How do you stepup voltage without a transformer or battery cel?


Joined Jan 17, 2007
so a battery cell would not work then?
A battery cell, connected to the appropriate voltage-booster circuit would work just fine. BUT its amp output would be limited by the circuit's efficiency and the battery cell itself. That means that in the end you'd get less available power due to unavoidable losses in efficiency in the circuit.


Joined Apr 28, 2012
There are USB outout boost converter modules.
Various small ICs are employed.
Some can be hacked for higher than 5v output.
Others are limited, like the MCP1640 has a 6.5 or 6v limit.

Charge pumps are used because small SMD ceramics are spot cheap.
Typically TFT modules use them.
MAXIM RS232 ICs have charge pumps inside.

Theres nothing wrong with coils just they cost more (typically).

If you talk 1A range dont even think about a charge pump.

If you need 12v at large currents you should use appropiate battery. Or deal with increased cost.

A coil does exactly what you want. After current flow is stopped, it reverses, and voltage increases as long as possible by the stored energy dumped into the load. If its resistance is high, voltage can increase by a great factor.

Capacitors must be connected in series after charging using semiconductor switches. Charge pumps typically are used upto some 10mA.

Roderick Young

Joined Feb 22, 2015
If you were planning to use a battery as part of the design, that might work, but you would be subject to the limitations of the battery. For example, if you had a 5V supply, and decided to connect 5 "D" cells to it, you'd get about 12 volts, but the batteries would still run down. If using batteries is ok, then might as well just use a 12V battery.

It would also be relatively cheap to just get a 12-volt supply in the first place, instead of trying to convert a 5V one. Just a few dollars on eBay.

Thread Starter


Joined Apr 6, 2016
"but the batteries would still run down" whats puzzleing me is when the batteries are run down will they act as an open circuit, or a short curciut, or would they still influence the voltage level? would they act like a resistor?


Joined Jan 17, 2007
"but the batteries would still run down" whats puzzleing me is when the batteries are run down will they act as an open circuit, or a short curciut, or would they still influence the voltage level?
I am going to assume that you're talking about what happens when the batteries are depleted when they're used with a boost converter. What happens is that the converter can only work with a minimum amount of input voltage (see the link to the module in my previous post, look at its specs), and when a battery starts to wear off, it's voltage goes down. How fast that voltage goes down depends on the type of battery (alkaline, lithium, etc) but they all run out eventually. After the battery reaches the module's low input threshold, the module will normally waver (have an erratic output voltage) or it will simply cut off, depending on its design.
The battery itself will drain current from other batteries in the system if it's installed in parallel with them and those other batteries still have a larger voltage than the one we're talking about. That's why it's normally not a very good idea to use batteries in parallel. A slight variation in voltage between them will result in current flowing from one battery to the other, both trying to reach equilibrium. And there are other issues involved when working with batteries in parallel. But I suggest you take a look at this site, for more information.


Joined Apr 28, 2012
Ohms law and Kirchhoff distribution laws are fundamental to understand eletrical circuits.

Chemical battery is a charge pump. When becoming depleted it looks like the internal resistance increases.
You can observe the battery recovers a little but very soon the lamp etc. goes out again.
You can often inject a little charge by heat or electrical current.
But that will be very limited amount.

Its still useful to know when you only need for instance to use a keyboard for a password.

A deeply discharged lead battery can be tricked with 2 9v batteries in series.
Before, the charger will reject it. After just a minute, theres enough charge the electronic charger will be able to start charging.

Yet you cant recharge or recover disposeable batteries infnitely.
At some stage, it will be finished, the internal structure destroyed.

Heres something weird.

I had a D cell connected to a joule thief. Some day the LED was already out.
Then there was a thunderstorm. They are rare here in Ireland and usually short.
Like an hour or so before, the LED started lighting up a little.
That was really weird!

Then you have things like

A chemical battery of course lives by the difference between its two poles. Over time, the chemicals migrate and corrode the carrier and mix, as well they dry out. When it is all corroded and mixed up and dried out, no charges can be pumped anymore.

Theres really such a thing as a completely dead battery measuring 0v. It means it has chemically decomposed.


no they dont usually become High-z or open circuit. For some time, they will take on a small amount of charge. They are not designed rechargeable, sure. And its not recommended at all to put them in a charger for long time. They will heat up and chemicals may leak. Alkaline may even explode.


Joined Apr 28, 2012
A dcdc converter needs a minimum current capability even to start up. For instance due to its capacitors. It also needs a feed of its quitescent current.

So the dcdc stops working even if the battery is not yet totally dead.


Joined Sep 24, 2015
I'm going out on a limb here - I'm not sure I understand your question, so I will try to answer what I THINK you're asking: I think you are saying you have a 5 volt supply (not battery). If you ADD something like a D Cell battery you'll get 6.5 volts (roughly). You ask what happens when the D Cell battery goes dead. If that's what you're asking then here's my answer:

Your D Cell battery ADDS 1.5 volts to the circuit. When it goes dead then you've lost that additional 1.5 volts. The 5 volt supply will still supply 5 volts but the D Cell will merely conduct the 5 volts through. However, it WILL induce some resistance to the 5 volts. I don't know how much internal resistance it will have so I can't say to what level it would affect your circuit, but it WILL reduce the amount of current. Not voltage, current.

A dead battery doesn't act like a short, nor does it act like an open circuit. In a flashlight, you have two (or more) batteries. They combine their voltage to produce a higher voltage. Typically they go dead at about the same rate, though one may drop a little faster than the other just out of the sheer fact that no two batteries are absolutely identical. Close, maybe, but not identical.

Your circuit will continue to see 5 volts but will lose the 1.5 volts the battery added. And the current will be reduced.

Now: If I totally missed the point - well, sorry. I tried my best.