How Can I Hack This Buck-Boost Converter To Get Higher Than Rated Current Output ?

Thread Starter

Mod91

Joined Sep 4, 2018
11
I need a DC-DC buck-boost converter to get around 80 - 120 Watts of power. This LTC3780 based DC-DC buck-boost converter seems to fit the bill, however its current output is maxed out at 10 Amps. Is there a way to override its max current limit without triggering shut-off ?

# At max I need no more than 6V - 6.5V (typically 3V - 5V) in any case but high Amps up to 25 Amps for up to 5 seconds at a time is needed.
# Input power from 4 lithium-ion batteries in parallel-series with each battery (3.6V nominal) capable of 20 Amps of continuous discharge.
# For electronic cigarette (e-cig) mod.

This buck-boost converter has 15 Amp circuit protection fuse onboard.


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Alec_t

Joined Sep 17, 2013
11,997
Is there a way to override its max current limit without triggering shut-off ?
I doubt it. Over-current shut-off is there for a reason. Components such as the inductor and the semiconductor switching devices could misbehave or fail if they pass excessive current. I'd be looking for a different converter, more closely matched to your application.
 

MrAl

Joined Jun 17, 2014
8,124
I need a DC-DC buck-boost converter to get around 80 - 120 Watts of power. This LTC3780 based DC-DC buck-boost converter seems to fit the bill, however its current output is maxed out at 10 Amps. Is there a way to override its max current limit without triggering shut-off ?

# At max I need no more than 6V - 6.5V (typically 3V - 5V) in any case but high Amps up to 25 Amps for up to 5 seconds at a time is needed.
# Input power from 4 lithium-ion batteries in parallel-series with each battery (3.6V nominal) capable of 20 Amps of continuous discharge.
# For electronic cigarette (e-cig) mod.

This buck-boost converter has 15 Amp circuit protection fuse onboard.


View attachment 160158
View attachment 160160

Hello there,

You could override the current limit, but you'd be taking a very very big chance on blowing the components going up to 250 percent of the rated current limit. If it was only maybe 10 percent or something it would probably work, but 250 percent is 2.5 times the current limit so the chance of failure is much higher.

Taking a quick look at the data sheet, i see the IC in question is actually a controller not really a stand alone switcher, so the current handling depends on the external switching transistors (typically four of them) and the inductor. That means the key components here are the four transistors and the inductor, and possibly the output cap(s).
The transistors have a rating for resistance that is usually selected based on proper operation and efficiency. If they were chosen with very high efficiency in mind, then there is a good chance they would survive short term over currents. They also have a resistance, and with higher current that would mean higher temperatures, and that would mean more cooling time between over current surges to give them time to cool down.
Unfortunately, one of the key components is the inductor, and that not only has a current rating is also has a current rating where the inductance starts to drop off. Once the inductance starts to drop off, the current in the inductor shoots up more and that causes the inductance to drop even more which then causes the current to shoot up more and so within the same time frame that it normally works well at normal currents, the current could shoot up so much that it destroys the transistors very rapidly.

With all this in mind, the two components to change first would be the sense resistor and the inductor, then hope for the best. At 250 percent over current though even the transistors should probably be upgraded, but with only a 5 second duration they may survive if given enough cool down time.

So without seeing actual ratings of anything like the transistors and inductor, most likely you'd have to change at least the inductor and hope for the best if you can afford to buy another unit if this one blows out.
If you really want to gamble, change the sense resistor, then stand back as you test it :)

If you can get the part numbers of the transistors and inductor we can make a much better estimate of what you need to do to get that 250 percent over current working.
Another consideration might be the circuit board itself which has track widths appropriate for the original intended current level.
 

Dodgydave

Joined Jun 22, 2012
9,935
Ok if want to make a bonfire, then pins 3, 4 are the Current Sense, so just short them out to override it........(cue mission impossible music)
 

dendad

Joined Feb 20, 2016
3,810
You could try 3 of these boards with their outputs all fed via a power diode each to the load. As one current limits, the others will take up the load. Try to adjust the volts to be the same on each supply. The voltage will have to be higher by the diode drop. It s not ideal but can work ok for some loads. You may not need diodes actually so...

Another way is to have a 0R05 resistor in line with each supply instead of the diode. The small resistor will help the supplies share the current. The resistor can in fact be same lengths of connecting cable.
But as pointed out above, no more current without inductor, FETs, diodes and caps changed. Probably other hacks too.
 

dendad

Joined Feb 20, 2016
3,810
Oh, I forgot to ask about the batteries.
Have you any charge control implemented? Just sticking Lithium Iron batteries on a high current supply is a great way to make a bomb.
 

Thread Starter

Mod91

Joined Sep 4, 2018
11
So without seeing actual ratings of anything like the transistors and inductor, most likely you'd have to change at least the inductor and hope for the best if you can afford to buy another unit if this one blows out.
If you really want to gamble, change the sense resistor, then stand back as you test it :)
How can determine or calculate to get the right type of induction coil ?
 

Thread Starter

Mod91

Joined Sep 4, 2018
11
Can i ask, if your using batteries, and you require 5 to 6V, why do you need this buck-boost converter anyway?
To use it as a power regulator.

Lets say I want to vape my electronic cigarette at 100 Watts of power and the electronic cigarette's resistor coil (vaporizer) has resistance of 0.2 Ohms then I can dial the voltage to 4.47V and draw 22.36A of current output, theoretically.

Likewise, I could dial the voltage to 4V to draw 20A of current to get 80 Watts of power at the same level of resistance in my vaporizer coil, theoretically.

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Thread Starter

Mod91

Joined Sep 4, 2018
11
You could try 3 of these boards with their outputs all fed via a power diode each to the load. As one current limits, the others will take up the load. Try to adjust the volts to be the same on each supply. The voltage will have to be higher by the diode drop. It s not ideal but can work ok for some loads. You may not need diodes actually so...

Another way is to have a 0R05 resistor in line with each supply instead of the diode. The small resistor will help the supplies share the current. The resistor can in fact be same lengths of connecting cable.
But as pointed out above, no more current without inductor, FETs, diodes and caps changed. Probably other hacks too.
Always wondered how commercially available electronic cigarettes manage to handle these insane levels of current discharge while being amazingly tiny when it comes to physical size.

This DNA 200 electronic cigarette chipset is rated at 200 Watts at up to 50A of continuous current discharge. It has some sort of miracle buck-boost converter, I guess.

dna200.jpg
 

MrAl

Joined Jun 17, 2014
8,124
How can determine or calculate to get the right type of induction coil ?
Hi,

This is for an E-cigarette? I'd be very careful with this as people have gotten their faces blown half off messing with high powered batteries and circuits for their E-cigs. There are currently new laws being considered to regulate that industry better too for various reasons. If you have ever seen a battery or transistor or capacitor blow up then you know how dangerous power supplies and Li-ion batteries can be.

The best way to find out the value the inductor is to find it on the board and see what value it is. That would tell you the value in 'Henries'. You could then look for a part that has a higher current rating.

Sometimes the manufacturer of the part (the controller in this case) provides a reference design or several reference designs. You could also check to see if you could find one for high current like you need. They often provide part numbers for those designs so you might get lucky here. If not, the data sheet may also provide more information about inductor choices. If not, then we have to apply general principles to calculate the value, but you should realize that this would be a 'new' design and new designs are always checked with a scope and several meters and different load values and input voltage ranges to make sure they work properly. If something goes wrong or is not exactly right during the first run usually the transistors blow out and have to be replaced before the next test can start. That means you might need another set of transistors and it's a bit of work to swap them out in most cases too.

Is there any chance you can find a power supply that can already do the current level you need? That would be a much faster and easier road to success. I have a strong feeling that would be a much safer route too in this case because of the type of end application we are talking about. In fact, after learning about the intended use here i would have to strongly recommend this instead of any kind of modification unless you have some industry experience doing these kinds of circuits as they can be very dangerous if not done just right and/or not tested thoroughly.
 
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Thread Starter

Mod91

Joined Sep 4, 2018
11
This is for an E-cigarette?
Yes, as mentioned in my very first post of this thread.

I'd be very careful with this as people have gotten their faces blown half off messing with high powered batteries and circuits for their E-cigs.
Likely due to using lithium-ion batteries that have broken or damaged wrappers or due to using lithium polymer (LiPo) batteries in high power electronic cigarettes. Lithium Polymer (LiPo) batteries are far more unsafe than Li-ion batteries at equivalent power levels. It is the reason why Lithium Polymer (LiPo) batteries are rarely found in commercially sold high power electronic cigarettes.


The best way to find out the value the inductor is to find it on the board and see what value it is. That would tell you the value in 'Henries'. You could then look for a part that has a higher current rating.
What is Weber/ampere ?
How are Weber/ampere, Ampere (current), and Henry (inductance) related ?


Is there any chance you can find a power supply that can already do the current level you need? That would be a much faster and easier road to success. I have a strong feeling that would be a much safer route too in this case because of the type of end application we are talking about.
I have been using for a while a set of 20A max continuous discharge batteries (Samsung 25R) in my commercially sold high power electronic cigarette. My commercially sold electronic cigarette has a max power output of 228 Watts at up to 50A according to the manufacturer's rating.

For my DIY electronic cigarette mod, if I use these same 4 X 20A max continuous discharge batteries in parallel-series to get 120 Watts, each battery will experience only about 12A of current draw, which is far below the max continuous discharge rating of Samsung 25R when the resistor (vaporizer) of my DIY electronic cigarette has 0.2 Ohm of resistance, theoretically.

120.jpg

 

MrAl

Joined Jun 17, 2014
8,124
Hi,

Sorry if i missed that E-cig note.

Inductors are usually specified in Henries (H). 1 Henry is equal to 1 Weber per Ampere.
The inductor will most likely be specified in Henries though, or micro Henries (uH where u=micro).

The relationship is:
v=L*di/dt

where
v is the voltage across the inductor,
L is the inductance,
di/dt is the rate of change of current with time.

In the case of converter circuits, it is convenient to solve for di/dt:
di/dt=v/L

and even for di alone:
di=v*dt/L

and here we see one of the main operating points, where the current increase (di) is dependent on the applied voltage (v) and the inductance value (L) and the time the voltage v is applied (dt).
 
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