Adjustable Voltage Step-up (0.7-5.5V to 2.7-5.5V)

Introduction:
A voltage step-up is a circuit that increases the voltage. It can be AC/AC, AC/DC, DC/AC or DC/DC.
This voltage step-up is a DC/DC adjustable voltage regulator. Usually a voltage regulator is fed by a higher input than output voltage, for example 9V IN to 5V OUT. This circuit will take a low voltage (down to 0.7V) and step it up to adjustable 2.7-5.5V. Since it is a regulator, the output voltage will stay constant regardless input voltage (0.7-5.5V), as long as output voltage is higher than input. It cannot step-down, only step-up. Are there any ICs that can do both?

I build it for usage in this project:
http://www.instructables.com/id/Battery-Charger-Powered-by-Fire/

Applications:
This is a typical circuit for a battery-powered USB-charger, for example two AA batteries (3V) to power 5V USB. There are tons of DIYs how to create that. They are often hard-specified to 5V output power. This construction can be used in a range of other applications. Many electronic devices work within 3-5 V and often you want to power them by low voltage power sources.

Some input examples:
• Solar panels
• Generators
• Batteries
• Peltier elements (TEC/TEG modules)
Some output examples:
• ICs
• Displays
• Motors
• LEDs
• Battery/USB charger
Costs:
All components cost me about 7€ but it's totally possible to find cheaper components!

If you want to construct this one, then continue reading!

Next Step: Specifications
Remove these ads by Signing Up
 1-40 of 43 Next »
arjunmenon1 month ago

You mentioned -

Since it is a regulator, the output voltage will stay constant regardless input voltage (0.7-5.5V), as long as output voltage is higher than input.

Shouldn't it be that "as long as the input voltage is higher than output"?

By going what you mentioned, so if I have a 2.4V battery cell, I can power a 5V load, even if the cell drops to 1.8V after sometime? It will still output 5V and have the load running.

Thanks.

arjunmenon1 month ago

You mentioned -

Since it is a regulator, the output voltage will stay constant regardless input voltage (0.7-5.5V), as long as output voltage is higher than input.

Shouldn't it be that "as long as the input voltage is higher than output"?

By going what you mentioned, so if I have a 2.4V battery cell, I can power a 5V load, even if the cell drops to 1.8V after sometime? It will still output 5V and have the load running.

Thanks.

Joohansson (author)  arjunmenon1 month ago
A better description is "as long as input is lower than output". 1.8V should be fine as long as the output current is not too high. I think I have a graph you can look at or it was in one of my other instructables.
Fragmaster1 month ago

If I needed to supply up to 1A output at 5.5V and a supply (battery) of 3V minimum, would this circuit be able to handle it? the Efficiency graph only goes up to 500mA, so I think I would need another regulator to do the job. Do you know of a variable output regulator that would work?

shina-ct2 months ago

i constructed the circuit on a breadboard, but on testing, it only produces 0.8V output for an input range of 1-6V. Please how do I know if the issue is with the connection or if the ic is damaged. Thank you.

Joohansson (author)  shina-ct2 months ago
Hard to tell. I will have to do a verification of the circuit. I will write here again when/if I have time to do that.
2 months ago
Ok. Thank you.
idrees.rasheed10 months ago

Hi, I am trying out this project. I was able to find all the parts except the Max757C. Please help me as to what to do? I was able to find 756C. What should be the circuit changes and is there any equivalent IC. Please help me. I really need more information. Thank you in advance.

Thom Kouwen11 months ago
You might have already found a converter that can boost, and buck. In case you haven't, you should take a look at this: http://en.wikipedia.org/wiki/Single-ended_primary-inductor_converter . It's called a SEPIC converter, it can even go from buck to boost, or vica versa, without an interuption in the output power, you won't even notice when it crosses the input voltage.

Thom
Joohansson (author)  Thom Kouwen11 months ago
Thanks, I might try that circuit some day =) Looks easy to build.
lraggio11 months ago
Hello! Thanks for this instructables :)
I've a question: Why on the scheme on pin 6-2 there's a P1 and R1 but on the "shopping list" and the board scheme there's also P1R1 and a P1R2?
Thanks!
Joohansson (author)  lraggio11 months ago
Hi,
I had to think a bit and now I remember! It´s actually in the description in some way. In the layout i have simplified P1 to make the drawing easier to understand. But P1 in the drawing consists of P1+P1R1+P1R2 in serie. P1R1+P1R2=57Kohm and P1=100Kohm. That create the correct interval of 57-157kΩ instead of 0-100Kohm if you just use P1.

That make sense?
Samuel kos1 year ago
Joo where did you happen to find theses maxim 757 ICs I cannot find them anywhere
Joohansson (author)  Samuel kos1 year ago
https://www.elfa.se/elfa3~se_sv/elfa/init.do?item=73-076-31&toc=0&q=max757
1 year ago
Thank you very much for the link! ( is it in German?)also would the product ship from Germany or where ever it comes from ?
Joohansson (author)  Samuel kos1 year ago
It was Swedish. You can choose other countries:
smartablet1 year ago
Hello, I appreciate your contributions but my question is, it possible to build a DC-DC step-up with input voltage of 6V/ 2Amh to out-put of 12V/7Amh if your answer is yes, email me kingjoejoe29@yahoo
jimvandamme1 year ago
You're exceeding the maximum rating on the boost switch output (pin 8, LX). Somewhere between 5 and 10,000 volts, it will fry. At 5.5V out, you're probably OK but if somebody wanted higher voltage, it would be cool to put an extra stage in there to take the extra voltage so you could get more out.
Hi, if you want to make it a tiny bit more effective for low output current, replace the 1N5817 diode for some other with lower junction capacitance (such as 1N4937 or 1N60).
Generally:
Standard silicon diodes are great for non-critical use. (usable as HF varactors, low Q)
Schottky diodes are good for use as high power rectifiers and varactors (they're great, but have a bit higher ESR than serious varactors, so do zener diodes).
Germanium point-contact diodes are great for RF equipment and low-drop rectifiers. Low capacitance. Not very good as varactors cos' they have high parasitic reverse current.
1 year ago
not quite...
this is not an RF application and forward current is simply too large for an envelope detector diode like 1N60 even for a very light load. this is SMPS step up circuit and diode here has to handle large current spikes. the IC used in this converter can operate up some 500kHz and actual operating frequency is likely lower. therefore effects of capacitance are negligible, specially when 1N5817 has capacitance of only some 110pF. this means that at these frequencies, reverse bias impedance is in the kOhms - large enoguh to be neglected in a circuit operating at mere few volts (loss is V^2/R or up to some 10mW). even if 1N60 was perfect, improvement here would be at bet those 10mW. much greater benefits in terms of efficiency are due differences in diode forward voltage drop. diode here has to handle large current spikes. since 1N60 datasheet does not show characteristis above 500mA, let's assume that current is something smallish, say 300mA. At 300mA 1N60 has massive forward voltage drop of 0.7V, 1N4937 is even worse at 0.9V and recommended 1N5817 has only 0.35V. power loss here is ~210mW, 270mW and 105mW so among those options, 1N5817 wins hands down even at such low current (in real use differences are even greater). This explains why 1N5817 is used in the reference circuits in the datasheet of many SMPSs including MAX757.

Comparison characteristics of voltage-current characteristics:
http://www.use.com/da353f27d64e1c3bf83e
1 year ago
Well, I am aware of the rather low current capabilities of the aforementioned diodes. By saying "for low current applications" I ment current in the order of units of milliamperes, not hundreds. I am saying this from my experiences from making many dc/dc converters for old 30V varactors.
Look at what I mean here: http://fhs-consulting.com/aa1tj/Diamp.html
1 year ago
varactors operate in reverse bias and practically draw no current. here we have a DC/D converter that is operating as a battery charger. what would be exact conditions where efficiency 1N60 is better? (input and output voltage, output current)
billgeo1 year ago
Very nice project as an exercise!
But it must be noted that you can get a better made and cheaper step-up regulator
from eBay (china), and there is also a constant current version
Joohansson (author)  billgeo1 year ago
Thanks, yes there are lots of alternatives regarding the step-up, both better and worse. Key aspect though was that I wanted to build everything myself.
1 year ago
Yea, it is great to design and built this things!!!
Its a great DIY exercise!

keep up the good work man!
Edgar1 year ago
Nice!
Went to my Blog:
ross_valusoft1 year ago
You have not shown the P1 wiper connection.
Joohansson (author)  ross_valusoft1 year ago
Could you please explain a bit more what you mean?
1 year ago
The pot (P1) has 3 connections. The middle one is called the wiper. Your circuit diagram shows that the wiper is not connected to any other part of the circuit. It is open circuited.Only the two ends of the pot are connected. Therefore turning the pot has no effect on the output voltage according to the circuit diagram because the value of P1 is a constant.
Joohansson (author)  ross_valusoft1 year ago
Ahh now I understand, I looked at the wrong picture (the component placement). The layout is changed, thanks!
tmorsch1 year ago
Looks like a very useful instructable to me.
I just wish i knew how to solder.
:-(
TwinMustang1 year ago
In theory, reversing which side of the transformer (inductor) has the feed power would create a step down from the same unit.

At least, it should. I've never tried it, but the principle is sound.
Joohansson (author)  TwinMustang1 year ago
Interesting.. I wonder if that works with this unit.
Moabyte1 year ago
You've got the wrong two of three terminals of P1 connected. As shown it is not a variable resistor between the two connected terminals.
Joohansson (author)  Moabyte1 year ago
I´m sorry but I cannot find what´s wrong. Maybe it depends on what potentiometer you use. My P1 use the middle pin as variable and I only use one of the other pins. I don´t need the third pin.
aking141 year ago
Search LM2577 on ebay, sort by price lowest to highest, and look around the \$12-15 area and you'll get 5 at <\$3 each..
Like \$13.11 shipped for 5 from czb(xxxx)..