Understanding 'Series / parallel '.

[sevenisthenumber]

I would like to understand series and parallel better.

What happens when you put two 1k resistors in series?
What happens when their parallel?

What about two 1uf caps in series?
in parallel?

What about two 1n4002 diodes in series?
in parallel?

I would appreciate if someone would help me understand what happens.

THANKS!

[tcobretti]

OK.

Two resistors in series = their values are added.  1k +1k = 2k
Two resistors in parallel = their values are averaged.  You multiply the two resistors' values, then divide by the sum.  (1k*1k)/(1k+1k)=500R

With caps, you do the opposite.  In series, their values are averaged, while in parallel they are added.

Diodes I'm not sure about, so I'll let someone smarter answer that one.

[aron]

Here is the answer for diodes: ( I searched google for: two diodes in series )

http://sci.tech-archive.net/Archive/sci.electronics.basics/2007-08/msg00460.html

Check out this too:

http://www.diystompboxes.com/wiki/index.php?title=DIY_FAQ

[96ecss]

Here is a great free program you can use to help you with calculations like that. It's called Electronics Assistant.

http://www.electronics2000.co.uk/download.htm#assistant

Dave

[sevenisthenumber]

Thanks!!!


On the diode question:

so diodes are different voltages? Why are they used for clipping? Do higher voltages create more distortion or less, (led vs germanium)?

[oldrocker]

Keep in mind that with capacitors in series if they're polarized then it works different.  When two polarized caps have their negative leads connected together in series and if both caps are of the same value for example they become  non-polarized and they equal the size of only one of the capacitors.  For example if two 10uf polarized electrolyic caps are connected in series by thier negative leads they become a non=polarized 10uf cap.  And of coarse if they were two 10uf non-polarized caps connected in series they would equal 5uf.

[R.G.]

What happens when you put two 1k resistors in series? What happens when their parallel?

Resistors are devices that cause a voltage drop when a current flows through them. That's one implication of Ohm's law, in the form
V = I*R. The value of R tells you how many volts per ampere flows. One ohm is one volt per ampere. One K is one volt per milliampere.

Two resistors in series have to have the same identical current in them, so the voltage V is just the sum of the voltages across each resistor.
That is, V = I*R1 +I*R2. Since I is the same, V = I*(R1+R2); so the equivalent resistance of two resistors in series is the sum of the resistors.

Two resistors in parallel have the same voltage across them. In algebra form, V = I1R1 =I2R2. If we want to find the equivalent resistance, we can compute the equivalent resistance as V = Itotal*Req. Itotal is I1 +I2, the sum of the two resistor currents. So
V = (I1+I2)*Req.
rearranging, Req = V/(I1+I2)
replacing I1 and I2 with V/R1 and V/R2, Req = V/(V/R1+V/R2)
Cancelling out the V's in numerator and denominator, Req = 1/(1/R1+1/R2) which is an ungainly form.

I work with the inverted form a lot: 1/Req = 1/R1+1/R2.  A term of the form 1/resistance is a conductance. It makes sense that for resistors in parallel, each conducts some current, so the conductances add. Today's calculators with a 1/x button make this the easiest way to deal with parallel resistors.

But that's no how everyone memorized it. So let's multiply both sides by R1*R2;  R1*R2/Req = R2 +R1
Rearranging give the form everyone wants: Req = R1*R2/(R1+R2).

So in series the resistances add. In parallel the conductances add.

What about two 1uf caps in series?in parallel?

Capacitors are really AC animals. At DC, all we can talk about is how much charge they store and how fast they store/release it. Capacitance is defined by how much charge is stored per volt, or C=Q/V. If we have two caps in parallel, they each store whatever charge they would have stored at that voltage, so the total charge stored is
Qtotal = C1V+C2V = V (C1+C2)
So Ceq = Qtotal/V = V*(C1+C2)/V = C1+C2

So parallel caps have the same total capacitance as one cap with the sum of the capacitances.

Caps in series are tougher to understand. If you put a voltage V across two caps C1 and C2, then the two caps start storing charge. There can be no DC current flowing after things settle down, because caps are insulators for DC. Not only that, the charge that's stored in one cap has to equal the charge stored in the other cap, because the two connected middle terminals can't support a DC voltage - they're tied together. The only what that can happen is if equal numbers of charge carriers go into the + side and out the - side of the two in series. The connected middle terminals just let charge equalize between the caps.

So the charge on C1 is Q1=C1*V1 and Q2 = C2*V2. The total voltage is Vtotal = V1+V2 = Q1/C1 + Q2/C2 and since Q1=Q2, Vtotal = Q/C1+Q/C2

We know that C = Q/V, so we divide both sides by Q to get Vtotal/Q = Cequivalent = 1/C1 + 1/C2

This looks like our friend the resistances above. The same steps in algebra lead to Ceq = C1*C2/(C1+C2)

A little thought gets us to a capacitance being a number which represents a conductance; a smaller cap lets little current flow, a bigger cap lets more current flow. It begins to make sense that caps in parallel add - that's how conductances work.

What about two 1n4002 diodes in series? in parallel?

Just like above, we look at what each component does by itself, then arrange them and compute the result.

With a voltage positive on the anode with respect to the cathode, a diode lets almost no current flow until it gets perhaps 0.6V across it, then current flows essentially unlimited by the diode. There is some small resistance, but it's trivial unless you're building high current power supplies. With the voltage reversed, only a tiny leakage current flows until the voltage is so big it literally breaks across the insides of the diode. This is the "inverse voltage" or "blocking voltage" of the diode. A 1N4002 is rated for 100V blocking, meaning that they will block at least 100V. How much more? No one knows without testing.

So if you put two diodes in series A to K, no current flows one direction until you put about 1.0-1.4V across it, then the diode does not limit the current. When you reverse the voltage, no current flows until both of the diode reverse voltages are exceeded. We know that will be more than 200V, but we can't tell how much more. Which one lets go first depends on the magnitude of the DC leakage currents in the diodes. The leakier one leaves its brother holding more of the voltage in the inverse of the ratio of the leakage resistances.

[R.G.]

so diodes are different voltages? Why are they used for clipping? Do higher voltages create more distortion or less, (led vs germanium)?

Diodes have different forward voltages, depending on the material. Germanium is about 0.2-0.3, Schottky is about 0.4, silicon is about 0.5-0.7, LEDs come in voltages from about 1.2V up to maybe 4V.

It's impossible to say whether higher voltage diodes create more distortion. The amount of distortion all depends on the relative size of the signal you're driving it with compared to the diode voltage. It's the amount of overdrive that matters. 100V signals will clip massively on any of these. 100mV signals won't clip on any of them.

[Kerly]

Back onto the capacitor subject of this all....
How reasonable would it be more me to try and add up the values of a 33pf and (4) 4.7pf caps in order to get 51pf?
Considering space and everything
If I had a 100pf it would be much easier to just put the 4.7pf and the 100pf in series, but I do not :[

[R.G.]

How reasonable would it be more me to try and add up the values of a 33pf and (4) 4.7pf caps in order to get 51pf?
Considering space and everything

You could do that. You would get 51.1pF nominally. Each cap you put in there has a 5% or 10% tolerance, so it would be no better, and no worse than a 51pF for value. It would be five times as big, of course, since all caps under 100pF are about the same size, roughly. Making do with what you have is a time honored tradition to the electronics hacker.

If I had a 100pf it would be much easier to just put the 4.7pf and the 100pf in series, but I do not :[

Actually, it would not. 100pF and 4.7pF in series would be 1/((1/100pF)+(1/4.7pf)), or (100*4.7)/(100+4.7) pf, which comes out to 4.489pF.

Capacitor values are  AC conductances, the reciprocal of resistance. One ohm is defined as one volt per ampere. One Farad is one ampere-second per volt, or proportional to amperes per volt, the inverse of resistance. So where resistors add in series, capacitor add in parallel. In parallel, you add the reciprocal of resistance (1/R) and then take the inverse of that sum.

[Ice-9]

Here is a link to a calculator that will automatically give you the capcitance of series /parrallel configs.  R.G's reply is the best though as he actually gives the maths of how i works, and once this is understood then it adds to your own understanding of things.

http://www.electronics2000.co.uk/calc/calccsp.php

[R.G.]

Thanks, Ice.

My quibble with that and the reason I keep posting the equations is that I have the view that unless one can do the calculations manually on their own, using a calculator to accomplish something that's not well understood serves as a roadblock to further learning of those concepts. This is not a jibe at you; just my own private hot button. Everyone should LEARN, not be limited to knowing which button to push. In my mind, you earn the right to push buttons by knowing exactly what the buttons do and being able, in theory at least, to go replicate what the button does all on your own effort.

Did I ever tell you how to refine iron ore?   

[Ice-9]

Yes R.G. I totally agree as i said, your reply with the maths will, in the long run make someone actually understand  more rather than blindly geting the answer. But the calcs do come in useful as well.

Oh and its ok about the iron ore as i can already turn lead into gold in my workshop.

And thanks for the recomendation of the Art of Electronics. i managed to get it and working with it now.

[DougH]

I apologize if R.G. mentioned this as I didn't read through his replies completely.

One thing that may help you visualize series/parallel behavior are Kirchoff's Laws, in particular the current law:  "The algebraic sum of current into any junction is zero"

http://physics.about.com/od/electromagnetics/f/KirchhoffRule.htm

Between Kirchoff's laws and Ohm's law (v=i*r) you have two powerful tools that can help you solve simple networks.

[petemoore]

V = I*R. The value of R tells you how many volts per ampere flows. One ohm is one volt per ampere. One K is one volt per milliampere. 
  I have a serious problem with that. Namely nomenclature and symbols. Where can I study how to read math?
* = multiply = X
  = = = which = equals
  I've got that.
  Punching in the numbers here, more importantly the 'milli's, K's and Mega's decimal points have always been an effective stumbling block, and when I get up I've lost all direction.
Two resistors in series have to have the same identical current in them, so the voltage V is just the sum of the voltages across each resistor.
That is, V = I*R1 +I*R2. Since I is the same, V = I*(R1+R2); so the equivalent resistance of two resistors in series is the sum of the resistors.
 
  This looks real neat whenever you type it, I can't seem to assign actual numbers to these equations because where's the decimal point go, and how does it relate to milli, megan and killi, I wish I could get and erasable blackboard in here...
Two resistors in parallel have the same voltage across them. In algebra form, V = I1R1 =I2R2. You've lost me here, how does the algebra form work here, I is for current, R is for resistance, when next to a number how does that work?
  That's enough math for me to try to get back to today...by tomorrow it'll probably be 90% lost again anyway...I think I've read a few simple equations accurately, I couldn't write much if my life depended on it.  If we want to find the equivalent resistance, we can compute the equivalent resistance as V = Itotal*Req. Itotal is I1 +I2, the sum of the two resistor currents.
  Right here I'm lost, I don't understand what is required, what 'Req' means, it wrecked all attempts at following your thought train
So
V = (I1+I2)*Req.
rearranging, Req = V/(I1+I2)
replacing I1 and I2 with V/R1 and V/R2, Req = V/(V/R1+V/R2)
Cancelling out the V's in numerator and denominator, Req = 1/(1/R1+1/R2) which is an ungainly form.

I work with the inverted form a lot: 1/Req = 1/R1+1/R2.  A term of the form 1/resistance is a conductance. It makes sense that for resistors in parallel, each conducts some current, so the conductances add. Today's calculators with a 1/x button make this the easiest way to deal with parallel resistors.

But that's no how everyone memorized it. So let's multiply both sides by R1*R2;  R1*R2/Req = R2 +R1
Rearranging give the form everyone wants: Req = R1*R2/(R1+R2).

So in series the resistances add. In parallel the conductances add.


Quote
What about two 1uf caps in series?in parallel?
Capacitors are really AC animals. At DC, all we can talk about is how much charge they store and how fast they store/release it. Capacitance is defined by how much charge is stored per volt, or C=Q/V. If we have two caps in parallel, they each store whatever charge they would have stored at that voltage, so the total charge stored is
Qtotal = C1V+C2V = V (C1+C2)
So Ceq = Qtotal/V = V*(C1+C2)/V = C1+C2

So parallel caps have the same total capacitance as one cap with the sum of the capacitances.

Caps in series are tougher to understand. If you put a voltage V across two caps C1 and C2, then the two caps start storing charge. There can be no DC current flowing after things settle down, because caps are insulators for DC. Not only that, the charge that's stored in one cap has to equal the charge stored in the other cap, because the two connected middle terminals can't support a DC voltage - they're tied together. The only what that can happen is if equal numbers of charge carriers go into the + side and out the - side of the two in series. The connected middle terminals just let charge equalize between the caps.

So the charge on C1 is Q1=C1*V1 and Q2 = C2*V2. The total voltage is Vtotal = V1+V2 = Q1/C1 + Q2/C2 and since Q1=Q2, Vtotal = Q/C1+Q/C2

We know that C = Q/V, so we divide both sides by Q to get Vtotal/Q = Cequivalent = 1/C1 + 1/C2

This looks like our friend the resistances above. The same steps in algebra lead to Ceq = C1*C2/(C1+C2)

A little thought gets us to a capacitance being a number which represents a conductance; a smaller cap lets little current flow, a bigger cap lets more current flow. It begins to make sense that caps in parallel add - that's how conductances work.


Quote
What about two 1n4002 diodes in series? in parallel?
Just like above, we look at what each component does by itself, then arrange them and compute the result.

With a voltage positive on the anode with respect to the cathode, a diode lets almost no current flow until it gets perhaps 0.6V across it, then current flows essentially unlimited by the diode. There is some small resistance, but it's trivial unless you're building high current power supplies. With the voltage reversed, only a tiny leakage current flows until the voltage is so big it literally breaks across the insides of the diode. This is the "inverse voltage" or "blocking voltage" of the diode. A 1N4002 is rated for 100V blocking, meaning that they will block at least 100V. How much more? No one knows without testing.

So if you put two diodes in series A to K, no current flows one direction until you put about 1.0-1.4V across it, then the diode does not limit the current. When you reverse the voltage, no current flows until both of the diode reverse voltages are exceeded. We know that will be more than 200V, but we can't tell how much more. Which one lets go first depends on the magnitude of the DC leakage currents in the diodes. The leakier one leaves its brother holding more of the voltage in the inverse of the ratio of the leakage resistances.