DC-DC converter design

[electronicsLearner77]

For a long time i am thinking of designing circuits on my own. So, just as beginning circuit i am trying Dc-Dc circuit and converting the 20V to 5V, because i come across dc-dc conversion very frequently. Below is the circuit i tried

i have drawn a resistor divider network and after calculations, i know if the ratio is R1/R2 is 3:1 i can get 5V. So, my questions are
a. Is my design correct?
b. What are the additional things i need to take care, like on what basis should i select R1 and R2?
c. What are the other better methods, if you can suggest i will try to draw them and simulate.
(As a first step i do not want to google and get myself confused)

 

[bertus]

Hello,

The output voltage in a resistor voltage divider will not be stabel when loaded.
The lower the value of the load, the lower the output voltage will be.

Bertus

 

[Harald Kapp]

a. Is my design correct?

Basically: yes, but (as @bertus stated). Do the math yourself and see:
For any load resistance R_load calculate the effective resistance of R_parallel = R_load parallel to R2, then re-calculate the output of the voltage divider made from R1 and R_parallel. See what Bertus means?

What can you do to remedy this: You need to decouple the load from the circuit setting the output voltage (your voltage divider). This can be done in various ways, starting with a simple zener diode,, moving on to transistorized circuits and not ending with a sophisticated control loop and power amplifier. Here's a (imho) good overview to get you started.

 

[ratstar]

If you use large resistances for the voltage divider, wouldnt it handle more load on the output loop? but is there a disadvantage in doing that?

Also, what if you went to a capacitor after the voltage divider, then added the load to the discharge of the capacitor instead of using the line directly.

 

[Harald Kapp]

If you use large resistances for the voltage divider, wouldnt it handle more load on the output loop?

No.

but is there a disadvantage in doing that?

Yes, do the math as shown.

Also, what if you went to a capacitor after the voltage divider, then added the load to the discharge of the capacitor instead of using the line directly.

What is it about capacitors that you are so obsessed with them? A capacitor parallel to the load will help reduce voltage drop if the load creates a short current spike. It will help by no means for the static voltage drop across R1 due to load current.

 

[ratstar]

If you output to the capacitor from the voltage divider, then you make another loop from the capacitor, thatll be like a voltage follower, the circuits load wouldnt affect the functioning of the voltage divider.

I actually dont know for sure, I should try these things out before I post em.

 

[bertus]

Hello,

Putting a capacitor on the output of the voltage divider will delay the voltage drop, but can not prevent the voltage drop,

Bertus

 

[Harald Kapp]

I actually dont know for sure, I should try these things out before I post em.

How true.