Help with Modifying DC-DC boost Converter

Thread Starter

JDoozer

Joined Mar 17, 2016
13
Hello,

I recently posted a question concerning a different boost converter. I shopped around and found a similar one (that wouldnt take a month to ship from china). Please see the photo attached.

I have a few questions regarding this boost converter,

I put 2.5-3 volts DC into the converter, and automatically 12 volts dc comes out. I find that this drains electricity too quickly. I was told that you can modify this converter using resistors/ changing the resistors labeled R1 and R2. I would like for the output to only be 5.5-6 volts.

1) How do I modify this boost converter to only output 5.5-6 volts? (please be as detailed as possible, I am so new to circuity that i bought two converters just in case i messed up)
2) Would reducing the output voltage reduce the rate of electricity being consumed by the converter?

Please note that i am a beginner in circuitry. Maybe some of the terms i used were incorrect, I am sorry about that. Any help would be greatly appreciated.

Regards,

J Doozer
 

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Thread Starter

JDoozer

Joined Mar 17, 2016
13
you need to alter the feedback sense, first what is the ic number like Lm/ or Ti..... chip or what?

Im sorry, but... huh? what does LM/ or Ti mean?

This is the part with more information from the website:

http://tinkersphere.com/power/1008-dc-dc-boost-module.html


Specifications:

  • Start Voltage: 1.5V
  • Input Voltage: 2V - 6V DC
  • Output Voltage: 5V - 20V (rated for 12V)
  • Max Power: 6W
  • Maximum Efficiency: 90%
  • No Load Efficiacy: 3 - 15mA
  • Maximum Input Current: 1.5A
  • Absolute Minimum Input Current: 1.5V
  • Start Up Current: 90mA
 
No need. There's a lot you can see by looking at the PCB and deduction.

R1 and R2 are obviously the feedback resistors. You can tell because they've got more numbers on them meaning they're precision resistors (probably 1%). I can read R1, it's marked "3322" which means 33.2K, and R2 is "1873" (I think) which would be 187K. You can even see in the photo that the "bottom" of R1 goes to ground at the right side of the output capacitors, the top of R1/R2 goes to the "FB" pin of the IC (feedback), and the bottom of R2 snakes over to the left side of the capacitors which is the output. Classic voltage reference feedback.

Working backwards, the R1/R2 voltage divider has a ratio of 33.2 / (187 + 33.2) = 0.150, and 0.150 x 12V = 1.8V which is reasonable for a reference voltage. Substituting 56K for the 187K gives you an output voltage of 1.8 x (56 + 32) / 32 = 4.95V. You could replace R1 with a 56K resistor and it would give you about a 5V output. Or solve the equation 1.8 x (R1 + 32.2) / 32.2 = 5.0V for R1 and find a 1% precision resistor value close to it.

Or better yet, bridge R1 with a 100K-200K potentiometer and you can have an adjustable output in the range of the 5V you wanted. If I misread one of the resistors, the above calculations may be off, but with a potentiometer on R1 you'll be able to dial in the voltage that you want.
 

Thread Starter

JDoozer

Joined Mar 17, 2016
13
No need. There's a lot you can see by looking at the PCB and deduction.

R1 and R2 are obviously the feedback resistors. You can tell because they've got more numbers on them meaning they're precision resistors (probably 1%). I can read R1, it's marked "3322" which means 33.2K, and R2 is "1873" (I think) which would be 187K. You can even see in the photo that the "bottom" of R1 goes to ground at the right side of the output capacitors, the top of R1/R2 goes to the "FB" pin of the IC (feedback), and the bottom of R2 snakes over to the left side of the capacitors which is the output. Classic voltage reference feedback.

Working backwards, the R1/R2 voltage divider has a ratio of 33.2 / (187 + 33.2) = 0.150, and 0.150 x 12V = 1.8V which is reasonable for a reference voltage. Substituting 56K for the 187K gives you an output voltage of 1.8 x (56 + 32) / 32 = 4.95V. You could replace R1 with a 56K resistor and it would give you about a 5V output. Or solve the equation 1.8 x (R1 + 32.2) / 32.2 = 5.0V for R1 and find a 1% precision resistor value close to it.

Or better yet, bridge R1 with a 100K-200K potentiometer and you can have an adjustable output in the range of the 5V you wanted. If I misread one of the resistors, the above calculations may be off, but with a potentiometer on R1 you'll be able to dial in the voltage that you want.


Would doing this modification reduce the amount of electricity the entire module drains from the batteries?
 

Alec_t

Joined Sep 17, 2013
11,406
The battery current drain depends on the load you apply to the output.
Voltage out (5V?) x current out = Effficiency (80%?) x battery voltage x battery current.
 

Dodgydave

Joined Jun 22, 2012
9,206
Use a multi turn preset like this, i calculate its around 77.8K resistance for 5v.

Ideally measure the voltage at the junction of the resistors, to see what it is and put the preset across the resistor that is connected to the output voltage.
 

Thread Starter

JDoozer

Joined Mar 17, 2016
13
Use a multi turn preset like this, i calculate its around 77.8K resistance for 5v.

Ideally measure the voltage at the junction of the resistors, to see what it is and put the preset across the resistor that is connected to the output voltage.

Do i have to remove the R1 or R2 resistor entirely?
 
Use a multi turn preset like this, i calculate its around 77.8K resistance for 5v.

Ideally measure the voltage at the junction of the resistors, to see what it is and put the preset across the resistor that is connected to the output voltage.

Dave,

How did you calculate 77.8K? I calculated closer to 56K above, based on back-calculating the reference voltage.
 
JDoozer,

Use a 200K potentiometer in case I've calculated incorrectly. Do not use a slide potentiometer, use a rotary one. Slide pots are too unwieldy. if you're going to mount it, the slide pot is fine. I like Dave's idea of using a multiturn trim pot (small and compact, easier to set the desired voltage).

Connect two wires to it. The potentiometer has (functionally) three terminals. If it has more than that, multiple ones are connected to each other or else they go to the frame. There are only three functional terminals. Two are the end terminals, and one is the "wiper", the changing terminal. Connect one wire to one of the ends, it doesn't matter which, and the other wire to the wiper. Solder the end of one wire to one end of R1, and the other wire to the other end of R1, effectively "bridging" it (going across it) in parallel.
 
12v/5v is a reduction of 2.4, so i reduced the 187k resistor by that, its just a guess better with a multiturn pot.
Nice try but there's another term in the calculation. The ratio of a voltage divider is R1 / (R1+R2). It doesn't change by either resistor directly proportionally. If you rearrange the terms to be a multiple of the reference voltage, the equation for the output voltage is Vref * (R1 + R2) / R1. If you simplify the terms, the (R1 + R2) / R1 can simplify to R1/R1 + R2/R1, or 1 + R2/R1. So there's an extra term in there, the "1+", that changes the proportionality. If you substitute in the full equation, you get:

Vout = Vref * (1 + R2/R1)
or expanding it to:
Vout = Vref + Vref * R2/R1.

(NOTE: WRONG EQUATIONS. SEE POST #21)

There's that extra term again -- "Vref +". So the output voltage isn't directly proportional to the resistor divider.

That's why I back-calculated the Vref from the original voltage and resistors, and then used the calculated Vref to calculate the new resistor for 5V.
 
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The battery current drain depends on the load you apply to the output.
Voltage out (5V?) x current out = Effficiency (80%?) x battery voltage x battery current.
JDoozer,

That was a good answer from Alec. I would just add as an overview that power (less efficiency) is preserved on both sides of the power converter because of the use of the inductor to transform the voltage. So if the voltage is going up from 3V to 5V, the available current will go down on the 5V side.

Or in other words, whatever current you draw at 5V will draw proportionally more current from the 3V battery supply. And that proportion is expressed in Alec's equation above, proportional to the ratio of the voltages (3/5) and the efficiency (~80%). That gives you the reduction in available current going from the 3V side to the 5V side, and you invert it to get the proportionally increased current draw from the battery.
 

Dodgydave

Joined Jun 22, 2012
9,206
It would be interesting to see what the minimum output voltage will be if the ref feedback pin is shorted to the output.
 
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Thread Starter

JDoozer

Joined Mar 17, 2016
13
Thank you all very much. This was very helpful. Unfortunately, i cant move forward until the potentiometer is delivered (from the UK) in about 2 weeks. Currently, if i understand this conversation correctly, i need to remove R2? from the converter so that i can replace is with the potentiometer.

Regards,

JDoozer
 
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