1.0uf replacment capacitors?

[tedzed1]

Hello again everyone, from time to time i come across some 1.0uf waxies in some of my repairs, but as yet i have not replaced any, due to what i have found the high cost of replacements, would it be o.k. to use two 0.47uf in parallel, as i have no problem sourcing these at a good price?

regards paul

[FRANK.C]

Hi Paul
It will be perfectly fine to use two 0.47 uF in parallel in place of a 1 uF capacitor.

Frank

[tedzed1]

Hi frank, i had an idea it would be o.k., many thanks for confirming it

regards paul

[DAVEHALL]

Originally Posted by tedzed1

Hi frank, i had an idea it would be o.k., many thanks for confirming it

regards paul

Ted - caps in parallel - add values direct , but ensure voltages are at least as high as needed individually. Caps in series -ct = (c1*c2)/(c1+c2) ,inverse of resistors -where resistors in series -add values ,resistors in parallel
RT=(R1*R2)/(R1+R2).

Hopeful not appearing to show off ,just passing on some hard earned theory .

[tedzed1]

Thanks Dave, had something like this explained to me before, but struggle with some of the equations, especially caps in series, that's why i need the help of you fine gentlemen (and ladies) at times, i am the sort of person that you can explain something to, five or six times, and not understand, then i could be lying in bed one night, and the penny will drop, and other times, i can work it out in seconds

regards paul

[ppppenguin]

Originally Posted by tedzed1

.... but struggle with some of the equations, especially caps in series...

The equations that Dave Hall gives are correct but not the best ones for understanding what's happening. Let's do resistors in series.

Start with ohms law: V= IR or R=V/I. Think of it as the voltage pushing current through the resistor. Put some Rs in series and you'll need more voltage to push the current. So just add them: R = R1+R2+....

Now resistors in parallel. The same voltage is across each resistor but the currents will be different, depending on each resistor. Take R=V/I and turn it upside down. 1/R = I/V. The inverse of resistance, 1/R, is often called conductance with the symbol G. So G = I/V. We have already seen that for parallel R the voltages are the same but the currents differ. So we can add all the conductances. G = G1+G2+.... But G = 1/R so

1/R = 1/R1 + 1/R2 +....

Which is the basic equation for R in parallel. Also easy to do on a calculator that has an invert (1/x) key. With a bit of tedious algebra you can show that this is the same as the (R1xR2)/(R1+R2) equation. This equation was easier to use in pre-calculator days but I rarely use it now.

There are similar derivations for inductance and capacitance. The important thing to remember what is physically happening. If you put caps in parallel they can store more charge so the resulting C is bigger than either on its own. So add them C=C1+C2+... But resistors in parallel are going to pass more current than any on its own so the total resistance is lower. Hence the parallel formula.

Hope this helps.

[DAVEHALL]

As said -But ,there's another way of looking at problem ,and possibly understanding it ) .
Series resistors - the current is constant - and is the (by Ohms Law) the voltage divided by the TOTAL resistance ( = R1+ R2 + .........)
Parallel resistance - the current will split ( think of it as a large water pipe supplying several smaller ones of differing sizes).The voltage is constant , so current through each resistor is voltage divided by resistance) .
So total current = v/r1+v/r2 +......
But total current = v/r .
so v/r =v/r1+v/r2 +............
Divide by v ............
We get 1/r = 1/r1+1/r2 + ..........

Capacitance - series- work out the impedance at a set frequency - using z=1/2*pi*f*c .( current is constant)

So Zt = z1+z2+............

so 1/2*pi*f*ct = 1/2*pi*f*c1+ ...........


for parallel capacitance -voltage is constant .

Current through each capacitor =v/z.

z =1/2*pi*f*c

So Itotal = I1+I2+............

=V/Z1+V/Z2+.............

=== V*2*PI*F*C1/1 + V*2*PI*F*C2/1+ .........

DIVIDE BY V,2,PI,F ----WE GET ...CT = C1+C2+.......

Inductors - we need to get into J notation due to angular differences caused by resistance and inductive reactance .So I'll pass.

Apologies for lack of ideas on my previous post - I just assumed ( wrongly ) that all had the basics of electronic theory .

Hope that might help .

[DAVEHALL]

Now back to original problem -and possibly I've got a neat solution - if you can find a few discarded telephone master ( and it has to be master ) sockets . The standard capacitor fitted to these is 1U8 @250v ,and they're good quality . Put two in series and ,Hey presto - that's 0.9 uf -within 10% of 1uF

[julie_m]

Another way to deal with capacitors in series:

Recall the general formula for capacitance,

C = ε * A / d {1}

where
C = capacitance in Farads,
ε = dielectric constant of the material in F/m,
A = area in m² and
d = distance between plates in m.

Let's take a special case: Two capacitors with the same dielectric (so ε is the same) and the same area of plates. So the only difference between them is the distance between the plates. Now

C1 = ε * A / d1 {2}
and
C2 = ε * A / d2 {3}

When you put these capacitors in series, the plates in the middle are joined together, and so will be at the same electrical potential; meaning, we can pretend they are not there. So we just have the two outer plates of area A, separated by a dielectric layer of thickness d1 + d2, giving

C = ε * A / (d1 + d2) {4}

Substituting {2} and {3} in {4} we get

C = ε * A / (((ε * A) / C1) + ((ε * A) / C2))
C = ε * A / (ε * A * (1 / C1 + 1 / C2))

and if we can cancel the ε * A and rearrange a bit, we get

1 / C = 1 / C1 + 1 / C2 {5}

This in fact holds even when the plate areas and dielectrics are not the same. There is no way to tell any of ε, A or d from capacitance measurements alone: we effectively discarded all that information when the capacitor was sealed up inside its encapsulation

If you're feeling particularly masochistic, you can prove it works even with different values of ε and A.

[DAVEHALL]

AJS -between us we've either got folks running for cover ,or running for an education centre . Hopefully the latter . Next topic could be why at resonance an LC circuit is either open circuit or virtual short ( well theoretically ) .And then we'll have to look out for the cyber ducking stool or cyber burning stake .

[Billy T]

No need for maths, that is unless you like banging your head against the wall, or enjoy the thrill of finding out that you were out by a factor of 10. There are free R/C/L calculators available on the web that will do it for you in a trice. In fact, there are more calculators out there than you can shake a stick at, and for just about anything you might need.

Lets you get on with the job without wasting time, and you can also get to use up some of those odd values that you may have lying around.

Seek and ye shall find...............

Cheers

Billy

[AlanBeckett]

The snag with calculators, in all forms, is that unless you know what you're doing, or at least know approximately what the answer should be, you can get all sorts of nonsensical answers without knowing it. I'm also nervous about books and articles which adopt a 'non-mathematical' approach
Alan

[Herald1360]

Don't forget that a quick sanity check for R's in parallel and C's in series is that the resulting value has to be less than the lowest value component in the group.

[Billy T]

Originally Posted by AlanBeckett

The snag with calculators, in all forms, is that unless you know what you're doing, or at least know approximately what the answer should be, you can get all sorts of nonsensical answers without knowing it. I'm also nervous about books and articles which adopt a 'non-mathematical' approach
Alan

May have happened to you with inferior programming, but the calculators I have used are 100% reliable and unequivocal in their display.

I also have believe that anybody who is capable of restoring a radio, whatever their backgound, will quickly learn that the value produced by resistors in series increases, and in parallel reduces. They will also know that the inverse applies to capacitors, i.e that series reduces and parallel increases.

Similarly, power rating for equivalent wattage resistors in parallel adds, and in series remains the same. Dissipation per resistor is another matter of course, but there are calculators to solve that one too.

The key attribute of dedicated calculators is that they save time, and eliminate errors, especially for people like me whose maths lessons were taken when the abacus was still evident in primary school and log-tables reigned supreme for the cognoscenti.

Cheers

Billy

[AlanBeckett]

Billy,
I wasn't implying that the calculators made errors. In fact I can't remember an instance when that happened. It's people that make errors, typically either during data input or using arithmetic functions. The problem is that without an idea of what the answer should be they have to then believe the nonsensicle answer.
Alan

[Billy T]

Originally Posted by AlanBeckett

Billy,
I wasn't implying that the calculators made errors. In fact I can't remember an instance when that happened. It's people that make errors, typically either during data input or using arithmetic functions. The problem is that without an idea of what the answer should be they have to then believe the nonsensicle answer.
Alan

Yes, I understand that, but in my opinion, all that a standard calculator will do is allow the mistakes to be made faster, through incorrect input of values and/or factors which in turn leads to gross errors. A dedicated component calculator removes much of the risk because you enter the actual component values in capacitance, inductance or resistance units. No arithmetic functions or result interpretation are required at all. The mathematical examples provided above make my head spin and there's no way on earth I'd choose them over a simple dedicated software calculator

In terms of R & C, calculators are really only needed for R in parallel and C in series, and as a simple rule of thumb, the resultant of two C in series will always be less than the lower of the two values, and the same applies for R in parallel.

The same principles apply to inductance in series or parallel, LC tuned circuits or RC time constants etc, dedicated calculators make it easy.

I left school 44 years ago with minimal maths, but I managed to pass all my electronics examinations and even ended up as a senior examiner myself for many years and having marked several hundred papers from trade students I have seen all the errors that can be made.

If you simplify the calculation (as dedicated calculator programs do) by only requiring two (or three) values to be entered, then clicking/pressing 'Enter' to get the answer, the chances of coming up with a nonsensical answer are much lower.

Your mileage may vary........

Cheers

Billy

[Station X]

For resistors in parallel or capacitors in series I always use "Total value equals product over sum".

So for 10 ohms in parallel with 50 ohms, the product is 10*50 or 500 and the sum 10+50 or 60. Divide 500 by 60 and you get 8.33.

[geofy]

Yes but it is easier with a calculator to add reciprocals 1/x

So (10 reciprocal 1/x) 0.1 + (50 reciprocal 1/x) 0.02 = 0.12 reciprocate (1/x) = 8.33

And (10 reciprocal) 0.1 + (50 reciprocal) 0.02 + > any number of reciprocals, reciprocate = total

[Station X]

I guess we all do what comes easiest to us. The large digit calculator on my benchtop doesn't have a reciprocal function.

For resistors in series or capacitors in parallel it's hardly worth using the calculator or even pen and paper.

For resistors in parallel or capaciors in series we're generally saying I have a component value X and I want to add a component value Y to bring its value down to value Z. In that case the use of reciprocals is appropriate.

It's probably better to get a component of the correct value to improve relaiability.

In many cases we're using components of similar value in order to get half the value or twice the value. That's exactly what post #1 is proposing, but we've been side tracked into a lot of what I consider to be over complicted mathematics.

[geofy]

It would be better to select a suitable component rather than putting them in series or parallel wherever possible unless an unusual value is required.

An ordinary calculator can still do reciprocals by;

Number , divide, = =

So 10 / = = 0.1

And 0.1 / = = 10

Geof
reciprocal
foeg