How many Farads?

Thread Starter

Lightium

Joined Jun 6, 2012
196
I calculated 4 amps per sec for 6 sec equals 24 coulombs.
And then 24 coulombs divided by 12 volts equals 2 farads.

Did I do anything wrong? When I simulate with LTspice I get less current then what the math says.
 

WBahn

Joined Mar 31, 2012
30,293
4 amps per sec would mean a current that is increasing at the rate of an additional 4 A every second.

Did you really mean this, or did you mean just 4 A for 6 s?

Where is this current coming from and going to?

What is setting it at 4 A?

It would be very helpful if you at least provided a schematic, and even better if you attach your LTSpice file.
 

Thread Starter

Lightium

Joined Jun 6, 2012
196
I am trying to replace an SLA with a super capacitor. I only need it to run at 12 volts 4 amps for six seconds, buy my simulation is wack compared with the math.
 

ErnieM

Joined Apr 24, 2011
8,382
OK, let's start with some fundamentals. The current to voltage relationship for a DC current is:

Current = I = Capacitance * (change in voltage) / (change in time)

I will note (change in voltage) as dV, and the change in time as dT, or:

I = Capacitance * dV / dT Solve for Capacitance:

Capacitance = I * dT / dV (THE equation)

You state you have a 12V system drawing 4 amps for 6 seconds. Placing these numbers into THE equation yeilds:

Capacitance = 4 amps * 6 seconds / 12 volts = 2 farads

BUT....

12 volts is what the cap starts with, and it ends with zero. You need to select a smaller dV to leave a working voltage on the cap to run your circuit after the time expires.

So I'm going to guess your circuit will run just fine at 10 volts. With this value we get:

Capacitance = 4 amps * 6 seconds / 2 volts = 12 farads

Yikes! That's a lot of farads. You may want to rethink this project.
 

Thread Starter

Lightium

Joined Jun 6, 2012
196
OK, let's start with some fundamentals. The current to voltage relationship for a DC current is:

Current = I = Capacitance * (change in voltage) / (change in time)

I will note (change in voltage) as dV, and the change in time as dT, or:

I = Capacitance * dV / dT Solve for Capacitance:

Capacitance = I * dT / dV (THE equation)

You state you have a 12V system drawing 4 amps for 6 seconds. Placing these numbers into THE equation yeilds:

Capacitance = 4 amps * 6 seconds / 12 volts = 2 farads

BUT....

12 volts is what the cap starts with, and it ends with zero. You need to select a smaller dV to leave a working voltage on the cap to run your circuit after the time expires.

So I'm going to guess your circuit will run just fine at 10 volts. With this value we get:

Capacitance = 4 amps * 6 seconds / 2 volts = 12 farads

Yikes! That's a lot of farads. You may want to rethink this project.

The math looks deep, thank you.
 

ErnieM

Joined Apr 24, 2011
8,382
This is calculus.
Well, since neither dV nor dT have been defined as function, but in fact as constants, there is no need for calculus here. It is a nice special case.

Also there is no need to scare the OP away from gaining a bit more understanding of his specific issue and also some general knowledge.
 

BobTPH

Joined Jun 5, 2013
9,264
I didn’t write the formula. I was responding to the TS’s comment that “The math looks deep.” Why are you insisting that we should not call a differential equation calculus?

If you wanted to do it without calculus, you should have used:

ΔV / Δt
 

Thread Starter

Lightium

Joined Jun 6, 2012
196
4 amps per sec would mean a current that is increasing at the rate of an additional 4 A every second.

Did you really mean this, or did you mean just 4 A for 6 s?

Where is this current coming from and going to?

What is setting it at 4 A?

It would be very helpful if you at least provided a schematic, and even better if you attach your LTSpice file.
Your right about that. I misspoke, I meant 4 Amps for 6 sec.
 

k1ng 1337

Joined Sep 11, 2020
1,014
Your right about that. I misspoke, I meant 4 Amps for 6 sec.
This simulation shows a 15F capacitor charged to 12V sourcing an active load of 4A. This assumes the load is able to compensate for the constantly falling supply voltage by drawing a steady 4A. After 6 seconds, the final capacitor voltage is 10.4V.

If you need more headroom, increase the starting voltage or the capacitance as both have the same effect. Post your calculations so we can figure out where you went wrong.

Untitled.png
 

BobTPH

Joined Jun 5, 2013
9,264
Between the load and the capacitor is a DC to DC converter. I am trying to simulate an Ion Thruster rocket with a 6 second thrust duration.
And you didn't tell us that?

Boost, buck or buck boost? What input and output ranges? Constant current or voltage?
 
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