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Increase the current of a boost IC

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Jordan Haining

New Member
Hello,

I am wondering if there is a way to increase the output current of a boost IC? Will a outboard pass-transistor work like it does with a linear regulator? I have found a suitable I2C boost regulator IC that operate in my desired voltage range the problem is it has a max I of 300mA but I need 20A. How could I do this.

Thanks Jordan
 

ronsimpson

Well-Known Member
Most Helpful Member
I2C boost regulator
Do you really need I2C?
I need 20A
You need an external transistor. You need an IC that is designed to drive an external MOSFET. ICs that have an internal power MOSFET can not be used like you want.

Please: What part do you like?
I need all the numbers to help you.
Input voltage, output voltage. output current (20A)
The more information the better.

It will be 8 hours before I can get back but some will probably help.
 

Jordan Haining

New Member
The part number is LT3582, the input voltage is 5v and the output voltage will be between 3.5v and 10 or 12v. What I was wanting to do was use a 12v 15A power supply and a zener to power the IC. I will read the data sheet further to see if it has an internal mosfet.

I suppose is doesn't need to be I2C, I just want it to be controlled by a microcontroller and it must be accurate. That IC steps in increments of 25mV which is perfect.
 

Les Jones

Well-Known Member
Most Helpful Member
This is a BOOST regulator so the output voltage must be above the input voltage. If you need the output voltage to go down to 3.5 volts it's input would need to be about 3.0 volts. this would mean that when the output was 12 volts at 20 amps the input current would be over 80 amps. (This would be true with any boost regulator.) You also say that you want to power the circuit from a 12 volt 15 amp power supply that is dropped down to 5 volts (Or for it to work 3.0volts.) using a zener diode. If we consider the 5.0 volts case and a maximum output of 12 volts at 20 amps then you need more than 12/5 x 20 amps = 48 amps. If the zener was in shunt mode it would need to be able to dissipate 5 x 48 = 240 watts and it's series resistor would need to be able to dissipate 7 x 48 = 336 watts. If the zener was used in series with the 12 volt supply then it would need to dissipate 336 watts. I have never seen a zener with such a high power rating. (It would be possible to create one using a smaller zener and power transistors.) No matter what you do it is not possible to get 12 volts at 20 amps from a power supply of 12 volts at 15 amps. Using your boost regulator method you would need an input current of over 80 amps at 3 volts. I think this method is a non starter.

Les.
 

Jordan Haining

New Member
This is a BOOST regulator so the output voltage must be above the input voltage. If you need the output voltage to go down to 3.5 volts it's input would need to be about 3.0 volts.
Sorry I meant 5.5V not 3.5 volts that was a typo, I also should have made it clear that I knew I wouldn't get 12v @ 20A from a 12V 15A. What I am trying to do is turn a 12v power supply into a 5-12v power supply that is mircocontroller controlled and has a single output. I wasn't looking at how I could make the boost converter supply the current on its own but if I could get an external transistor to supply the extra current. I do expect the max current to go down as voltage goes up.


If you already have 12V why do you need "and 10 or 12v". What current?
The reason I said 10v or 12v is because I was aware that I already had a 12v line but didn't want to have to unhook my wires from one terminal and connect to another to get 12v. I was thinking of going 5.5v-10v and using a relay to switch between 5.5v-10v from the IC and 12v supply voltage since at 12v I will already have 15A which is sufficiant at that voltage. I also want more current then the average regulator supplies. so was wondering if I could use an external transistor to increase the current my regulator id supplying. I am going to look for an I2C buck converter to step my voltage down instead.

Edit: I guess 20A is pretty high I would even be happy with 10A.
 
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Jordan Haining

New Member
Be you like I2C and Linear Tec;
Looking at the datasheet for this IC it has the same problem as the I2C buck converters I have founf. It has a max voltage much lower than I want. In this case 5.4v. I haven't purchased a power supply yet so I could even find one with a higher voltage. I am doing the research before I purchase anything.

Edit: I do however have my old ATX bench top power supply that has a 5v rail and 12v rail I could just use that.
 
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ronsimpson

Well-Known Member
Most Helpful Member
I think you have a good idea. Use a off line power supply to get away from the power line. Then use a safe voltage that will not kill you.

I would use a 24 to 48 volt power supply. Then buck down to 5 to 12 programmable.
If you use 12 or 11V supply then you will need to "boost/buck". Buck=5V and Boost=12V. This can be done but it takes more parts.

I could not find a good I2C buck IC. I will look more.
 

ronsimpson

Well-Known Member
Most Helpful Member
Something like this? The two feedback resistors could be replaced with a digital pot. The digital pot is controlled by the micro.
upload_2017-3-12_21-0-53.png
 

ronsimpson

Well-Known Member
Most Helpful Member
Synqor has other modules. I was trying to find something that is programmable and not too hard to use. I remembered using several modules that had programmable V and I.
 

Jordan Haining

New Member
I was checking out their website, they have a lot of options for all sorts of applications. I am looking forward to building this. To power it should I just buy a 40V power supply or use a suitable transformer and diode bridge?
 

ronsimpson

Well-Known Member
Most Helpful Member
I find I can get a 24V or 48V or something "brick" power supply for good prices. I like the Idea the input is 110/220 AC and the output is 40V regulated and current limited. A transformer + diodes is not current limited and not regulated.
 

Jordan Haining

New Member
Sounds good, thanks for the help this route is way better then what I originally had planed. I was reading the datasheet and it seems pretty easy to adjust the voltage and current.
 

Jordan Haining

New Member
Sounds good, thanks for the help this route is way better then what I originally had planed. I was reading the datasheet and it seems pretty easy to set the voltage and current.
 
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