TL494 synchronous buck converter LOW output voltage

[codemaster11]

that's picture of an inverter based on TL494 & HIGH LOW side mosfet driver IR2101. i modefied this picture for my
synchronous buck converter placed a ferrite inductor instead of transformer. use IR2110 as driver while output side is
as like shown in following picture.

feedback loop of the converter is as shown in above picture of TL494 and IR2101. frequency of the converter is set to 45KHz & duty cycle is greater than 90%.
these are my reference picture of my schematic diagram. i m facing one problem that is "the converter max output voltage is half of the input voltage
where i need more output voltage than half" Vin 16.5vdc but max Vout = 8.25 volt . the feedback loop of TL494 works upto 8.25 volts mean it decrease below this
level but not going to increase above 8.25 volts changing feedback resistors but nothing happens to increase output voltage. here what to do increase the output voltage.

 

[Easy peasy]

Indeed - as the Vin goes up - so do the sw losses ! on the mosfet

 

[Easy peasy]

14V out at 23ADC continuous is really pretty good for a newbie ! ! !

--- Updated Feb 14, 2024 ---

Please post a picture of the converter !

 

[codemaster11]

after replacing low side Mosfets with shotckey Barrier diode. i did not test the converter with 100Vdc input to test efficiency at 48Vdc output.
only test it with one 550Watt solar panel with DC IN voltage = 41Vdc , with DC IN current = 8.5Amp total I/P power = 348watt & output power Vdc- output = 13.1vdc & Iout-dc = 23.2Amp. after replacing lowside mosfets with diode the converter is heating a little with colling fan expected that it can be continuously
run at 50 to 60Amp dc with Vout = 12 to 14vdc.

 

[Easy peasy]

Very nice !

 

[codemaster11]

I see equation for buck converter efficiency in an article on google that is D = Vout/(Vin x efficiency) seeing this equation tell that higher efficiency require lower D (duty cycle ) but for high Vout higher duty cycle required dropping efficiency according to equation.

 

[Easy peasy]

Ah. no - what that equation is trying to say, very clumsily is:

Vout = D. Vin. Efficiency, so as eff goes up - so does Vout. If eff = zero, Vout = zero. If eff = 1.0, Vout = Vin . D

 

[codemaster11]

now it is correct that Effi = Pout/Pin While duty cycle D = Vout/Vin. with this converter i 'm trying to run a 400watt water pump that is 48Vdc & 8.3Amp with two 550watt 48v in series by adjusting the converter output voltage to 48Vdc but the converter efficiency drop at this output voltage Vin from solar panel drop to 46volt when solar have VOC = 96vdc and converter output drop to Vout = 31volt from 48vdc

 

[Easy peasy]

you are asking people to solve your issues with NO information on the real hardware you are using - water pumps draw high current when they are starting - you seem to have overlooked this in your design calculations.

 

[codemaster11]

according to my calculation P = 400 watts , V = 48 Vdc , I = 8.33 Amp. mpp of two solar panel is exist at 76% of the VOC. that is Vmpp = 72.9 mpp
for 48Vout at this Vmpp the D = 48/73 = 66%. the converter require stable 73 Vdc at the input in loaded condition. each panel is rated for 13 ampere at short circuit current. that's why i'm thinking that two 550 watts solar panel in series that has total of 1100 watts power is sufficient for driving this water pump. the ferrite toroid ring has an
outer D = 40mm , iner D = 23mm and H = 11mm.
Number of turns is = 11Turns (swg 22 x 6).
Fsw of the converter = 45KHz and
unloaded Duty-cycle = 62%.

here i don't know how much current the motor is driving during starting while the motor is rated for 8.5Amp in loaded condition.
i didn't have further knowledge to solve the problem.

 

[BradtheRad]

Your solar panels appear adequate to the task. I suspect you're losing power in wiring to the converter, or in the converter itself.

Simulation of expected performance (based on your parameters).

Notice by definition no power is produced when you measure solar panels either open circuit (max voltage) or short circuit (max Amperes). My solar panels were rated 12V and produced 23V with no load.

I purchased a 24 VDC well pump. Testing it in a bucket I found it accepted a wide range of supply voltage (drawing a couple Amperes). If it were at the bottom of 40 feet of water then current draw would have been greater.

For testing purposes you can install a simple resistive drop. It wastes power and dissipates 150 or 200 W of heat (example, small space heater). Take readings of voltage and Amperes at your panels and your pump. Notice it only takes series 2.7 ohms to reduce 70V to 48V at 8A. Similar parasitic resistance could account for your own observations.

 

[D.A.(Tony)Stewart]

DC motor starting current is based on DCR of the coils so it is basic Ohm's Law. V+/DCR=Istart. If you cannot measure this remember that DC Motors may be estimated to draw 10x the rated current 8.5 A or 85 A and AC motor surge tends to be 5x 8.5 rated current = 42.5, which will exceed full solar current of your PV's for series panels rated for 13 A.

Parallel panels would work better for driving the motor.

The motor current should be ramped up like a VFD to not cause the Vmpt for the solar input to go out an efficient range. This will start in full direct sun at Vmpt = 82% Voc and decrease towards 60% for most PV panels and the middle range might be 72%. Is your 76% due to the solar angle off-peak summer season instead of 82% best case?

So I recommend you use current sensing and Vmpt to regulate the motor current. I have made a lot of assumptions but this is based on my experience. Can you use PWM to regulate the motor current to start without dropping below Vmpt ? Current sensing will be useful for linear feedback, voltage feedback is a 2nd order effect with less phase margin for regulation. A derivative on current sense to complete the PID loop filter will add stability with an RC snubber or an AC motor film run cap across motor filter to reduce EMI noise may assist in the optimum design.

An alternative is to see if you can deliver the same DC output current but ramp up the voltage on motor startup from say 25% to final.

Do you have a battery to provide high starting current for the inverter. I'm afraid I do not see a full system block diagram with interface specs, so I may be making false assumptions.

 

[codemaster11]

high inrush current at startup which is 10 x rated current is dropping converter output voltage.
the converter is working in voltage mode. i think the motor require a soft start circuit at the motor input
to reduce inrush current or a pwm soft start. can it work if soft start implement in TL494 circuit?

 

[Easy peasy]

read the data sheet - it shows how to implement a soft start !

 

[D.A.(Tony)Stewart]

https://www.ti.com/lit/an/slva001e/slva001e.pdf?ts=1708816293030

 

[Easy peasy]

Here is the info from the TL494 data sheet