# What capacitor do I need for 20V, 10A for only 300µs?

1. Jul 1, 2015

### Tomas1337

Im really a beginner when it comes to designing electrical circuits and only have the ability to replicate cicuit designs. I have no idea how to choose and power up a capacitor the way I want it.

So what I need is a capacitor than I can charge up using a low voltage (around 5V) that can then supply with 20V with a device sucking up about 10A but only need to sustain it for 300µs.

In this equation though, isnt the delta time the time it takes to charge it up? And is my Vs to be 5V or 0V?
I plan to supply the source the power to the capacitor by using a MOSFET which activates by getting a signal from a micro controller. The micro controller only activates the mosfet for the time needed for the capacitor to charge up? Am I making any sense?

Also, other considerations in helping me design this would be awesome.

2. Jul 1, 2015

### Staff: Mentor

You are planning to charge the capacitor from a 5V source. So in a short period of time you can bring the capacitor's voltage to 5V.

I don't understand where the 20V fits in. Please explain that.

3. Jul 1, 2015

### Tomas1337

I want to be able to charge the capacitor to 20V using a power supply of 5V.

4. Jul 1, 2015

### Staff: Mentor

Perhaps you'd better first explain what circuit or device is going to be powered by 20V and needing 10A for a short interval, so we can decide what is the best way to approach this.

5. Jul 1, 2015

### Tomas1337

The device is an electropermanent magnet. To switch it on and off, I need to run a coil with that power and it only takes a switching time of around that interval.

6. Jul 1, 2015

### phyzguy

Perhaps the first question you need to think about is how you plan to get 20V from a 5V supply. What do you have in mind? Some type of charge pumping scheme? An inverter-transformer-rectfier scheme? Something else?

A second question is how far the 20V can decay by the end of the 300 uS interval. If you have a capacitor charged to 20V, and you start sourcing current from it, the voltage will decay. How high does the voltage need to be at the end of the 300 uS? 10V?, 15V?, 19V?, 19.9V?..... This will determine how large the capacitor needs to be.

7. Jul 1, 2015

### Tomas1337

Hi Phyzguy!

At first I was thinking of using just a simple DC-DC Boost converter with a Inductor and a PWM signal from the Micro Controller (Arduino). But now I dont think I can do that because of the 10Amp requirement. I plan to source the power from a typical lithium ion battery which is just rated for low amps.
So yeah, I guess my first question should have been, Is there anyway I can generate the 20V and 10Amps using a small battery of 5V at around 2500mAh. I know this typically breaks the law of Conversation of energy but the trade off is I only need this power for 300uS and the next time I'll need it is in the next minute or so. I was hoping to use that break time to charge the capacitor back up to what I need it to be.

The decay during the 300uS from the research paper im copying showed that it only had a decay to about 14V at the end of the interval.

8. Jul 1, 2015

### the_emi_guy

Don't worry, you are not breaking any conservation laws.

Check out the Linear Technology LT3751

9. Jul 1, 2015

Break this into two problems - the size of the cap needed, and how to charge it.

I'll bet a small inductor, diode and a switch -- you could do it manually! - just not very precisely!

10. Jul 1, 2015

### Jeff Rosenbury

You might look up RC time constant. It will help you figure the capacitor's size.

Is this to run a relay? Relays can be tricky and you should read some app notes, possibly these.

Consider using a IGBT transistor switch for your 10A. You should be able to switch it from the arduino.

11. Jul 1, 2015

### phyzguy

The DC-DC Boost converter should work OK. You don't need it to source 10A. If you need 10A for 300 us once per minute, the average current is only
10 Amp * 300E-6 sec/ 60 sec= 50 microamps. Given that you want the capacitor to source 10A and discharge from 20V to 14V in 300us, you should be able to figure out how big a capacitor you need. Give it a try.

12. Jul 1, 2015

### Staff: Mentor

Before making decisions about the power arrangement, it would be prudent to check the figures you are using. You have a many-turn solenoid wound on a magnetic core? Applying 20V to it for 0.3msec is not a problem; that's easy. Whether the current is going to even reach the level you are hoping for is a separate matter altogether. Do you know some electrical characteristics of that electromagnet?

You seem to be referring to a published design, what is the URL where we can read further?

13. Jul 2, 2015

### Tomas1337

Hey, the magnet assembly is not an electromagnet but an electro permanent magnet. Here is the paper title, you might find it interesting: Electropermanent Magnetic Connectors and Actuators: Devices and Their Application in Programmable Matter

14. Jul 2, 2015

### Tomas1337

Thanks! In Choosing a Capacitor I tried some initial computations and have come up with using a 300uF Capacitor with a 2 Ohm Resistor with an initial voltage supply of 20V. (Its really gonna be 5V but for the sake of computing a capacitor I used 20V).

If im right this will give the capacitor 10A and will be able to discharge 20V to 19.9V in 300uS (which is fine). Is it as easy as this for choosing the capacitor? This doest feel right. From these values, I also computed that it will fully charge the capacitor in 3mS. Will it really be able to get that much current in 3mS?

Time to move on to figuring out how to supply it with 20V from 5V?

15. Jul 2, 2015

### meBigGuy

If you draw constant current from a capacitor, the voltage drop linearly.

The basic capacitor equation you want is I = C dv/dt where dv is the change in voltage, dt is the time you discharge for, and I is the discharge current. Oh, and C is the capacitance.

For example, let's say it is OK to decay to 15V in the 300uS, and that I will remain constant at 10A (which isn't realistic since the voltage is dropping, but this gets us close in a simple way).

10A = C * 5V/300us so C = 10A * 300us/5V = 600uF

If you can only drop 2.5V, you need twice the capacitance.

Again, that is for constant current. If you are discharging into a constant resistance, then the equation is different since the current drain gets smaller as the voltage drops.

The voltage decay into a resistor is exponential, which you can read about here.
https://en.wikipedia.org/wiki/RC_circuit

Reverse the above equations to determine the charge times.

16. Jul 2, 2015

### meBigGuy

Here is a PDIP part that can do 65ma at 20V http://www.ti.com/lit/ds/symlink/tl499a.pdf
Or, are surface mount parts OK?

You never said how fast you really need to charge it. If you need more current, you may need to go to discrete switches.

17. Jul 2, 2015

### Tomas1337

Hey! Thanks for the help, really. I all humbled by you guys.

The time it takes to charge it is not really important application wise because once I release that charge, i will only need it again in about 1 min or more. But from the equations isn't the discharge time somewhat related to the charge time? So if I need the discharge time to be about 300uS my charge time will be about 6mS?

Also, theres something that has been unclear to me about capacitors and current and Ive been looking everywhere for an explanation. How do you make that 65mA turn into 10A? Do you just let it flow to the capacitor until it reaches 10A? If so, how do you control it?

18. Jul 2, 2015

### DrZoidberg

Imagine you have a big water tank at a height of 5m above the ground and a bucket placed at 20m above the ground. Then you have a pump that gets water from the tank and puts it into the bucket at a rate of 65ml per second. After 100s you will have 6.5 liters of water in the bucket. If you then empty the bucket within 0.65s that gives you a flow rate of 10l per second.

19. Jul 3, 2015

### phyzguy

The discharge time and the charge time are totally unrelated. Following up on the water analogy, the time you take to fill the bucket is completely independent of the time you take to empty it. You can fill it with a slow trickle that will take a whole day to fill it up and then empty it in a fraction of a second. It's the same with the capacitor. The charge time is dependent on the current you put in, and the discharge time is dependent on the current you pull out. You can make these be pretty much whatever you want. You control it with some sort of switch ( a relay or a semiconductor switch) that switches the capacitor from the charging circuit to the discharging circuit.

20. Jul 3, 2015

### meBigGuy

The water analogies are good. Fill it with a drip over 2 days, and then turn it over and empty it almost instantly.

If you look at this capacitor equation it relates the rate of change in voltage (dv/dv, volts/second) to the current I for a given capacitance C. If I is small, V changes slowly If I is large, V changes rapidly.

I = C dv/dt OR dv/dt = I/C (So, the bigger C is for a given current, the slower the voltage changes)

The other equation that help make this clear is Q = CV. The charge Q (in coulombs, or 6.24 x 10^18 electrons) on a capacitor is proportional to the voltage. You can pile the charge up slowly, and dump it as fast as you want. The faster you dump it, the higher the current had to be (since current is the flow of charge where 1 amp = 1 coulomb per second.

You can calculate how fast 50ma will recharge the capacitor just by a simple ratio. 300uS * 10A = X * 50ma so X = 60ms.
(10A to 50ma is a ratio of 200 so it will take 200 times longer to recharge than it took to discharge, given constant currents)

I should be clear that this is all based on constant currents, not constant resistances. But I think you understand how to deal with the exponential decays associated with RC circuits.

The TL499A is a neat part for your boost since it has internal switches and comes in a DIP package. But there are LOTS of boost regulator controllers to choose from.