Not able to get high current from buck boost (sepic) converter made using TL494

Thread Starter

logesh0304

Joined Dec 23, 2023
6
I am trying to build a buckboost converter for converting 5-40v input to 5-35V output at 3A max to use it for many purposes by usign TL494. I am using a SEPIC topology as i want a non inverting output. I simulated the circuit in proteus. The circuit works fine when i connected a 500 Ohm load, I was able to get output of 5 to 35 volts at 100kHz frequency.

But when I used a 5 Ohm load, the voltage drops from 35v to 2v, because it can't able to give enough current, The maximum i get is somewhat around 400mA

I think the problem is in buck-boost stage, because I tested the buck-boost stage seperatly by giving a PWM signal, it has the same result.

I used inductors with 82uH and capacitors with 22uF, Shottkey diode and IRF540 MOSFET, I also change the shotkey diode with a high current one, but no results.

Here the simulation results,

with 500 Ohm load
no_load.png

with 5 Ohm load
high_load.png

waveform of L1 current, C1 voltage and PWM at 5 Ohm load
high_load_graph.png

But, when i changed the capacitor values to 1000uF, inductor values to 220uH and frequency to 5kHz i get 2A for 10v, but still not 3A

1. So, what i need to do to get desired output, what are the parameters needed to tweak?
2. Do i have to operate the buck-boost stage at resonant frequency to get max output?
3. What limits it form providing high current?
 

Thread Starter

logesh0304

Joined Dec 23, 2023
6
What does the waveform look like at the point C1(2)?
What is the voltage rating of the schottky diode?
The blue line on the graph is the voltage across c1. I used a generic schottky diode, but I replaced one 3 parallel 1N4007 each capable of 3A. But results were same.
 

Ian0

Joined Aug 7, 2020
9,826
The blue line on the graph is the voltage across c1. I used a generic schottky diode, but I replaced one 3 parallel 1N4007 each capable of 3A. But results were same.
I mean the voltage between C1(2) and ground.
1N4007's won't work at 100kHz. You will get low output, because of the reverse recovery time.
 

Thread Starter

logesh0304

Joined Dec 23, 2023
6
I mean the voltage between C1(2) and ground.
1N4007's won't work at 100kHz. You will get low output, because of the reverse recovery time.
This time i used 1000uF cap and 220uH inductor with 10kHz frequency, I was able to get 2A for 10v max for 5 Ohm load output set to 35v. I also replaced generic shottky diode with 45v 15A one.

Here's the voltage of c1(2) and ground. (The purple line)
c2.png

I want to know what limits the output current, afterall it is just a simulation.
 

Ian0

Joined Aug 7, 2020
9,826
One important point about the diode on a SEPIC: The minimum Peak Inverse Voltage of the diode is |Vin|+|Vout|, so for 35V in 40V out you need a diode rated at 75V minimum.

It looks to me as though the PWM waveform is correct for the output that you are getting, but not for the output that you want.
If the output is lower than expected, then the feedback voltage would be substantially lower than the reference voltage. If the feedback voltage and reference voltage are about equal then it is doing what it should (although that may not necessarily be what you want)

The inductor saturation current will be what limits the output current - the point at which the inductor cannot store any additional energy. Very few simulators simulate inductor saturation!
 

Papabravo

Joined Feb 24, 2006
21,225
The other thing to be aware of is the immutable rule of DC-DC conversion schemes. It is: "The output power will always be less than the input power. In some cases, it will be a great deal less". You may have stumbled on such a case, but we have no chance of evaluating that possibility without some additional component information.
 

Thread Starter

logesh0304

Joined Dec 23, 2023
6
One important point about the diode on a SEPIC: The minimum Peak Inverse Voltage of the diode is |Vin|+|Vout|, so for 35V in 40V out you need a diode rated at 75V minimum.

It looks to me as though the PWM waveform is correct for the output that you are getting, but not for the output that you want.
If the output is lower than expected, then the feedback voltage would be substantially lower than the reference voltage. If the feedback voltage and reference voltage are about equal then it is doing what it should (although that may not necessarily be what you want)

The inductor saturation current will be what limits the output current - the point at which the inductor cannot store any additional energy. Very few simulators simulate inductor saturation!
Yeah, the feedback voltage is less than the reference, you can see than on the voltage probe along the -ve input of error amplifier.
But that's due to output can't able to give enough current for given load for desired voltage.

As i am using an ideal inductor the saturation current doesn't matter right?
 
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