# Getting more voltage from a battery

#### Mahmoud Athab

Joined Feb 28, 2016
4
Hi,
Is there a way to get more voltage from a battery? say 200 Volts from a 20 V battery or so. I know this question seems inconsistent but I only need the voltage for a few milliseconds. As in storing a huge charge and then discharging it very quickly to activate a solenoid.

#### WBahn

Joined Mar 31, 2012
30,293
Use a DC-DC converter.

#### #12

Joined Nov 30, 2010
18,224
quickly to activate a solenoid.
More likely you want to quickly activate a car thief.

Anyway, use the battery to run a booster of some sort. Start with an oscillator and use the inductance of a coil to get an inductive kick. Use that to charge a capacitor and monitor the voltage to turn the oscillator off when it gets to 200 volts.

#### Mahmoud Athab

Joined Feb 28, 2016
4
More likely you want to quickly activate a car thief.

Anyway, use the battery to run a booster of some sort. Start with an oscillator and use the inductance of a coil to get an inductive kick. Use that to charge a capacitor and monitor the voltage to turn the oscillator off when it gets to 200 volts.
lol, actually it's for the robocup competition since we need a solenoid to kick the ball toward a direction.

I could use something really small in size, have you got any references that I could read? anything will be helpful.

#### Alec_t

Joined Sep 17, 2013
14,407
I could use something really small in size
Unfortunately you are likely going to need a large capacitor to store enough charge to power the solenoid. What is the solenoid spec? Do the maths to see how much energy each kick will require.
Personally, I'd be looking to use a DC motor mechanism rather than a solenoid.

#### #12

Joined Nov 30, 2010
18,224
I'm not on the, "unfortunately" team, but I have to work today. Publish the specs on the solenoid and see who bites.

#### AnalogKid

Joined Aug 1, 2013
11,196
Depending on the solenoid, you might get away with a simple inductive kicker similar to what is in an ignition system.

ak

#### ian field

Joined Oct 27, 2012
6,536
lol, actually it's for the robocup competition since we need a solenoid to kick the ball toward a direction.

I could use something really small in size, have you got any references that I could read? anything will be helpful.

You could pinch the photoflash PCB out of a disposable camera.

Most common types use a single AA cell, the secondary is grounded to the base of the oscillator so the charging current augments the feedback pulses - it works hard into a discharged capacitor to charge it fast, but throttles back as the voltage comes up. Good battery economy, but kept topped up ready to go.

#### Tonyr1084

Joined Sep 24, 2015
7,991
Here's a solution for consideration. However, you will need to convert your DC into alternating current for this to work. It's a voltage quadrupler. Throw in a transformer and you can reach even higher voltages. However, it doesn't provide significant current, so you'll have to take the voltage and store it in a large capacitor then give your solenoid a sudden kick in the bricks to make it boot the ball. I've never built one of these, but I have the circuit laying around for a while.

#### Mahmoud Athab

Joined Feb 28, 2016
4
Hi guys and thank you for the replies. my teammates finally provided me with the specs: right now they are using a 7.5 Volt Lithium Ion laptop battery, and we need 40 Volts and 25 Amps for the solenoid. We are flexible on the battery but it needs to be relatively small. I know that we might have to change the design and use something else instead of the solenoid, but I really hope that this would be or worst case scenario.

#### Papabravo

Joined Feb 24, 2006
21,305
Here's the thing. When going from a lower voltage to a higher voltage you lose power. What does this mean?
The output power you require is:

$$40 \; \text Volts \times \; 25 \text Amps = 1000 \text Watts$$

If your conversion process was 80% efficient, then the required input power is 1250 Watts.
So:

$$\frac{1250 \; \text Watts}{7.5 \; \text Volts} \approx 167 \; \text Amps$$

At that rate your batteries will be rapidly drained and rendered useless.

#### Mahmoud Athab

Joined Feb 28, 2016
4
Here's the thing. When going from a lower voltage to a higher voltage you lose power. What does this mean?
The output power you require is:

$$40 \; \text Volts \times \; 25 \text Amps = 1000 \text Watts$$

If your conversion process was 80% efficient, then the required input power is 1250 Watts.
So:

$$\frac{1250 \; \text Watts}{7.5 \; \text Volts} \approx 167 \; \text Amps$$

At that rate your batteries will be rapidly drained and rendered useless.
Woooo! does this also happen when drawing this kind of power for only a few milliseconds?

#### AnalogKid

Joined Aug 1, 2013
11,196
1000 watts for only "a few" milliseconds is not the same as a continuous load. The total energy needed for 5 ms activation is 5 ws, a very reasonable amount that can be achieved with a beefed up version of the camera strobe approach mentioned above. The two big questions are 1) how often does the circuit need to discharge? 2) What storage device can deliver 25 A peaks?

If we start with a capacitor, and say arbitrarily that the output voltage can sag from 40 V to 35 V over 5 ms but the current remains constant, then solving the capacitor equation for a constant current yields this:

ec=it --> Voltage times capacitance = current times time. Rearranging:

C = (25 x .oo5) / 5 = 25,000 uF, not an outrageous number. A 75 V cap will last longer than a 50 V cap, but also has a higher output impedance that might attenuate the peak current available or reduce the risetime of the pulse. BTW, what are you using to switch the output pulse? Does the output need a shaped falltime, or can the cap just discharge into the solenoid until it is discharged?

ak

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#### Alec_t

Joined Sep 17, 2013
14,407
If the solenoid has considerable inductance (as is likely), the current won't immediately rise to 25A. You might need to factor that in.

#### Bernard

Joined Aug 7, 2008
5,784
Might give a passing look under Completed Projects; Wendy's Completed Projects Index; Solenoids; I Made a Solenoid.

#### Bernard

Joined Aug 7, 2008
5,784
On ebay I found several DC-DC convertors covering 6 V & up input to 80 V out,& 5V- 52V out.

#### MrSoftware

Joined Oct 29, 2013
2,212
Why not grab a few LiPo batteries used for flying toys. Fully charged is 4.2v/cell, discharged is about 3.7v/cell, so a 5s battery will give you a peak of 21v, two in series give you from 37v to 42v with hundreds of continuous amps on tap, and they're relatively light weight. You can use an efficient circuit to reduce voltage for your control circuits. You'll never be waiting for voltage to build up. And if you run your motors on a higher voltage, they would potentially be more efficient since you would have less heat loss to current.

#### Tonyr1084

Joined Sep 24, 2015
7,991
I'm starting to think you want to power a spacecraft using just one 9V battery. You're really looking for too much power out of too weak a source. And in converting that source to a higher potential does come at a cost. Depending on the materials you're using, you can see anywhere from a few percentage points loss all the way up to 20, 30 or even as much as 50% loss.

Nothing comes for free, and in trying to up the power source, you often give away more than you get in return.

Why not take one of those muscle stimulators (I think they're called "Tem's machines") and use that to charge a capacitor. Then see what kind of voltages you can store and how much power you can dump in an instant. It's kind of like a taser, where you use a couple AAA batteries to deliver a numbing 50KV shock. You can GET the voltage up there, but the POWER won't be there.