Usually the starting point is the expected minimum dc input voltage at full load, which might correspond to the end point of battery discharge.
If that were to be (say) 40v for a 48v lead acid battery, the turns ratio would be calculated for 40v/330v allowing just a little a bit extra for the expected voltage drops at full load plus dead time at maximum full duty cycle.
So your turns ratio might end up being close to 8.5:1.
At anything more than 40v the duty cycle reduces accordingly.
The low input voltage maximum full duty cycle condition is also used to calculate the required number of primary turns at the desired flux density and frequency.
*edit*
3C90 material is fairly old and not optimum for 100Khz, but anyhow the recommended flux density at that frequency for 3C90 is 100mT
At 40v input and 5uS on time:
6 primary turns works out to 90.5mT
5 primary turns 108.6mT
Either should work.
The conventional method is to run the duty cycle right up to maximum at the minimum design input voltage, whatever you decide the minimum input will be.
I assumed a typical lead acid battery in my example, but obviously that need not be the case.
You mean to say that my second approach is the correct one?
2) The second method I followed was based on the duty cycle value. I took Minimum voltage =42V and the max duty cycle for it to be=0.7. I took the absolute value of the duty cycle to be =0.81 to compensate for the transformer.
I used the equation (Vin*D=(Vout+Diode drop)*n).=(42*0.7=(330+0.4)*n)
By doing these calcultions, I got the turns ratio of n=Ns/Np=11 And according to its further calculations the turns become as Np=5 and Ns=56.
First issue is that you will never be able to run a high power inverter direct off solar panels without using a battery.
The problem is that solar panels are current sources, the voltage will vary hugely with the current drawn, and the maximum usable power from solar input is only available over a fairly narrow voltage range.
Another worry is that inverters are negative resistance loads. As the input voltage falls, duty cycle increases and so does input current.
The result is a highly unstable system that can never work.
A battery is a necessity to stabilise operating conditions, and maintain a somewhat constant load for the panels, and a reasonably stiff voltage source for the inverter.
Try to imagine how smoothly a single cylinder gasoline engine might run with zero flywheel mass, and you get the general idea.
Once you have "something" to stabilise the operating voltage (a battery is suggested) you design your inverter operating voltage around the battery type and number of cells.
The only duty cycle you are interested in is the absolute maximum at the minimum design input voltage.
At voltages higher than minimum, the duty cycle will obviously self adjust via feedback to be lower, but the actual figure is of no interest.
I agree, 3 turns is pushing the limit.
Four turns is much more sensible.
You are going to drop some voltage across your switching mosfets, and the resistance of all the internal wiring, as well as the transformer.
These might be around a volt (2.5% conduction loss) in a good design.
Or might be as high as two volts (5% conduction loss)
So something like 330/39 8.46:1 ratio
Or 330/38 8.68:1 ratio
You need to round up to full turns, and if you split the secondary into two halves it will be an even number.
So 4 Turns to 36 turns sounds ideal.
First wind 18 turns, then the foil 4 turn primary, then another 18 turns on top.
My 2 cents worth.
Np=E*10^8/(4*F*A*B)=3.26
This is correct as far as you went. But you have to cover the full voltage range from 39vdc to max charge rate 55vdc? To be safe use 4 primary turns.
Your low voltage is 39vdc your secondary voltage is 330vdc, a rule of thumb is to add 5% for duty cycle loss. 330vdc x 5% = 347vdc
347vdc/39vdc = 8.9 ratio
4 primary turns x 8.9 ratio = 35.6 turns = 36 turns Minimum.
You may need one more turn for misc losses, diodes, copper losses, ect.
you will get 330v at full maximum duty cycle at the minimum design input voltage.Ok. Got it.
But kindly tell me that at what value of duty cycle will I get 330V? How can I calculate that?
you will get 330v at full maximum duty cycle at the minimum design input voltage.
If that is 40v then it will work right down to 40v.
At voltages higher than 40v, the feedback reduces the duty cycle to hold the output constant at 330vdc.
At 48v the input has increased by 48/40 = 20%
The duty cycle will decrease by 20% from maximum.
At 52v the input has increased by 30%
The duty cycle will decrease by 30% from maximum.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?