Will this circuit be efficient ? I'm thinking of the input supply to be LIPO 11.1V 2200mAH for output of ~6V and 4A

 

Input current = 19.7A DC???

 

For goodness' sake don't try that one out, especially with a rechargeable battery. I'm not an expert in this sort of thing but this circuit does not look even vaguely correct. I fear you are likely to start a fire or worse.

Please wait for someone better informed to advise you.

 

Need more info on what you're trying to do. I see stepping 12v supply down to 6v, but I don't see what the 6v current load is. This would be a lot of trouble for a lightbulb. Those transistors will handle a few amps of current - if heat sinked - but the other components would be toast.

 

I set the load to be 6V 24W as an example. The lightbulb is just a test load for loads like a servo or dc geared motor

 

Input current = 19.7A DC???

I know it's high but not as high as when I used MOSFET. It went up to >50A which was totally absurb.

This happened when I step down the voltage to 6V. On 12V output, everything's fine (input current to be 4A). Since this is the case, how do i pick 6V out of the 12V output ?

 

Do you really need 4A current output? That's quite a bit.

Why don't you consider using a 6v sealed lead-acid battery instead of an 11v+ Li-Po?

 

I know it's high but not as high as when I used MOSFET. It went up to >50A which was totally absurb.

This happened when I step down the voltage to 6V. On 12V output, everything's fine (input current to be 4A). Since this is the case, how do i pick 6V out of the 12V output ?

OK, I have a better idea of what is going on here. There is about 16A flowing to ground via a BZV55-C6.V2 Zener diode.
That is a half-watt device and so should not be run beyond about 80mA continuous. Unfortunately, the Zener is fed via a low-resistance path from the input supply. There is nothing to restrict the current to a safe value.

The circuit may seem to "work" in SPICE, but in real life you would see smoke. The overall efficiency is only about 10%, and your circuit would dissipate nearly 200W because of the huge current being wasted to ground. Even a competently designed linear regulator should give you around 50% efficiency in an 11V to 6V application.

The bottom line is that you need either to get a better idea of the principles behind circuit design, or to get hold of a reliable ready-made circuit. I am not your man for that, perhaps someone else on the forum could help.

Please do not try to use circuits like the one you have posted in practice. If you do, there is liable to be an accident.

 

OK, I have a better idea of what is going on here. .....
The bottom line is that you need either to get a better idea of the principles behind circuit design, or to get hold of a reliable ready-made circuit. I am not your man for that, perhaps someone else on the forum could help.

Please do not try to use circuits like the one you have posted in practice. If you do, there is liable to be an accident.

Please do not worry. I have no intention of putting the zener diode there to drop the voltage. I am concern of the safety of the battery and myself fiddling with high current as well.
Using a resistor to drop will generate too much heat on the resistor itself and considerable power loss. Anyway, do u have some products ready-made circuits you mentioned to recommend? Preferably to be able to withstand high output current.

On the SLA batt, would it be bulky and heavy?

 

An SLA battery would be larger and heavier than a Li-Po battery.

What are you attempting to do with this power supply? Are you certain of your output current requirements?

 

Yes , I'm sure of the current output requirement. I had tested the load on dc power supply (the big ac to dc regulator type) in my school lab. The load (servo motors) would not run on 6V 2A supply. But it could run perfectly on 5V/5A spec.

However, I would re-check again tomorrow.

 

OK, now you're changing things again.

Previously, you were saying 6v, 4a. Now you're saying 5V/5A.
The power requirement is nearly the same (24W for the former vs 25W for the latter), but it's still a change.

 

. But it could run perfectly on 5V/5A spec.

Then just pull a used power supply out of a trashed computer. As long as the PSU is still working, 5A at 5V is exactly what these things were designed for.

 

OK, now you're changing things again.

Previously, you were saying 6v, 4a. Now you're saying 5V/5A.
The power requirement is nearly the same (24W for the former vs 25W for the latter), but it's still a change.

I'm sorry to confuse things around. The motors have operating conditions between 4.5V to 6V. Since before I posted this thread, there was no way to run the motors on the 6V with only 2A. So,my option was only that 5V/5A supply, decided to lower the voltage in order to use higher current. Even running perfectly, I'm only running like 2/3 of my loads. Running at full load will not work even on the 5V/5A. Thus, I like to start from ground and take things slowly. First, powering the 2/3 loads and then find other methods to power the rest of the load.

Now, I'm thinking of using higher voltage ~6V and current hovering between 3 to 4A as it gives more power at the output.

 

The SIMPLEST way is to match the power supply voltage to the max voltage needs of the motor in question.

Your motor in question needs 6 volts max, and you seem to need the max output since 5volt 5 amp wasn't enough.

ANSWER: a 6 volt lead acid battery, will power that motor and can be charged with cheaper circuits, than building a fancy step down supply just to use Li-Po. If you need Lithium batteries for the weight decrease in model applications, then USE A MOTOR that is compatible, and you will realize greater efficiency overall.

The method you are pursuing is called a 'kludge' and means you are taking what is available and forcing it fit your needs. That is OK, as long as it gives you what you want or need. If you want MAX power and efficiency then you need to change something as outlined above.