I'm just wondering now
Many videos and sites have made similar examples of this circuit.
Where is my problem?
All the parts have been changed with different brands and quantities, but the problem is still there!!!!!!
uc3843 buck-boost
- Thread starter mrh586
- Start date
Many videos and sites have made similar examples of this circuit.
Where is my problem?
All the parts have been changed with different brands and quantities, but the problem is still there!!!!!!
A couple of things to check:
1. transformer polarity connections. Make sure the phase connections are the same as on the schematic.
2. It seems that at a switching frequency of 25Khz that the inductance should be around 150uh, not 10uh.
Thankful
I have checked the connections many times.
Is it possible to raise or lower the frequency instead of increasing the capacity of the inductor?
And besides, I have tested the core from 10 to 40 microhenries, shouldn't there be some change in the problem between these two?
No. That is insignificant compared to 150uh.
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Usually Flyback is used when load isolation is needed. That said, a septic topology provides isolation but is more efficient. So the choice between the two, IMO, is the septic topology. However, the assumption here is that isolation is needed. If it isn't, then just use a standard buck-boost topology and off the shelf inductors, and save yourself some pain...
There's no isolation with a SEPIC circuit.
I stand corrected sir.
Thanks.I don't like the wikipedia description though, no mention of coupled inductors.
Please include a schematic example of what you mean
Interesting to note that if you remove the capacitors, it does become an isolated flyback converter, but the current flow is completely different, as, with the capacitors, the input current can be continuous.I stand corrected sir.
I don't like the wikipedia description though, no mention of coupled inductors.
Yes that's right
I removed the parallel inductor capacitors and tested them.
But when the load is connected to the circuit, a lot of pressure is applied.
As for the circuit, when I lower the frequency (especially by increasing the resistance between pin 4 and 8 of the IC), the MOSFET heats up less, and I have less voltage drop.
But the inductor whistles when the load is connected
When the frequency increases, the sound does not come out of the inductor and the MOSFET heats up
I think that at the same time as lowering the frequency, the number of turns of the inductor coil should be increased
Assuming the inductors are ~10uH...
Try raising the switching frequency to 100khz.
R=17.4k C=1n
Thankful
I will definitely do it and announce the results
Please explain to me the formula for calculating the frequency based on the inductor, if it is not too complicated
This information can be found in Hart, Daniel W., Power Electronics, Chapter 6, §6.6, pp. 222-226
This is a freely downloadable e-book in PDF format. The output voltage for a buck-boost converter is given by:
\( V_{o}\;=\;-V_{s}\left(\cfrac{D}{1-D}\right) \)
where D is the nominal duty cycle for open loop operation. To operate in continuous conduction mode, the inductor current must remain positive. To determine the boundry between continuous conduction mode (CCM) and discontinuous conduction mode (DCM) we can use the following:
\( (Lf)_{min}\;=\;\left(\cfrac{(1-D)^2R}{2}\right) \)
which we can rearrange to get
\( L_{min}\;=\;\cfrac{(1-D)^2R}{2f} \)
where f is the switching frequency and R is the load resistance across which the output voltage is developed. This result is easy to verify with a simulation.
This is a freely downloadable e-book in PDF format. The output voltage for a buck-boost converter is given by:
\( V_{o}\;=\;-V_{s}\left(\cfrac{D}{1-D}\right) \)
where D is the nominal duty cycle for open loop operation. To operate in continuous conduction mode, the inductor current must remain positive. To determine the boundry between continuous conduction mode (CCM) and discontinuous conduction mode (DCM) we can use the following:
\( (Lf)_{min}\;=\;\left(\cfrac{(1-D)^2R}{2}\right) \)
which we can rearrange to get
\( L_{min}\;=\;\cfrac{(1-D)^2R}{2f} \)
where f is the switching frequency and R is the load resistance across which the output voltage is developed. This result is easy to verify with a simulation.
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maybe lost in previous posts but..,
What is your specification?
Vin min=
Vin max=
Vout max=
Vout min=
Iout max=
I am really grateful
But I don't know if the information I give you is what you want.
I asked the formula because currently my core is 30uH.
The input of my circuit is a 27V 15A power supply
And I want the output from 3 to 60 volts (of course, currently with R1=1 k, the output is about 40 volts), I don't know how to calculate the current.
Once before, when I wanted to increase the frequency with 1n capacitor, as soon as I connected the load to the output, the MOSFET burned, but the resistance was 8.2 k
It is not possible to build or test your design if you don't have a design specification to work from.
It indicates you have no idea what you are building. So how can anyone help if you don't know what your building?
Think about these parameters and provide values:
Supply voltage input:
Vin min=
Vin nominal= (mandatory)
Vin max=
Required output voltage:
Vout max=
Vout nominal=(mandatory)
Vout min=
Required output current:
Iout min= (we'll figure this out, but it will be a few hundred milliamps for CCM)
Iout max=(mandatory)
Please provide practical values for the above. We cannot continue without you providing the values marked "mandatory".
