Simply put a reference voltage into the FB (Feed-Back) Pin from your controller,
super easy, very efficient.

And I have to ask .......
Why do you want to control a Power Supply Voltage with a Micro-Controller ????
.
.
.

for making my own variable constant current and constant voltage supply

 

A Micro-Controller is of limited usefulness in a Bench-Power-Supply.
Unless you just have to have "Up-Down" Buttons instead of Knobs.
Usually you will simply want 2-Knobs on the front Panel, one for Voltage, and one for Current,
along with 2-Indicating Meters, or 3, if you want to actively display the actual Current being used.
You could also have "Range-Switches", with the Knobs being only a Fine-Adjustment.

When building anything, you need to specify all of the parameters that you expect you will need.
"There's more than one way to skin a Cat"
Please be specific in your expectations.
ALL information is important,
if you don't really know exactly what you need in terms of performance,
then please provide the Ideas that you have been thinking about regarding this pursuit.
And someone else here, or I, can point you in the right direction.
.
.
.

 

okay guys i made some changes. when i test the first curcuit its efficiency is about 87% here it is

And with buck converter efficiency is about 82% ?? how is that possible
what am i doing wrong here guys

There are a couple things to talk about here.

First is the MOSFET that looks to be a good low Rds_on type which you want if you need high efficiency. Second is that back catch diode which i am assuming is a Schottky type with a decent current rating maybe 1.5 times the max output current. The inductor is 100uH so you should get decent operation at around 50kHz and that is where the frequency is high enough to allow smaller components but not too high such that other problems creep in like skin effects and turn on and turn off transient times that end up being too comparable to the on and off times themselves. However, for loads as low as 1 Ohm you will need a very low ESR inductor for high efficiency.

But then we come to the drive circuit. That's the bipolar AND the 1k resistor gate to drain.
When the MOSFET is turned on, the small bipolar transistor might provide a decent amount of current into the gate of the MOSFET. When the transistor turns off though, the only thing left to turn the MOSFET off is that dinky 1k resistor. We could do a little equation that shows that as the MOSFET equivalent gate charge is energized by the transistor it 'probably' happens pretty fast, but as the transistor turns off the 1k takes a long time to remove that gate charge so that the MOSFEt can turn off fully. In the mean time, the MOSFET has both sizeable current through it and a decent amount of voltage across it and that combination leads to power dissipation that gets lost in the MOSFET and that reduces efficiency as well as heat sup the MOSFET. Since the MOSFET is a small package the max dissipation, all things considered and a typical mounting scheme, is probably only around 1 watt. That means the MOSFET could even overheat and blow out just due to that 1k resistor used for the turn off period.
The thing is, when using MOSFETs in high efficiency power products you have to use a really good drive scheme that really delivers a good amount of current to the MOSFET gate. A min is probably around 500ma but 1 amp would be a good idea as a min. If you look above you will see i quoted the word "probably' even when turning the mosfet on, and that is because even that transistor may not be good enough to get high efficiency.
So this leads us to what is called a MOSFET gate driver IC chip which is made specifically for driving MOSFETs so that they turn on and off very sharply. You will need to look this up and see what one you might want to get. If you dont like that idea then at least go to a two transistor drive that provides one transistor to turn on and one to turn off, that is probably a minimum requirement but wont be as good as a specialized mosfet driver chip.

Next, that 10uOhm input resistor beings up the question of what exactly is the input source, where is that coming from. Is it a lead acid battery or some other power supply? That has some bearing on the efficiency also as well as the general working of the circuit. You are going to need some input capacitors also of the low ESR type. They should be mounted close to the circuit.

Lastly i see you want to use a microcontroller to control the output and possibly regulate it.
The best bet here is to develop the circuit using pure analog control that depends on an analog input voltage first, then later add the microcontroller. The uC then is used to vary the analog control voltage. This gives good control although you will have to work on the speed of response, and it is also a good idea to have analog current limit so that it can work very fast when it comes to detecting and acting on over currents on the output.

We can talk more about these and other things that come up. In the end you should get a good circuit out of it and i think you will understand these circuits much much better too and that will be rewarding in and of itself too
Since you are obviously interested in theory too, it might interest you to know that the buck circuit has the simplest control law for the output voltage relationship to the pulse width. This is probably the first theory to learn with these circuits. The simple relationship is:
Vout=Vin*D
where D is the fractional duty cycle (like 0.50 is 50 percent and D<=1).
This tells you, given all ideal components, what duty cycle you need for a given output.
Since this ignores all losses, we always incorporate feedback to make up for the variations that come in the practical circuit, but that is the starting point in the theory of the buck circuit.

 

There are a couple things to talk about here.

First is the MOSFET that looks to be a good low Rds_on type which you want if you need high efficiency. Second is that back catch diode which i am assuming is a Schottky type with a decent current rating maybe 1.5 times the max output current. The inductor is 100uH so you should get decent operation at around 50kHz and that is where the frequency is high enough to allow smaller components but not too high such that other problems creep in like skin effects and turn on and turn off transient times that end up being too comparable to the on and off times themselves. However, for loads as low as 1 Ohm you will need a very low ESR inductor for high efficiency.

But then we come to the drive circuit. That's the bipolar AND the 1k resistor gate to drain.
When the MOSFET is turned on, the small bipolar transistor might provide a decent amount of current into the gate of the MOSFET. When the transistor turns off though, the only thing left to turn the MOSFET off is that dinky 1k resistor. We could do a little equation that shows that as the MOSFET equivalent gate charge is energized by the transistor it 'probably' happens pretty fast, but as the transistor turns off the 1k takes a long time to remove that gate charge so that the MOSFEt can turn off fully. In the mean time, the MOSFET has both sizeable current through it and a decent amount of voltage across it and that combination leads to power dissipation that gets lost in the MOSFET and that reduces efficiency as well as heat sup the MOSFET. Since the MOSFET is a small package the max dissipation, all things considered and a typical mounting scheme, is probably only around 1 watt. That means the MOSFET could even overheat and blow out just due to that 1k resistor used for the turn off period.
The thing is, when using MOSFETs in high efficiency power products you have to use a really good drive scheme that really delivers a good amount of current to the MOSFET gate. A min is probably around 500ma but 1 amp would be a good idea as a min. If you look above you will see i quoted the word "probably' even when turning the mosfet on, and that is because even that transistor may not be good enough to get high efficiency.
So this leads us to what is called a MOSFET gate driver IC chip which is made specifically for driving MOSFETs so that they turn on and off very sharply. You will need to look this up and see what one you might want to get. If you dont like that idea then at least go to a two transistor drive that provides one transistor to turn on and one to turn off, that is probably a minimum requirement but wont be as good as a specialized mosfet driver chip.

Next, that 10uOhm input resistor beings up the question of what exactly is the input source, where is that coming from. Is it a lead acid battery or some other power supply? That has some bearing on the efficiency also as well as the general working of the circuit. You are going to need some input capacitors also of the low ESR type. They should be mounted close to the circuit.

Lastly i see you want to use a microcontroller to control the output and possibly regulate it.
The best bet here is to develop the circuit using pure analog control that depends on an analog input voltage first, then later add the microcontroller. The uC then is used to vary the analog control voltage. This gives good control although you will have to work on the speed of response, and it is also a good idea to have analog current limit so that it can work very fast when it comes to detecting and acting on over currents on the output.

We can talk more about these and other things that come up. In the end you should get a good circuit out of it and i think you will understand these circuits much much better too and that will be rewarding in and of itself too
Since you are obviously interested in theory too, it might interest you to know that the buck circuit has the simplest control law for the output voltage relationship to the pulse width. This is probably the first theory to learn with these circuits. The simple relationship is:
Vout=Vin*D
where D is the fractional duty cycle (like 0.50 is 50 percent and D<=1).
This tells you, given all ideal components, what duty cycle you need for a given output.
Since this ignores all losses, we always incorporate feedback to make up for the variations that come in the practical circuit, but that is the starting point in the theory of the buck circuit.

View attachment 231735

wow thank you alott that R3 resistor is just for me to measure current in the input
what im wondering is why that first circuit is looks same efficiency like second buck curcuit

 

wow thank you alott that R3 resistor is just for me to measure current in the input
what im wondering is why that first circuit is looks same efficiency like second buck curcuit

I dont think that is anything to worry about unless the inductor is saturating. Where did you get the inductor? The main point is to get the MOSFET driven better that should improve things a lot. We might need more info on your input source too though.

The inductor has to have the right ratings too.

 

I dont think that is anything to worry about unless the inductor is saturating. Where did you get the inductor? The main point is to get the MOSFET driven better that should improve things a lot. We might need more info on your input source too though.

The inductor has to have the right ratings too.

i meanthe reason i converted it to buck converter circuit is for efficiency like crutschow said . bu my first curcuit without inductor and diode is still showing same efficiency what is the differance then

 

i meanthe reason i converted it to buck converter circuit is for efficiency like crutschow said . bu my first curcuit without inductor and diode is still showing same efficiency what is the differance then

Yeah and i mentioned several things that could cause that.

The first is the inductor. If the inductor saturates it acts like a regular piece of wire and that would mean you essentially have the same circuit as before. That is why we have to get the specs of the inductor before anything else.

 

Yeah and i mentioned several things that could cause that.

The first is the inductor. If the inductor saturates it acts like a regular piece of wire and that would mean you essentially have the same circuit as before. That is why we have to get the specs of the inductor before anything else.

i understand that what im actually asking is how could firs circuit be that much efficient like 87% really ? with that cicuit its should dissipate alot of heat which is decreases efficiency alott not like im seeing here 87% percent right ? i mean if this circuit is 87% percent efficient then why should i bother put inductor and diode in there . maybe something is wrong about that efficiency calculation of mine ? isnt 87% efficiency is too high for this circuit ?

 

actually you are great teacher now i understand so in order to make stable voltage with cap in curcuit i have to put a feedback for the Pulse signal right
i just want to make the curcuit give me the variable voltage i want by controling PWM signal Duty cycle

Boost circuits are there to take advantage of inductive voltage spike and hold that value with a capacitor. That is more efficient than a usual regulator, but you're mis-understanding PWM. PWM doesn't control voltage, it controls current. Width of pulse delivers more or less current. You need to have a handle on Ohm's and Watt's laws before you go designing circuits you don't understand.

This will help:

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

There are other (easier) ways to make a voltage controller. LM317T, for example.

 

i understand that what im actually asking is how could firs circuit be that much efficient like 87% really ? with that cicuit its should dissipate alot of heat which is decreases efficiency alott not like im seeing here 87% percent right ? i mean if this circuit is 87% percent efficient then why should i bother put inductor and diode in there . maybe something is wrong about that efficiency calculation of mine ? isnt 87% efficiency is too high for this circuit ?
View attachment 231779

A buck regulator is considered to be more efficient then say a linear regulator but that is the most common case and not always true. There can be times when the linear is equal to or better than the buck, but it is rare to see this because the input output voltage differential is usually high. When it is not high however, a linear regular can be more efficient.
So what this means is that the input to output voltage differential plays a part in the efficiency.

With your original circuit however, you have another thing to consider because it pulses. That is the output ripple voltage. The output ripple voltage will be terrible without an inductor and that is something that is usually undesirable except in cases where the load does not care about ripple.

So you should look at your load and see if it requires a good smooth output voltage. If it does, you will need an inductor.

Another consideration is the ripple current. This is important in order to protect the output filter capacitors. They have ratings too, and if the peak current is too high they can overheat and becomes damaged. If they start to become high impedance, the filter action will decrease and thus the output will be very choppy.

Any other questions about why we use an inductor?

 

A buck regulator is considered to be more efficient then say a linear regulator but that is the most common case and not always true. There can be times when the linear is equal to or better than the buck, but it is rare to see this because the input output voltage differential is usually high. When it is not high however, a linear regular can be more efficient.
So what this means is that the input to output voltage differential plays a part in the efficiency.

With your original circuit however, you have another thing to consider because it pulses. That is the output ripple voltage. The output ripple voltage will be terrible without an inductor and that is something that is usually undesirable except in cases where the load does not care about ripple.

So you should look at your load and see if it requires a good smooth output voltage. If it does, you will need an inductor.

Another consideration is the ripple current. This is important in order to protect the output filter capacitors. They have ratings too, and if the peak current is too high they can overheat and becomes damaged. If they start to become high impedance, the filter action will decrease and thus the output will be very choppy.

Any other questions about why we use an inductor?

oh yeah thank you soo much now i UNDERSTAND because of you i will now build a circuit with feedback and all and test it i hope you guys will help me more in future questions thank you so muchh

 

oh yeah thank you soo much now i UNDERSTAND because of you i will now build a circuit with feedback and all and test it i hope you guys will help me more in future questions thank you so muchh

You are welcome. I would suggest getting the circuit running properly first before adding feedback.
There may be an issue with the inductor and you want to get the right one first.
If you can find the specs of that inductor we could determine if it is a good choice or not.

 

You are welcome. I would suggest getting the circuit running properly first before adding feedback.
There may be an issue with the inductor and you want to get the right one first.
If you can find the specs of that inductor we could determine if it is a good choice or not.

okay thanks )