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

[minnie]

   What you need is probably the square root of the wattage
divided by the resistance. HAND!

[tinman]

As opposed to what? 

Consider a .01R CVR in series with a 10R load being fed 100VDC.

Load dissipates very close to 1000 watts.  CVR dissipates very close to 1 watt.

Surely you don't think the CVR must be a 1000 watt resistor.

The 84 watts you mentioned in the post I commented on had nothing to do with CVR requirements.

PW

I love how you guys re arrange things so as they suit your need to look right.
The fact is,the example you have above has nothing at all to do with the calculations i made with the posted data by Luc--and my power calculations of the power being dissipated by the .1 ohm resistor are correct in both cases,and where in this case,the power being dissipated in Luc's setup was dependent on the power rating of the used resistor. 

It would be great if you guru's could use the examples that relate to the test being carried out,in stead of going off in a different direction to show us all how you !!can!! be right in other situation.
We all know this,but how about sticking with the subject and DUT being investigated.

Brad

[gyulasun]

Ok I'm doing something wrong here but I'm not sure what. I must have butchered woopys circuit some how http://youtu.be/GCE84y7gIrs

Hi Jimboot,

One question:  why is your multimeter in the AC voltage mode?? 

It should be in DC voltage mode, no?

Gyula

[picowatt]

I love how you guys re arrange things so as they suit your need to look right.
The fact is,the example you have above has nothing at all to do with the calculations i made with the posted data by Luc--and my power calculations of the power being dissipated by the .1 ohm resistor are correct in both cases,and where in this case,the power being dissipated in Luc's setup was dependent on the power rating of the used resistor. 

It would be great if you guru's could use the examples that relate to the test being carried out,in stead of going off in a different direction to show us all how you !!can!! be right in other situation.
We all know this,but how about sticking with the subject and DUT being investigated.

Brad

Timman,

I was commenting on this statement you made:

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 "rearranged" nothing.  The 84 watts you mentioned has nothing to do with CVR requirements as you appear to suggest.

PW

[tinman]

Timman,

I was commenting on this statement you made:

I "rearranged" nothing.  The 84 watts you mentioned has nothing to do with CVR requirements as you appear to suggest.

PW

It has quite a lot to do with it as far as Lus DUT go's.
Do you know the resistance of the two drive coils?
Dose he have them hooked in series or parallel ?.

Did you read my reply to you a few post back?

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.

Did i get that part right?.

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.

Did i get that part right?.

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.

And do you believe the above to be correct?

If it's a yes to all three,why are we having this conversation?,as i have also stated and corrected my wording in saying that i was referring to the 84 watts as being the power consumption of the device during the on time. Knowing all this,we can also get a pretty accurate measurement of what the resistance of the two drive coils are.

I think you guys spend more time going through post looking for mistakes others make(which are more often corrected later on than not),than you do looking for ways to help improve on what we are trying to achieve.



Brad

Brad