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24V to 5V Buck converter

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Okay. If the Vf is between 3.3V and 5V, whatever the Vf is, and if you are sure that you want Vout of 5V, then you need a resistor of (Vout-Vf)/20mA in series with the LED at the output. You have to realize that that resistor is going to contribute to some power loss and, consequently, drop in efficiency.

Now, let's discuss the reference voltage. For this purpose, I'd recommend a 2.5V zener diode. This means that you will have equal resistances in the output voltage resistive divider circuit.

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22k or 27k should do for each of the two resistors.

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Now, we need to select an inductance value. Which conduction mode do you prefer? CCM or DCM?
 

Rload = (5-3.3)/20mA = 85 ohm
Lcrit = (1-D)*Rload/(2*fsw)

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If you want a duty cycle of 25% = 0.25 at fsw = 20kHz,
then Lcrit = (1-0.25)*85/(2*20000) = 1.59mH.

your inductor should have lesser value than this for your buck to be in DCM all the time. However, if you need a duty cycle of about 25% then your L has to be close to Lcrit. The 100uH that you selected would divide Lcrit by about 16 times meaning that (1-D) would be divided by 16 too. This would make (1-D) = about 0.75/16 = about 0.047. This means that David would be about 1-0.047 = about 0.95.

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My recommendation, use Rload = 82 ohm for about 20.7mA load current; with a duty cycle of 25% = 0.25 at fsw = 20kHz, then Lcrit = (1-0.25)*82/(2*20000) = 1.54mH
Use 1.5mH.

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Oh! You even used L = 1uH. That was a major problem then.
 
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My recommendation, use Rload = 82 ohm for about 20.7mA load current; with a duty cycle of 25% = 0.25 at fsw = 20kHz, then Lcrit = (1-0.25)*82/(2*20000) = 1.54mH
Use 1.5mH.

Thank you for your response.
After simulating the circuit using the values you provided, i got the same previous result (Higher output voltage).

Buck.png

Have a nice day.
 

The answer is at your fingertips, review my comment in post #6.
 

Thank you for your response.
After simulating the circuit using the values you provided, i got the same previous result (Higher output voltage).

View attachment 153466

Have a nice day.

Please always show your simulation waveform along with the schematic.

I noticed that you didn't include your LED in series with the 82-ohm resistor at the output. I just simulated it myself. The duty cycle is off. The duty cycle that I got 5V is about 12.5% with the values calculated.

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Also remember to reference you gate drive voltage to the MOSFET source terminal.

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Maybe you should show how you are driving the MOSFET too.
 
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    FvM

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Hello there,
Read your question, you think you must not know **broken link removed**, your problem can be solved with lm2596 integrated circuit, give you a link you can go to see the detailed circuit diagram. Production is also very simple and convenient. Hope it helps you.

**broken link removed**
 

I noticed that you didn't include your LED in series with the 82-ohm resistor at the output. I just simulated it myself. The duty cycle is off. The duty cycle that I got 5V is about 12.5% with the values calculated.
Post #23 Lcrit calculation would be correct for an actual load resistance of 85 ohm. Connecting it as LED series resistor reduces the load current to a third, respectively threefold L required for continuous current mode.

It's an instructive exercise to find out the condition for CM versus DCM and the effect on buck transfer function. A practical buck converter should have pulse width regulation to be prepared for varying load.
 

See my simulation.

Buck14.PNG

Maybe you should look into your simulation again.

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Post #23 Lcrit calculation would be correct for an actual load resistance of 85 ohm. Connecting it as LED series resistor reduces the load current to a third, respectively threefold L required for continuous current mode.

I calculated for the 85-ohm load current considering 3.3V as the forward voltage of the load current.
Here:

(5V-3.3V)/20mA = 1.7V/20mA = 85 ohm.

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A practical buck converter should have pulse width regulation to be prepared for varying load.

Yes, definitely. But you have to design the power stage first before the control loop. You close the feedback loop depending on the frequency response of the power stage.
 

I calculated for the 85-ohm load current considering 3.3V as the forward voltage of the load current.
I know. My point is that 85 ohm can't be used as effective load resistance in the Lcrit calculation.
 

See my simulation.

View attachment 153478

Maybe you should look into your simulation again.

That was corrected by using another of mosfet (2N7002 instead of IRF54), now I should design the control loop, any suggestions are welcome.

I have another question : How to calculate the inductance value if the output is variable (i.e : from 5V to 24V)?

Thank you.
 

I calculated for the 85-ohm load current considering 3.3V as the forward voltage of the load current.

Sorry, I meant to say forward voltage of the LED here.


My point is that 85 ohm can't be used as effective load resistance in the Lcrit calculation.

FvM has a point here. This will correct the duty cycle somewhat. The equivalent load resistance should be 5V/20mA = 250 ohm.
Lcrit then gives 4.68mH. I used 4.5mH in the simulation and I got a 5-V output at duty cycle of 20% as compared to the expected value of 25%. This is far better than 12.5% we had with the 85-ohm used as load current in the previous Lcrit calculation.

Buck14.PNG

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… now I should design the control loop, any suggestions are welcome.
...
Thank you.

You should give more detail about your input range so we don't go back and forth later on. The calculation that we did before was based on a single value input voltage that you specified in post #1.

We will need the input voltage range. A 555 timer IC as controller is...

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I just looked at the LM2596 datasheet as suggested in post #27. It is a simple and inexpensive implementation. Maybe you should take a look at it.
 
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The input voltage is fixed at 24V
Output voltage : from 5V to 24V
Imax > 200mA
Frequency : 20kHz
Duty cycle : 0% to 100% (not necessarily)

Please suggest MOSFET to be used as a switch.
Note also that I will use a 555 timer to drive the MOSFET

Thank you.
 

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