Understanding Electronics Basics #1

[Muttley600]

The decimal thing has been bugging you. Glad to see you're getting past that snag.

TBH the caculator is making me think about it more than a multimeter display, so my logic finally made sense of it

No. Look a little more closely at the displayed circuit values. Don't forget what the battery is called.

Whoops, ok, RI= 5V / 0.0025mA = 2000ohms

ok, just humour me for a moment while my logical brain is in gear

If we can discount I from the equation & in post #381 you said V has nothing to do with frequency, that only leaves R from main three components & was understood to be playing a part in this, but then we went chasing after time constant & something about RC

What sets the frequency if we suddenly jumped to it being an input to find whatever we were looking at & what was it we were looking at?

 

[Muttley600]

Good catch.

ermmm, no it wasn't really was it, looking at it again was just another way of saying what is V when expressed as V=

Sorry **broken link removed**

thanks for putting up with me

damn, CBB you were right again, looking back, The frequency was not changed in the sims I did, you've just taught me to look more closely in future

I'm back **broken link removed**

I don't mind making mistakes while learning but for the life of me I struggle when I can't see where it's all going wrong

reading back, R is controlling the cap to get the required frequency as without R the cap is unstable & can change the frequency, R is just stabilising it at what ever frequency you decide on

so does this mean the Rc is just the output graph of the cap

so frequency isn't some strange word for wavelength but simply the frequency of the cycle or in english, how often the cycles are over time

 

[Muttley600]

V can be altered around circuit but I is spread even throughout circuit, so I is only ever a product of V & R, I see where your going with this, duly noted in folder

ok, have been playing with sim

View attachment 62349

'I' is still spread around circuit until AM4 where it has no choice but to come back together, AM1/2/3 are all 1A each but if I change to 10V AM4 reads 6A meaning 'I' reading does not stay stationary when V is altered but ohms does, it has no choice it is fixed

 

[KeepItSimpleStupid]

Only read this https://en.wikipedia.org/wiki/Wavelength to "Standing waves". No more.

 

[KeepItSimpleStupid]

He's trying to get you to see that AM1+AM2+AM3 = AM4, I hope. The "current" A4 divides between the 3 resistors.

 

[Muttley600]

He's trying to get you to see that AM1+AM2+AM3 = AM4, I hope. The "current" A4 divides between the 3 resistors.

you can see I'm trying **broken link removed** I was trying to see the difference between the two

in series 'I' is spread even through the circuit & in pararell, he was showing me two different concepts for a reason wasn't he

 

[KeepItSimpleStupid]

Yea, current and voltage division.

 

[Muttley600]

Yea, current and voltage division.

ok

in series - A are constant through circuit whereas V can be altered by R
in pararell - A are divided through circuit but V is unaffected by R

 

[Muttley600]

-3db is nothing more than 70.7% or 0.707 or √2 divided by 2 of the reference input. The frequency this occurs at for a resistor and capacitor in series is f=1/(2∏RC)

Frequency = 1 (complete cycle) / 2∏ (2 radians (1 radian = 180degrees))

EDIT: does that mean the -3 db rule means one cycle = 70.7% of whatever charge going into cap

RC stands for reactive capacitance which I'm thinking is something along the lines of: (in english of course)

low V = low resistance = high frequency (cycles moving faster)
High V = high resistance = low frequency (cycles moving slower)

so for 'I' low R means more time to energise cap
High R means less time to energise cap

Theres more to put to this

 

[Muttley600]

RC is just R * C or the product of R & C.

i.e. R = 10 M ohms, C = 1 uf
RC = 10,000,000 * 1E-6

Right, got it

 

[KeepItSimpleStupid]

You might not understand this, but here goes. Remember I said that units are important?

Well C = Q/V and R = V/I, so R*C = QV/IV and the V's cancel, so R*C = Q/I, but I = Q/t and when you solve, RC has units of time. Seconds to be exact.

Sorry, I pulled Q out of the hat. Q is charge.

 

[Muttley600]

You might not understand this, but here goes

you know me, I'll have a damn good go

R*C = Q/I, but I = Q/t

Resistor * cap value = charge / I

I = charge / time

and when you solve, RC has units of time. Seconds to be exact.

1 coulomb = 1A * 1sec
1 coulomb = 1farad * 1V

1F = A*s / V

is this all saying that 1V = 1A over 1s

 

[KeepItSimpleStupid]

R*C = Q/I, but I = Q/t

R*C = Qt/Q ; Q's cancel and thus thus the answer is t.

 

[Muttley600]

Does that mean a 1μF cap has a charge of 0.000 001A over 1s

hold on, don't answer for a sec

so 12V / 8A = 1.5ohms

1.5ohms * 1uF = 0.0000015A charge over 1s before oscillation goes other way

 

[KeepItSimpleStupid]

1C = 1F x 1V
1C = 1A x 1second

 

[Muttley600]

& this is before we even get to phase **broken link removed**

 

[Muttley600]

1C = 1F x 1V
1C = 1A x 1second

=1hz

think I need a cuppa

well, unless I've lost the plot, how do things get to MHz from uHz for radio frequencys or do you have to ratio them to the unit of s

 

[Muttley600]

Morning CBB
We finally have a bit of movement in this thick head of mine, I can nearly see what's happening now, just gotta work out why 'I' is so low compared to frequency then I should be able to relate to things.

But please continue with other stuff as well so it helps me understand

 

[cowboybob]

ok, have been playing with sim

View attachment 62349

'I' is still spread around circuit until AM4 where it has no choice but to come back together, AM1/2/3 are all 1A each but if I change to 10V AM4 reads 6A meaning 'I' reading does not stay stationary when V is altered but ohms does, it has no choice it is fixed

Yes. And yes. And, finally, yes.

But since you raised the parallel resistors issue, here we go.

Please do the following in the order given.

1. Thinking about the example above and the circuit current value you observed, what is the effective resistance of the the two parallel resistors?

2. In other words, if the two parallel resistors were replaced with a single resistor, what would its value be?

3. Do this using a math solution method FIRST.

4. If you're stilled stumped, or just wish to confirm your answer, construct (exactly) the below circuit and check its resistance.

5. View attachment 62365

6. Don't you find it curious that the combined resistance of the two parallel resistors is less than the value of the lowest value resistor?

 

[Muttley600]

I can see where we are heading at last & how it is relating to frequency, but ive ran out of time before I got to understand it properly.....sigh
Out on bike all day tomorrow, back Wednesday

Have a good day tomorrow guys, have a day off from my antics.lol

At least we can move forward on this when I get back now I know what we are looking at