only other resistance would be the switch, alligator clips and the 0.4 Ohm of the parallel MOT primary coils have.

Does it look any better?  or should I need to use heavier wire leads to replace all the alligator clips?

Thanks for your input

Luc

[picowatt]

I think it is the CVR not being able to handle the current,and so it is showing a way higher voltage across it than should be there. Looking at the video,i can see the yellow channel is set to 500mV per division. The yellow trace is the voltage across the bucking coils. Luc says the probe is set at 10x for the yellow channel(channel 1) If the probe is set to 1x on channel 1,then he has only about 1.2 volts across his coils. Now,if channel 2 is set to 10x on the probe,and 5 VPD on his scope,then he actually has only 700mV across the .1 ohm resistor. This being the case,that is still 7 amps at around 12 volts,which is still 84 watts. That CVR is definitely not rated at 84 watts. I would say 1,maybe 2 watts at most.

Tinman,

Just FYI, the power dissipated by a resistor is equal to the current flowing thru the resistor times the voltage across that same resistor.

For example, assuming DC current, a CVR with 7amps flowing thru it and 700mV measured across it is dissipating 4.9 watts (not 84).

PW

[gotoluc]

Here is the best scope shot after removing all the alligator clip and replacing it with no. 12 AWG.
I also added two 0.1 Ohms in parallel as CSR, so it's now a 0.05 Ohms CSR.

It makes quite the difference to reduce accumulated resistive losses. I've also included the original scope shot (last one) of before I started to reduce circuit losses so you can compare the difference of before and after.
Now the most resistive losses in the circuit now the two primary coils which are 0.4 Ohms connected in parallel bucking fields.
The parallel Inductance is 1.05mH with no I core, 1.18mH with I core at switch on position, 2.38mH with I core at switch off position and 4.85 with I core fully in between E cores.


Luc

[tinman]

Tinman,

Just FYI, the power dissipated by a resistor is equal to the current flowing thru the resistor times the voltage across that same resistor.

For example, assuming DC current, a CVR with 7amps flowing thru it and 700mV measured across it is dissipating 4.9 watts (not 84).

PW

yes-this i know PW.
But i was talking about the power flowing through the resistor,not the power being dissipated by the resistor. The power being dissipated by the resistor as heat is directly determined by the power being delivered to the load through that resistor. If the resistor is rated at 1 watt,then trying to dissipate 4.9 watts of power from a resistor rated at 1 watt is going to lead to wrong calculations.

Remember,my first calculations were made from Luc's first video where he said that channel 2(the one across the .1 ohm CVR) was set at 5 VPD and the probe at 1x.(see post 148) This gave us 7 volts across a .1 ohm resistor--that 1 or 2 watt resistor is now dissipating 490 watts of power during the pulse.
When Luc corrected  10x on the scope channel for the CVR,we then had an average voltage of about 700mV across the .1 ohm CVR during the pulse,and so now that 1 or 2 watt CVR is dissipating your 4.9 watts every pulse.

If the pulse is long enough,and the resistor is rated well below the power it is trying to dissipate,then incorrect(higher) voltages can be seen across that resistor,but the current flowing through it will not be as high as ohms law says it should be.

Brad