Understanding Electronics Basics #1

Have fun, Graham.

cowboybob said:

Have fun, Graham.

Click to expand...

Thanks CBB, weather looking ok, be nice to get out with the lads, having a comms test day with our new radios before France, hoping for better range

Morning CBB
Had good day out yesterday but radio check wasn't that great, line of sight at best, 1 mile with 4w radio over brow of hill or just about 2 miles along top of hill

back to what we were talking about, R in series is adding more R

R in parallel is divided, I see this bit clearly now

Just working on KISS's calculation in post #411, I'm seeing of you had equal R & V they would cancel each other out so you would get no I, so I'm trying to work out what you are trying to show me

So if I had 10V & 10ohm but the 10ohm were 2x 5ohm R in parallel, I would get less R even though values are the same because the R is halved by division, hence, allowing power to flow

On the frequency front, I can see where I was misunderstanding, I was thinking if 1A = 1Hz how do they get 1MHz, where do they get all that power from, but it works backwards doesn't it, as R increases so does frequency, so we don't need more power for a higher frequency, we actually need less but at the same time as we lift frequency we are also increasing R against I making it take longer, so this has something to do with charge times, where 't' comes into play


So now I'm working on the overall picture still & somehow I need to understand the link between it all, I can see the logic for the LOG graph, due to the small amounts & as we are working backwards, the graph is somehow the oppisite side of 0 but the right side of 0 too?

No, it must be above & below for that comparision to work, so somehow frequency is a hidden part of I

I can almost visualize it in front of me, so the longer it takes, the less time the cap has to charge, so the higher the frequency the lower I due to time taken

That's it, I can cleary see whats happening now if thats right so far

How we work it out, well thats another story that I may need your help with **broken link removed** least I can see whats going on now

Hope you enjoyed your day off from me, now back to all these dumb Q's I keep asking **broken link removed**

Mornin', Graham.

Be with ya in a minute.

No worries, having a chilled day, aching all over

The -3 db rule is sort of an average for AC = rms/equivilent to DC

I think I'm starting to get the picture now, a cap resists instant change, therefore it takes time to build the charge, so there are lots of different uses

I can see the filtering function, I'm starting to see the frequency function but I need to nail this down really to understand exactly what we are looking at, so frequency affects the output V via time which would alter the response time to anything on the end of it, i.e. phase, so if you wanted one part of the circuit to start before the other, you could use it like this or it could give you a different waveband etc etc as KISS was saying a variety of jobs are achievable, I just need to understand it now

So by being able to work out frequency gives us the conclusion for work available (VP) on the line should it be required but in a miniscule form

(Question for later while I remember it, does that mean an inductor would do the same thing for 'I' giving us delayed Amps available, only inductors collapse when powered off unlike caps)

back to what we were talking about, R in series is adding more R

Click to expand...

Yes. Resistors in series merely add togeher.

Just working on KISS's calculation in post #411, I'm seeing of you had equal R & V they would cancel each other out so you would get no I, so I'm trying to work out what you are trying to show me

Click to expand...

The formulas in #411 are not relevant to the task at hand (parallel resistors). They are relevant to circuits with "reactance" so disregard them for the time being.

So if I had 10V & 10ohm but the 10ohm were 2x 5ohm R in parallel, I would get less R even though values are the same because the R is halved by division, hence, allowing power to flow

Click to expand...

Sort of, but we need to clarify the math.

Let's start again with a basic two resistor parallel circuit:

View attachment 62417

To solve for the equivalent resistance for this circuit, we use the following formula (LateX isn't working):

R1 X R2 divided by R1 + R2. This can also be written: (R1*R2)/(R1+R2).

So, in this case: (10*40)/(10+40), or 400/50, which equals 8.

So in words, the equivalent resistance of a parallel, 2 resistor circuit consisting of a 10Ω and 40Ω resistors is 8Ω.

Try this in TINA with different value resistors, do the math and then confirm your answer with the multimeter.

It isn't so important to know why the formula is used as it is to realize that this is how it's done. And it works.

You might ask: "Why not use an 8Ω resistor, instead"? Well, when you can't find a certain value resistor (and an 8Ω resistor IS NOT a normal value [and neither is a 40, for that matter]), it's OK to create an "equivalent" value "network" circuit of multiple resistors to achieve that.

And since we've been investigating RC timing circuits, for a very specific RC time constant we might have to do just that.

One last example. Let's say you wanted an 18Ω resistor (also an odd value, as is 40Ω, but I wanted to keep the math simple). This is how to get there:

View attachment 62418

In other words, an equivalent 8 ohm parallel circuit (composed of 10 and 40 ohm resistors) in series with an 10 ohm resistor.

On the frequency front, I can see where I was misunderstanding, I was thinking if 1A = 1Hz how do they get 1MHz, where do they get all that power from, but it works backwards doesn't it, as R increases so does frequency, so we don't need more power for a higher frequency, we actually need less but at the same time as we lift frequency we are also increasing R against I making it take longer, so this has something to do with charge times, where 't' comes into play

Click to expand...

If it's alright with you, I'm going to address this issue in the next post.

You've got some of it right but some of it is not right. I'm going to start over with some examples that I hope will help separate the parts.

I'll get to post #425 shortly.

cowboybob said:

Sort of, but we need to clarify the math.

Let's start again with a basic two resistor parallel circuit:

View attachment 62417

To solve for the equivalent resistance for this circuit, we use the following formula (LateX isn't working):

R1 X R2 divided by R1 + R2. This can also be written: (R1*R2)/(R1+R2).

So, in this case: (10*40)/(10+40), or 400/50, which equals 8.

So in words, the equivalent resistance of a parallel, 2 resistor circuit consisting of a 10Ω and 40Ω resistors is 8Ω.

It isn't so important to know why the formula is used as it is to realize that this is how it's done.

You might ask: "Why not use an 8Ω resistor, instead"? Well, when you can't find a certain value resistor (and an 8Ω resistor IS NOT a normal value [and neither is a 40, for that matter]), it's OK to create an "equivalent" value "network" circuit of multiple resistors to achieve that.

And since we've been investigating RC timing circuits, for a very specific RC time constant we might have to do just that.

One last example. Let's say you wanted an 18Ω resistor (also an odd value, as is 40Ω, but I wanted to keep the math simple). This is how to get there:

View attachment 62418

In other words, an equivalent 8 ohm parallel circuit (composed of 10 and 40 ohm resistors) in series with an 10 ohm resistor.

Click to expand...

Thats clever isn't it, plain english I understand **broken link removed**

I did try it in Tina & have just done the math, it does work **broken link removed**

Works every time **broken link removed**

OK.

Now RC circuits. I'm going to build (edit) this post on the fly. When I'm through, I'll indicate that at the end of the post.

Basic 10VDC powered, 1uF cap plus 1MΩ resistor (RC time constant of 1 second), then a 1uF cap with a 500kΩ resistor (RC time constant of 500m seconds):

View attachment 62420View attachment 62421

The RC time constant is defined as "that point in the rise of the voltage across the capacitor where it reaches two-thirds (2/3, or 0.633 times) of the value of the power source (battery).

Note that the V value on the graph is 6.33 VDC (or 2/3 of the battery's 10VDC value) at that point where the rise time value is 1 second for the 1MΩ resistor, and 1/2 second (500ms) for the 500kΩ resistor.

Now these examples. The only difference is that the battery level has been raised to 20VDC.

View attachment 62422View attachment 62423

Notice how the RC time constant is identical to the 10VDC battery examples. The voltage level at which the RC time constant is reached is higher, to be sure, but the TIME that it took to get there is unchanged.

This is why it is called an RC time constant.

In other words, the voltage level of the power source (battery) has absolutely no effect on the RC time constant value.

You with me so far?

OK, I'm through with this post.

OK, now I'm REALLY through with this post.

The formulas for series and parallel resistors:

Rs = R1_R2+R3+.....Rn

1/Rp= 1/R1+1/R2+1/R3+...1/Rn (Try it!)

Now for fun, what I'll say is that for capacitors in parallel the formula becomes: Cp = C1+C2+C3+...Cn. So a 10uf capacitor in parallel with a 10 uF capacitor is 20 uf

and, not surprising for capacitors in series: 1/Cs=1/C1+1/C2+1/C3+....1/Cn

The formulas for R in series and parallel are just the opposite of C in series and parallel.

So ignoring V for a moment, I can see the time taken by R & cap is equal, so these are setting for the charge time & nothing else

so I was right in thinking R was the control of the cap, I just couldn't relate it to time

cowboybob said:

400/50, which equals 8.

Click to expand...

ok, that took an age but you are basically dividing V by A for ohms

however, the time it took to figure out why two 25ohm R's 400/50 = 12ohms
wouldn't work was because you need to be able to times V by ohms to be equal to the two R's
so if you work out you want a 8 Ohm resistor via 2 in parallel how are you working it backwards which is what I've been trying to do

Edit: it doesn't matter how I try, I simply cannot work backwards, so me knowing I need a 8ohm R wouldn't be a lot of use even knowing V & I, I can work back as far as needing two R's split but then get lost

The only thing I can conclude from today is that if you already have two resistors in parallel that you can work out V & I but not actually work it back to work out what actual resistors you need & R reading is fixed whatever V you have, changing V only alters I

KeepItSimpleStupid said:

1/Rp= 1/R1+1/R2+1/R3+...1/Rn (Try it!)

Click to expand...

1/RP = means total of R's in parrellel

What does the rest mean?

I take it Cn or Rn just mean total?

Latex doesn't work.

Rp means the resistance of the parallel combination.
The other gibberish is known as an infinite series, but you can solve it for any number of resistors (n) in parallel.
So if there are two, 10 ohm resistors.
1/Rp = 1/10+1/10
1/Rp = 2/10
Rp = 10/2 or 5

I solved it using fractions, you can solve it using decimals too.
1/Rp = 0.1+0.1
1/Rp = 0.2
Rp = 1/0.2
Rp =5

The formula would work for 100 resistors in parallel and 1,000,000 resistors in parallel. It doesn't matter. It will even work for 1.

Using 10 ohms: 1/Rp = 1/10; Rp = 10

Each resistor can have a different value.

You learned a while ago that R*C has the units of ohms * farads (Usually written as (ohms- Farads) and that reduced to the units of seconds (s). I mistakenly used t. Units that are multiplied together show up as what looks like subtraction. (ohm-cm) is one such unit.

KeepItSimpleStupid said:

So if there are two, 10 ohm resistors.
1/Rp = 1/10+1/10
1/Rp = 2/10
Rp = 10/2 or 5

I solved it using fractions, you can solve it using decimals too.
1/Rp = 0.1+0.1
1/Rp = 0.2
Rp = 1/0.2
Rp =5

The formula would work for 100 resistors in parallel and 1,000,000 resistors in parallel. It doesn't matter. It will even work for 1.

Using 10 ohms: 1/Rp = 1/10; Rp = 10

Each resistor can have a different value.

Click to expand...

That seems long winded, why wouldn't you just add them together & divide by the amount **broken link removed** or is it that you have kept it simple for me & having different value R's would give varying amounts each side depending on value

KeepItSimpleStupid said:

You learned a while ago that R*C has the units of ohms * farads (Usually written as (ohms- Farads) and that reduced to the units of seconds (s). I mistakenly used t. Units that are multiplied together show up as what looks like subtraction. (ohm-cm) is one such unit.

Click to expand...

Is this part of the math that I don't understand?

now we need to run over this 's' thingy again, you asked me if I could get 50 as answer to something

KeepItSimpleStupid said:

The length of one cycle is called the period and it's measured in time units, like 20 mS. 1/(period) is the frequency. So 1/20 mS is 50 Hz. See if you can get that answer on your calculator?

Capacitors change their behavior based on frequency. When capacitors are part of a timing circuit, the quantity R*C is usually involved. That quantity is called the TIME CONSTANT.

Click to expand...

found it, when you put this in calculator you get 0.05?

I think I've lost the will to live **broken link removed**

R is fixed & we are multiplying cap value (F) which isn't fixed (the time bit) so R gives us the control over how much charge the cap gets

no thats still not right, I must have been over this a million times but I can't see for the life of me how it's working

Graham,

]found it said:

So ignoring V for a moment, I can see the time taken by R & cap is equal, so these are setting for the charge time & nothing else... so I was right in thinking R was the control of the cap, I just couldn't relate it to time

Click to expand...

I think I'm starting to get the picture now, a cap resists instant change, therefore it takes time to build the charge, so there are lots of different uses

Click to expand...

Yes. And it takes the time dictated by the value of the resistor times the value of the cap to arrive at the RC time constant.

And notice KISS's note on parallel capacitor determination formulae.


So, now, we'll go on to a circuit and show how the RC constant is used to determine the timing of an event.

In this example, I'm using a simple, triggered "pulse generating" circuit that demonstrates how an RC circuit can control an event (in this case, the length of the pulse).

Most of the circuit is hidden, but the essential portion (the square positive pulse making circuit) that is relevant to this discussion is shown:

View attachment 62438

Note that, though not exactly 1 second, the pulse is close (2.5s to 3.75s, or a1.25s pulse).

Why not exactly 1s, as the RC formula predicts?

Well, here the planning world hits smack up against the real "let's make a circuit that does something" world.

The OpAmp has internal resistances (and capacitances as well). These values interact with the RC circuit and alter it.

But all is not lost. Here is how you "adjust" for these anomalies:

**broken link removed**

Note that I reduced R1 to 900kΩ. That sufficiently reduced the RC value to come into line with the desired 1s pulse width, given the unseen values within the OpAmp that needed to be allowed for.

Messy? Yes.

But then, most analog (as opposed to digital) electronics is a bit messy. Often, in order to get a circuit to work as you wish, compromises are made.

I'm going to post a simple pulse generating circuit later that you can alter to demonstrate how an RC circuit affects timing.

It's taking me some time to do it because there are a lot more set-up parameters that are needed to make it work. Not a lot of parts, but those few parts must be correctly configured.

Muttley said:

found it, when you put this in calculator you get 0.05?

Click to expand...

You missed something: 1/20 mS; What is a millisecond? 20e-3; therefore 1/20e-3 or 1/0.020 is (?) any you won't get 0.05.

cowboybob said:

You need to be able to keep this rule firmly set in your mind: under normal circumstances, neither Voltage (V), nor Current (I) can or will alter resistance (R). NEVER.

But the reverse IS true: resistance WILL alter both current AND Voltage.

Click to expand...

Hiya CBB
I totally get that bit, haven't read the rest yet, but rest assured I can see the bit you have shown me as plain as day **broken link removed**

I think it's the time thing & frequency I'm struggling with