# Calculating which capacitor for a project

• JonHumphries
In summary: How to Electronics" by K.C. Pong and I think it might be in there. Thanks for checking!In summary, my daughter and I are building a little robotic arm and I want to check my math before we start ordering parts. I'm calculating that we need a 5400 microfarad capacitor at 5v. Does this sound right?
JonHumphries
My daughter and I are building a little robotic arm and I want to check my math before we start ordering parts.
We are using a piece of flexinal (.11 ohms resistance per inch & 12 inches long, and requires 4amps per cycle to actuate with a cycle time of approximately 14 seconds) and want to use a capacitor to keep it constantly actuated.
I'm calculating that we need a 5400 microfarad capacitor at 5v. Does this sound right?

Thank you :)

No, that doesn't sound right.

You have a resistor of 1.32 ohms and you want to maintain 5.28 volts across it and only refresh the voltage every 14 seconds? (Is that right?)

I did a simulation on this and you would need a much larger capacitor to do this. Something like 30 FARADS. 5000 µF would be fully discharged in about 30 mS.

Thanks for the reply. Can you tell me the formula or program you used to do the calculations?

I use LTSpice which is a free simulator available from www.linear.com.

The formula for time constant is this:
T = C * R
where T is in seconds, C is in Farads and R is in ohms.
This is the time required for the voltage to drop to 36.8% of its original value (assuming a charged capacitor in parallel with a resistor).

So, if you had a 1.32 ohm resistor in parallel with a charged 0.005 Farad capacitor, (5000 µF)
the time required for the voltage to drop to a negligible voltage is about 5 time constants.

So, 5 times T = 5 * 0.005 * 1.32 = 0.033 seconds.

So C= T/R, C= 14(seconds)/1.32(resistance of actuator) = 10.6 Farads?

Yes, but you would not want the voltage to drop to 36.8 % of the fully charged value, (1.94 volts), so it would have to be even higher than that.

However, the charging current for the capacitor becomes very high if the capacitance is so large.

Maybe you could consider using small lead acid or NiMH batteries?

If this gets out of hand or there is another source for this stuff, please let me know :)

What causes the voltage drop?

Also, I was going to build this on a basic stamp which would provide 5v DC but you think I need something that can put out more amps?

JonHumphries said:
If this gets out of hand or there is another source for this stuff, please let me know :)

What causes the voltage drop?

Also, I was going to build this on a basic stamp which would provide 5v DC but you think I need something that can put out more amps?

Maybe you could draw a picture of what you are trying to do. Just draw it on paper and take a well lit and well focussed digital photo of it. Reduce it in size and then attach it to a post.

This is from Wikipedia:
Capacitance is the ability of a body to hold an electrical charge. Capacitance is also a measure of the amount of electrical energy stored (or separated) for a given electric potential. A common form of energy storage device is a parallel-plate capacitor. In a parallel plate capacitor, capacitance is directly proportional to the surface area of the conductor plates and inversely proportional to the separation distance between the plates. If the charges on the plates are +Q and −Q, and V gives the voltage between the plates, then the capacitance is given by

C = Q / V

The SI unit of capacitance is the farad; 1 farad is 1 coulomb per volt.

end quote.

V = Q / C
So, as the charge is removed by discharging the capacitor, the voltage drops.

okay I understand the voltage drop better. As discharge and recharge occur, Q and C are changing. So how did you get to 36.8%?

I attached our plan. Thank you very much for all your help :)

#### Attachments

• Muscle Wire project.jpg
32 KB · Views: 460
Final voltage = Initial voltage / e ^(t/RC).....see textbook for derivation of this.

If t = RC (one time constant) then 1 / e = 1 / 2.71828 = 0.367879 (which is abbreviated to 36.8 %). Then Final voltage = Initial voltage * 0.3678 for one time constant.

However, I'm not sure what is happening with your circuit.

(There seemed to be an extra wire from the bottom switch position to the negative line.)

But it looked like the capacitors would charge up and have no way of discharging. Maybe it is the way you have drawn the switch contacts. Is the electrical path as you have shown?

Do these muscle wires contract by themselves, or do they do it because you apply power at regular intervals? The only reference to Flexinal on Google was your post about it here.

I appreciate your time here, thank you for the formula. I haven't come across that one yet. I'm using "Foundations of Electronic Circuits and Devices" 4th edition which probably has all of this information but I don't always know what I'm looking for. I'll investigate what you've given me. Much appreciated :)

As far as the wiring is concerned the actual SPDT relay I have has two power input posts and two power output posts with a fifth post that actuates which power line is getting connected. This is because it can handle a much higher voltage than what is necessary to actuate it (so that later I can actuate with a pulse from the basic stamp microcontroller). Is this what you mean by an extra wire?

For the capacitors, I couldn't see which side was curved when I was drawing them so they may be backwards. The idea is that they are charged when the SPDT relay is on their path and once charged up, they hit the muscle wire with a pulse, causing it to contract.

Flexinol is made by a company called Dynalloy http://www.dynalloy.com/#Scene_1
I'm using the .020 diameter wire for this project because its what I have lying around. The wires contract when the appropriate current is applied to them. If they get too hot (from applying the current for too long) then they will snap. Here is a link to the tech sheets on them. http://www.dynalloy.com/TechData_US_Units.html

Thanks again! I've got some reading to do :D

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That circuit would give you one pulse when the capacitor charges, but there is no way of discharging the capacitor. So, it would only work once.

If you could make a different choice of wire, there would be advantages. For example, if you chose the 0.006 inch wire, it would have a resistance of 16.8 ohms per 12 inches.

You could switch power to this using your Basic Stamp to provide exact timing.
It would require a simple 2 transistor amplifier to do this and you would need one for each wire.

The current would only be 300 mA from a 5 volt supply. This can be increased by increasing the supply on the right in the diagram below. The specifications suggest a current of 410 mA, so you would need about an 8 volt supply to get this current.

[PLAIN]http://dl.dropbox.com/u/4222062/Darlington.PNG

This is an example of the sort of circuit you would need. The transistor types are fairly optional, although Q2 needs to be a power NPN transistor.

At the left is a circle which would represent the Basic Stamp. (If you are not committed to this, the Picaxe series of chips are much cheaper and use a similar language).
It would produce a single pulse of 1 second duration and this would apply a current to the wire. When the pulse goes away, the current stops.

I have shown the effect of varying the voltage from the Stamp, but you would simply supply a 5 volt pulse to one of the outputs of the Stamp. Only 1.2 mA would be drawn from the Stamp.

You can vary the pulse width with the programming of the Stamp chip.

You would need to take another pulse from a different output to a similar amplifier for another wire.

The 16.8 ohm resistor is your wire.

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vk6kro said:
I use LTSpice which is a free simulator available from www.linear.com.

The formula for time constant is this:
T = C * R
where T is in seconds, C is in Farads and R is in ohms.
This is the time required for the voltage to drop to 36.8% of its original value (assuming a charged capacitor in parallel with a resistor).

So, if you had a 1.32 ohm resistor in parallel with a charged 0.005 Farad capacitor, (5000 µF)
the time required for the voltage to drop to a negligible voltage is about 5 time constants.

So, 5 times T = 5 * 0.005 * 1.32 = 0.033 seconds.

Sorry to resurrect a thread, but.. you say that the time for the voltage to drop to effectively 0 is 5 time constants. Is it 5 time constants for any RC circuit? How did you know it was 5 time constants?

5 time constants means 0.368^5 or 0.0067 times the original voltage.

It is a bit arbitrary, but it means that the voltage has dropped below 1% of the original voltage, so maybe this was the origin of the choice of 5 time constants.

Because a capacitor never finishes discharging, some cutoff point has to be chosen where the remaining voltage doesn't matter any more. Less than 1% is as good as anywhere so that means 5 time constants.

It is applied to the charging of capacitors too. If the capacitor is more than 99% of the charging voltage, you can say it is fully charged and this takes about 5 time constants to achieve.

vk6kro said:
5 time constants means 0.368^5 or 0.0067 times the original voltage.

It is a bit arbitrary, but it means that the voltage has dropped below 1% of the original voltage, so maybe this was the origin of the choice of 5 time constants.

Because a capacitor never finishes discharging, some cutoff point has to be chosen where the remaining voltage doesn't matter any more. Less than 1% is as good as anywhere so that means 5 time constants.

It is applied to the charging of capacitors too. If the capacitor is more than 99% of the charging voltage, you can say it is fully charged and this takes about 5 time constants to achieve.

Ah, ok thanks.

## 1. What is a capacitor and why is it important in a project?

A capacitor is an electronic component that stores electric charge. It is important in a project because it can help regulate voltage, filter out interference, and provide a temporary power supply.

## 2. How do I determine the capacitance needed for my project?

The capacitance needed for a project depends on the specific requirements of the project, such as voltage and frequency. To determine the capacitance, you can use the formula C = Q/V, where C is the capacitance in Farads, Q is the charge in Coulombs, and V is the voltage in Volts.

## 3. How do I choose the right type of capacitor for my project?

The type of capacitor you choose depends on the requirements of your project. Some factors to consider include the type of dielectric material, maximum voltage and frequency, and physical size. It is important to research and consult with an expert to choose the most suitable type for your project.

## 4. What are some common mistakes to avoid when selecting a capacitor for a project?

Some common mistakes to avoid when selecting a capacitor for a project include choosing a capacitor with too low or too high capacitance, not considering the operating temperature and frequency, and not considering the tolerance and leakage current. It is important to carefully read the specifications and consult with an expert to avoid these mistakes.

## 5. Can I use multiple capacitors in a project and how do I calculate the total capacitance?

Yes, you can use multiple capacitors in a project. To calculate the total capacitance, you can use the formula Ctotal = C1 + C2 + ... + Cn, where Ctotal is the total capacitance and C1, C2, etc. are the individual capacitances of each capacitor. It is important to connect the capacitors in either series or parallel, depending on the desired total capacitance.

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