@WBahn
I'm going out on a limb, is this accurate?

V/L = Icc (seconds)

where V = (ΔV) Total voltage change
Icc = Constant Current (in Amps)
L = Inductance (in Henrys)

???

No.

First off, it is not dimensionally consistent, so we KNOW it is wrong.

V/L has units of amperes/second, while Icc·(seconds) has units of amperes·seconds.

The constitutive equation for an ideal linear inductor is

V = L di/dt

If you have a constant current, then di/dt is identically zero. So if you have a constant current in an ideal inductor, then the voltage across the inductor is zero.

If you have a constant voltage, then you also have a constant di/dt, meaning that the current is changing at a steady rate.

V/L = di/dt

Integrating both sides over a time period of ΔT therefore yields

V·ΔT/L = Io + ΔI

 

Icc(seconds) = C V

So my interpretation is:
Icc = Constant current (in amps)
C = Capacitance (in Farads)
V = ΔV (total voltage change)

Did I interpret that right?

Yes. And, (seconds) is the time it takes for the voltage across the cap to change by V volts (delta-V). It is one equation with four variables. To solve for any one of them, you have to know the other three. I use this to ballpark R-C timing ramps. Obviously it is not as accurate as the exponential equation, but it is math you can do in your head and is good for about 10% error.

ak

 

Math Check

Note: I have been meaning to post this much sooner, but I have been quite sick over the last week. So, because of my Illness I am typing this off line to avoid more confusion, especially from my side, a common problem, my abilities with math have been diminished, hence this post.


I am having trouble reconciling the formula Joey gave in post #9. Where he stated:

For some reason, it has always been easiest for me to remember:


It=CV


Rearrange as necessary.


This only works for constant I.

I have been given the following formulas in this post:

The math comes directly from the defining relation for the capacitor.


Q = CV


I = dq/dt = C dv/dt


So


dv / dt = I/C


Note that C has units of farads, which are coulombs/volt. I has units of amperes, which are coulombs/sec. So dividing current by capacitance leaves you with units of volts/sec (or, in general, voltage/time).


If you want to relate this to a fraction of a particular current (Vcc or any other), then you just need to multiply this by 100%/Vcc and you get


d%/dt = (100%·I)/(C·Vcc)

So to handle this I reached way back to my Algebra two days an started doing the math in the form of a proof.


c = Coulomb or 6.2415 × 10E18 electrons (This is actually unimportant for the use of the calculations, the point being that Coulomb is an actual quantity of charge, for the gross math the actual number is unimportant other than to know it exists.)

F = Farads
t = 1 second
d = Δ, Delta, one of my confusions (and I suspect it is for many people) is that it is actually a variable and can be treated as such for purposes of algebraic manipulation, I will use d in its place.
Note that unless specified all units are 1.


So it’s given that:


C = 1 Coulomb

A = Constant Current (in this case)

t = 1 sec


The SI definition of an Amp:

A = Ct

1C=1A/sec (One of the Basic definitions of Coulomb)

or

A= C/t Formula 1


Capacitance is

F = V/C Formula 2

Somewhere in the two assumptions above I may be screwing up. Variable names are a bear, as C is frequently used for Capacitance, but here it is coulombs. This is why I am trying to be extra careful with my definitions. It can be very confusing. So starting with:



A = C/t Formula 1

F = V/(A sec)


Joey999 said (I think.). Using my current definitions


It =CV formula3 (Joey999's assertion)


I think where my confusion is coming from in his formula


Assuming C=F(?)

Then

At = FV formula 4


If so I will redefine his formula using my variable standards.


It = F V
At = FV

Which yields

A = V / tF



Let’s try a different interpretation.

F=V/ C Formula 2

Merging formula1 we get:

F = V / At

A = V / (tF) which is the same as Joey's formula. Hypothesis proved.



I had trouble sleeping last Sunday (just starting to get sick) and was obsessing over this. My math skills have taken a major hit. By proving Joey’s formula, my understanding is increased. Thanks again folks. This is why changing what variables stand for can be a bad thing. Delta Δ is not obvious as a variable, so I will use d instead of using it. The way it is slung around is very confusing. Using C for Coulomb is also confusing when a beginner is used to using C for capacitance. Also not good. I say this because of people like @WBahn , who are teachers, and whose familiarity with the subject may not allow them to see where the confusions are. Getting my algebraic definitions down is also important for my understanding..

Now for step 2, How does this relate for a 555 charging time on an op amp integrator triangle wave generator circuit? This will be the subject for a different thread, just making sure I can understand the basics before moving on.

 

The SI definition of an Amp:

A = Ct

1C=1A/sec (One of the Basic definitions of Coulomb)

or

A= C/t Formula 1

These are contradictory. How can A=C/t and also be equal to C·t. You know at least one of them is wrong.

The definition of a coulomb is the amount of charge that is transferred each second by a constant one ampere current.

So

1 C = 1 A · 1 s

which can be arranged to yield

1 A = 1 C/s

In general, current is charge per unit time.

You are mixing units with symbolic variables. It should be I = dq/dt or, in the average case I = ΔQ/ΔT.

Capacitance is

F = V/C Formula 2

You are again mixing units and symbolic quantities. F is a farad, which is a unit of capacitance.

Somewhere in the two assumptions above I may be screwing up. Variable names are a bear, as C is frequently used for Capacitance, but here it is coulombs. This is why I am trying to be extra careful with my definitions. It can be very confusing. So starting with:

So stop using units like they are variables -- they aren't, they are units and need to be paired with a numerical scaling coefficient. If you mean one coulomb, then write it as '1 C'. If you mean capacitance, then use 'C'. Just like 'm' is generally used for mass, but '1 m' is one meter.

And Δ is not a variable (in the context it has been being used here), it represents a change in a quantity. For instance,

ΔQ = Q_final - Q_initial

I_avg = ΔQ/ΔT = (Q_final - Q_initial) / (T_final - T_initial)

 

You make that sound so easy. It's not for me. My mind keeps mixing them up. I will not use C for coulombs, unless I have formulas where it is required. Cb perhaps, to distinguish it? Doing basic algebra, I need some way of distinguishing the two. It has been around 45+ years for me, I'm trying to recover lost ground here. Thank you for helping (if I sound sarcastic, it isn't). I'm struggling with basic concepts. I think I have some understanding now, More than I did.
A recap

1F = 1V / 1Cb
This is correct, yes?

Then
1 Cb = 1A * 1 sec
Is this a correct statement?

Have I mixed any apples/oranges yet?
I'm stopping now to regain my composure and think this through.

 

You make that sound so easy. It's not for me. My mind keeps mixing them up. I will not use C for coulombs, unless I have formulas where it is required. Cb perhaps, to distinguish it? Doing basic algebra, I need some way of distinguishing the two. It has been around 45+ years for me, I'm trying to recover lost ground here. Thank you for helping (if I sound sarcastic, it isn't). I'm struggling with basic concepts. I think I have some understanding now, More than I did.
A recap

1F = 1V / 1Cb
This is correct, yes?

No. Let's consider if this makes sense.

If I have a large capacitor, that means that at the same voltage it will hold a lot more charge than a small capacitor, right? Conversely, if I put the same charge onto two capacitors, the larger capacitor will have the smaller voltage.

What does this equation require, though? If you double the capacitance (make it 2 farads on the right), then you have to double the left side as well, making it 2 V for the same 1 coulomb charge. Does that make sense?

The defining relationship for a capacitor is

Q = CV

Rearranging, you have

C = Q / V

The capacitance is equal to the charge divided by the voltage. You have it flipped.

Think in terms of the simple, fundamental concepts. I have a lousy memory, so I never just write down C = Q/V or V = Q/C. I always start by writing down (or at least mentally picturing) Q = CV because I can immediately check that this make sense because it is has become intuitively obvious to me that the charge on a capacitor increases if I increase either the capacitance or the voltage. If I need one of the other equations for what I am doing, well that's what algebra is for. The same is true for Ohm's Law; I always start with V = IR and rearrange as necessary. If I try to start with R = V/I, then I am going to mess it up some nontrivial fraction of the time. Fortunately, if I do, the fact that I am religious about tracking units means that I almost always catch that something is wrong immediately.

Then
1 Cb = 1A * 1 sec
Is this a correct statement?

Yes.

 

What is Q?
Is it?
Q = V*Cb
If so then
C = Q / V
C=V*Cb/V (I seem to have messed up again.)

This equation seems to be at the root of my mental hangup.


Sorry, but I'd rather ask, and appear stupid, that pretend I know what that is and prove I really am stupid. I suspect without knowing I am zeroing in on my comprehension issue.I am trying to keep C (capacitance) from C(Coulombs), Like I said the definitions of C is confusing me

William, I appreciate you taking the time and having the patience to teach an old tech who is past her due date. For the most part I can still communicate clearly, thank God for small favors. To think I used to be a lead tech for Collins Radio. <sigh>

 

A shot summary of capacitance---

Q is charge in coulombs.
C is capacitance in farads.
and, of course, V is the emf in volts.
Don't use Cb for charge as that's confusing to the rest of us (and I'm easily confused).

Thus (if my math is correct)
C = Q/V
V = Q/C
Q = C*V
and to help solve your original question
Q = I*t where t=time

 

A shot summary of capacitance---

Q is charge in coulombs.
C is capacitance in farads.
and, of course, V is the emf in volts.
Don't use Cb for charge as that's confusing to the rest of us (and I'm easily confused).

Thus (if my math is correct)
C = Q/V
V = Q/C
Q = C*V
and to help solve your original question
Q = I*t where t=time

Please understand, I am having trouble distinguishing C (Capacitance) and C (Coulombs). Could you please give me those formulas again using Cb for Coulombs?

The only thing I think I have down is
Cb = I * t(seconds)

Better be careful, lest my stupidity does a reverse osmosis, and I drag you down to my level.

Ah! so Q = Cb ??

I have some rereading to do with a fresh eye.
God,
Do I feel thick(dumb).

 

Ah! so Q = Cb ?

You have it, grasshopper.
Get rid of that Cb designation for coulombs and use Q, which is the standard designator for electrical charge.

 

OK, Thanks. It doesn't hurt to use a hammer to pound it in (except for my pride).

Thing I find interesting is it is a little like ohm's law, where the designation.(Ω,V,A)come together to create new designations. Q seems to have the simplicity Ohm's Law when taken as a whole. I have to admit my understanding of charge was/is weak at best, But like most idiots who don't know their idiots (TX highways have a lot of those types) I didn't know how much I didn't know. Thank you again.

 

Thing I find interesting is it is a little like ohm's law

Yes, but it relates voltage directly to charge and inversely to capacitance, instead of directly to current and resistance, as Ohm's law does.

 

I just love straightforward linear equations...

 

... I didn't know how much I didn't know.

We've all been there (for me, just last week). One of my favorite quotes, this has kept me sane, humble, and excited about learning for many years. The second part has a delightful ambiguity.

CHAP. XVII. The Master said, 'Yu, shall I teach you what knowledge is? When you know a thing, to hold that you know it; and when you do not know a thing, to allow that you do not know it;-- this is knowledge.

Confucius (September 28, 551 – 479 BC)

American translation:

Know that you know what you know, and that you don't know what you don't know.

ak

 

Ah! so Q = Cb ??

Only to the degree that in the equation

C = 2πr

that, somehow,

r = feet

But then

C = feet

as well. So does that mean that r = C?

In the equation C = 2πr the 'C' and the 'r' are symbol (or symbolic representations or symbolic variables, if you prefer), The first is a symbol representing the circumference of a circle, while the second is a symbol representing the radius of the circle.

Both represent measures of distance and so have units of distance, be it feet, meters, miles, microns, astronomical units, or light years. But we shouldn't replace the symbol in an expression with one particular unit that we use to measure that quantity. For instance, we we did that using the S.I. unit for distance, the meter, we would end up with

m = 2πm

The same with the equations we use to describe electrical phenomena.

V = I·R

That V is NOT the unit of voltage, it is a symbol representing the voltage difference across the resistor. The V is not 'volt' is apparent because neither I nor R are units of current or resistance, respectively. They are symbols representing the current through the resistor and the resistance of the resistor, respectively.

The same applies to

Q = C·V

The Q is the symbol representing the charge stored on the capacitor, the C is the symbol representing the capacitance of the capacitor, and the V is the symbol representing the voltage difference across the capacitor.

 

Cheat Sheet/Math review

OK, so I’m going to put the formulas I would like to remember (and the ones I never will) on this post. I will transfer a cleaned up copy on a different thread and bookmark it on my Index. Formulas I feel I should remember I will put an asterisk * after, ones that are not important to remember I’ll put a *- after it.


Q = 1C (Coulomb) *

1C = 6.2*10E18 electrons *-

Q = It *

Q = CE *

C = Q / E *


Q = 1 Coulomb (redundancy is good, so I’m told)

C= Capacitance (Farads)

E = Volts

t = seconds

I = Current (A, Constant Current assumed)


That pretty well covers it, I think.

Odd, when I boiled it down, it was much simpler than I had thought.

 

Q = 1 Coulomb (redundancy is good, so I’m told)

Not in this case. Q is any number of coulombs.

 

Yes all values start at 1, then are modified by the other values. which are rarely 1. 1V change fits the formula Q = C V nicely.

1C =1F * 1V, If you assume 1 then many of the definitions fall into place, it all works together.

1C= 1A * 1 sec, all of these values can be other, but their relationship doesn't change

 

Hello,

If you want to know how the capacitor reacts to a constant current then we have:
dv=i*dt/C

where
dv is the CHANGE in voltage,
i is the constant current,
dt is the CHANGE in time,
C is the capacitance.

For constant current dv is really delta_v and dt is really delta_t which simply means that the slope is constant.
This formula is easy to deal with because we just plug in the values on the right side.
Note we have to specify the CHANGE not the absolute level of the quantity because it holds for any place on the 'curve' when the curve is just a straight line.

The inductor is the dual of the capacitor, so just swap current and voltage and switch capacitance C from the cap formula to inductance L.
Starting again with:
dv=i*dt/C

replace dv with di and i with v and C with L and get:
di=v*dt/L

and here we are looking at a constant voltage because in the cap formula we had a constant current, and we solve for the CHANGE in current (di) with the CHANGE in time (dt).

Note all we had to do was swap some variables to get from the cap formula to the inductor formula.

 

Cheat Sheet/Math review

OK, so I’m going to put the formulas I would like to remember (and the ones I never will) on this post. I will transfer a cleaned up copy on a different thread and bookmark it on my Index. Formulas I feel I should remember I will put an asterisk * after, ones that are not important to remember I’ll put a *- after it.


Q = 1C (Coulomb) *

This would be like someone saying that they should remember that

t = 1 second

But 't' is just a symbolic variable that (in this context) represents some amount of time. In any given situation we could have

t = 1 second
t = 15.8 minutes
t = 0.734 years

or anything else that is a measure of time.

Similarly, Q is a symbolic variable that represents some quantity of charge. It could be

Q = 1 coulomb
Q = 15 electrons
Q = 17.5 doubly-ionized nitrogen atoms

or anything else that is a measure of electric charge.

1C = 6.2*10E18 electrons *-

The magnitude of one coulomb is equal to the combined charge of (about) that many electrons, but the actual charge of this many electrons is -1 C since electrons are negatively charged.

Also, it's not equal to that many electrons because negative charge is just one property of that many electrons. You could also talk about the mass of that many electrons but you can't talk about the mass of 1 C of charge because a coulomb has nothing to do with mass.