Buck converter with voltage compensator not working

Thread Starter

acheriti

Joined Apr 26, 2020
41
Hi guys,
I'm trying to design a buck converter with a PWM voltage feedback loop.I put an image of my circuit in the linked files. To do this design I followed the procedure proposed in the following document :
https://www.infineon.com/dgdl/an-1162.pdf?fileId=5546d462533600a40153559a8e17111a
I put the image of the topology in a below picture.
My buck has the following parameters :
Vin=10V; Vout=5V;
For a reason I'm not sure to understand, my circuit is not working. My output voltage is not going to 5V. Furthermore, the error voltage that goes out of the error amplifier always seems to be the rail as if this amplifier was not doing its job.
My PWM is thus always constant with a constant duty cycle.
My output results are also linked in a picture linked in this post.
Does anyone have an idea of what could be going wrong in my circuit ?

Thanks a lot guys !
 

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Ian0

Joined Aug 7, 2020
9,489
Hi guys,
I'm trying to design a buck converter with a PWM voltage feedback loop.I put an image of my circuit in the linked files. To do this design I followed the procedure proposed in the following document :
https://www.infineon.com/dgdl/an-1162.pdf?fileId=5546d462533600a40153559a8e17111a
I put the image of the topology in a below picture.
My buck has the following parameters :
Vin=10V; Vout=5V;
For a reason I'm not sure to understand, my circuit is not working. My output voltage is not going to 5V. Furthermore, the error voltage that goes out of the error amplifier always seems to be the rail as if this amplifier was not doing its job.
My PWM is thus always constant with a constant duty cycle.
My output results are also linked in a picture linked in this post.
Does anyone have an idea of what could be going wrong in my circuit ?

Thanks a lot guys !
1. Your reference signal (V2) should be a triangle wave
2. You need a high-side driver for M1, because U8 has a maximum supply voltage of 5V, and you need the output to be at least 20V with sufficient output current to charge the gate of M1.
3. The inputs to U8 are reversed (I think)
 

Thread Starter

acheriti

Joined Apr 26, 2020
41
1. Your reference signal (V2) should be a triangle wave
2. You need a high-side driver for M1, because U8 has a maximum supply voltage of 5V, and you need the output to be at least 20V with sufficient output current to charge the gate of M1.
3. The inputs to U8 are reversed (I think)
Hi buddy,
Thank you for your time & answers.
1. The ref signal is a triangular wave (applied to U7 pin-)
2. That's a good point. I will search for one thank you for the proposal.
3. I think they are good if I follow the guide linked in my first post. This user guide is done by Infineon Technologies

Thanks a lot !
 

Ian0

Joined Aug 7, 2020
9,489
Hi buddy,
Thank you for your time & answers.
1. The ref signal is a triangular wave (applied to U7 pin-)
2. That's a good point. I will search for one thank you for the proposal.
3. I think they are good if I follow the guide linked in my first post. This user guide is done by Infineon Technologies

Thanks a lot !
1. There isn‘t a U7. (Don’t forget some of us may be answering on iPads and they don’t do SPICE) and V2 is a pulse, not a triangle
3. If the final output goes high, then the output of the reference amplifier will go low, which is connected to the inverting input of the comparator, and that will tend to make its output spend longer high, which will make the output go even higher. So I think it is reversed.
 

ronsimpson

Joined Oct 7, 2019
2,890
Here it is.
Thanks for the file. It saves me time.
1) the voltage comparator is to be used at 5V and you are using it with a 12V supply. SPICE does not know that to do with that. SPICE has the output driving to 12V but in real life that will not happen. Choose a much higher voltage part and drive it with 15 volts or more.
2) Many people think that applying 0V and 12V to the gate of a MOSFET will turn it on. The FET does not care what the voltage is as measured from ground. The FET only cares what the Gate to Source voltage is. In this case the FET wants about 4.5V G-S to turn on. The Gate is at 12V, substrate 4.5V =7.5V is as high as the Source is going to get. You need to drive the Gate to 5V to 10V above the 10V supply.
3) The error amplifier cannot output more than 7V. Yet the ramp goes to 10V. I think the ramp should go 0 to 6V.
4) C2 is tooooooo small. Change to 1nF for now.
5) Logic is inverted. Swop the two input pins.

It seems to want to regulate at 4V.
Do these first. 5, 4, 3, and it works. It needs more work but that will get you going for now.
 

LowQCab

Joined Nov 6, 2012
3,935
Do You really need the extra complexity of this Circuit ?
Why did You choose this particular Circuit ?
How much Current do You need at 5-Volts ?
Do You have any particular physical size or cost constraints ?
Do You want to build your own Switching-Regulator from scratch ?
.
.
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