What does that bit there do? Learning to understand circuits.

[WGTP]

OK, I'll comment on the aspects I focus on for Tone Tweaking.

R3 - sets input impedance which in turn effects the high's getting into it.  100K would probably roll some off.  Pre-gain

C3 - I use this for some low freq roll off.  With R3 at 1M C3 at .1uf the low roll off point is around 1.6Hz, no effect.  With C3 at .001uf, the lows are rolled off around 160Hz.  Pre-gain

R5 - sets max gain when Drive is at minimum resistance/max gain.  In conjuction with C4 it sets the low freq roll off point for the op amp gain.  With C4 at .047uf it = 720Hz, (if I got my decimal in the right place) which is sort of typical.  The Drive pot lowers this as the resistance increases and gain decreases adding bass at lower drive levels.

C4 - see above.  I think C4 is very important in setting the "Character" of the distortion.  Halving it raises the roll off point to 1440Hz, decreases gain and makes it more Distortion like IMHO.  Doubling it lowers the roll off point to 360Hz increases gain and makes if more Fuzz like IMHO.  

R5 in conjunction with C4 allow the gain and roll off point to be adjusted and along with the diodes etc. determine the "Character" of the pedal.  Consider using switches here.

R9 - sets gain for op amp.  Some distortions use a pot here and have the R5 fixed.  Larger than 1M is not usually seen as it creates to much noise.  470K and 100K are other typical values.

C5 -  can be used to roll off post op amp bass, but I usually make it large enough to minimize bass loss at this point.  Bass reduction before the op amp is usually preferred for smoother distortion.

D1 & D2 - Almost limitless possiblities.  Again, switching recommended.  

C6 - High freg roll off.  as is .001uf = approx. 16Khz.  .01uf = 1600hz and might be more desireable, with any value inbetween tweaked to taste.

I see this as a 2 stage distortion.  At first the diodes will clip and when the Drive is high enough, the op amp will start clipping as well.  Different tweakers like different op amps.  Let me know if I screwed any calculations up.  

[Fret Wire]

C3 - I use this for some low freq roll off.  With R3 at 1M C3 at .1uf the low roll off point is around 1.6Hz, no effect.  With C3 at .001uf, the lows are rolled off around 160Hz.  Pre-gain

Another error on the schematic: C3 is .01uf.

Different tweakers like different op amps.

The 741, with it's low slew rate, and the biasing arrangement are definately part of the ckts "character". I get the feeling that it was taken into account way back when it was designed. Try using a TL071 and the 22k/22k/10uf divider from the 250 on the D+ with no other changes, and it feels/acts different.

[dano12]

Another error on the schematic: C3 is .01uf.

Both fixed now.

[R.G.]

If I don't mention it, I agree with the comments.

C1: We haven't seen this one in the previous schematics—a small value cap from the top of the middle of the voltage divider to ground. Purpose?

As noted in my article on bias networks at GEO, the capacitor forms a low AC impedance from the bias point to ground. This is the same function that the JFET biasing network I was describing in reply #31 works. It shunts noise from the bias resistors to ground and holds the DC level relatively firm in the face of signal wiggling the other end of the bias resistors.

C2, C3 and R4: What is this? It looks a bit different than our previous examples of the input stage that forms a high-pass filter and stabilizes the bias point. Hmm.... Is C2/R4 the high-pass filter, and if so, what does C3 do?
C2 is a low pass filter. It's there to shunt high frequencies to ground before they get into the circuit. What's the cutoff frequency?
No way to tell. The resistance that makes this a low pass filter is back inside whatever drives this. If you feed it from a direct guitar signal, then the 1nF cap combines with the cable capacitance to shunt treble to ground within the audio band because of the highly inductive nature of guitar pickups. In that way, it smooths the distortion by cutting the size of the treble components in the incoming guitar signal. It cuts treble more with more highly inductive pickups, like humbuckers.

General Question 1: For the Distortion + specifically, and pedals in general: does the opamp perform any of the clipping? Or is it all the diodes, or a combination of both?

Let's think about it.

For a diodes-to-ground clipper, there is a 10K resistor in series with the diodes. The output voltage across the diodes increases without the diodes having much to do with it until we hit the beginnings of the diode's knee. From that point on, a change of about about 200mV changes the diodes from not playing to sucking all the current they can. The opamp can only put out maybe 3.5V across that 10K resistor (4.5V minus about a volt that it can't swing near +9V or ground), so the max current the opamp could ever supply to these is about 350uA (3.5V/10K). The diodes reach their clipping at about 0.7V. The rest of the way to 3.5V changes the diode voltage not much at all. When the opamp stops providing current, the signal is already almost flat. Which contributes? I'd say the diodes are doing vastly more of the work of clipping than the opamp because the diodes are already almot flat-topping the signal when the opamp quits being able to push the diodes harder. The diode knee has aready contributed whatever softness to the edges that it can by the time the opamp can start clipping the signal, so the opamp clipping is hidden by the diode's having already limited.

Is opamp clipping desirable?

Good question.  What happens when an opamp clips?

Because the opamp has a huge gain that can be spent on making the output be exactly what the feedback network and input signal say it should be, the excess gain hides any internal changes that happen as long as the opamp has excess gain to hide changes. When an opamp clips, the output circuit is no longer able to pull the output high enough (or low enough) to match what the feedback network and signal say it should be. This amounts to a lower open loop gain. And the open loop gain is what is hiding the internal loss of gain! What happens is that an opamp runs out of gain with a bang. At some signal level, it can simply no longer keep up the facade that everything is OK inside, and the output signal simply flat-tops. It happens FAST. High feedback clipping thresholds are some of the sharpest-corner clipping you can get. This makes for synthesizer-sounding buzzy edges with lots of high harmonics. On mixed-frequency signals - like two or more note chords - you get mounds of intermodulation distortion.

Whether that is desirable depends entirely on your musical taste. It is not all that musical, but then neither is a cow bell, and I hear those in songs too. But it's not the soft, smooth, creamy distortion that a lot of people like.

What are the general design parameters for mixing opamp clipping with discrete diode clipping?

The D+ *is* discrete diode clipping IMHO. The diodes clip so much before the opamp that the opamp adds little. It's not zero, granted, but little.

R9/C4/R5/R6: Ok, it looks like this is the feedback loop. The signal passes from the output of the opamp, through R9 back into the inverting input of the opamp. Here's my guess at what is happening. The more of the output signal that is fed back into the inverting input, the less gain the opamp is going to produce. The R6 pot bleeds more of this inverted signal off to ground as you turn it up, hence more gain. Is this correct, and what is the purpose of R5 and C4?

C4 blocks DC from going through R6 and R7. So for DC purposes, the gain is alway unity, and the bias voltage is not multiplied by the AC gain. This cap forces the output DC voltage to be whatever DC voltage is on the + input, as long as the opamp can possibly make that true.
R9 is the feedback resistor and R5+R6 is the value of the input gain setting resistor.

A noninverting opamp will have a gain of unity (because the output always follows that + input, henca a gain of one) plus Rf/Ri. In this case, the gain is 1+(R9/(R5+R6)). R5+R6 can be as much as 1,004,700 ohms or as little as 4700 ohms, giving gains of 1+ (1) up to 1+213. So changing R9 changes the gain from 2 to 214 for all signals above the C4/(R5+R6) rolloff. That changes the output signal level, and that in turn changes how soon the diodes clip from the input signal and how hard they clip. Your description of R9/R5/R6 is correct. Changing the setting of R6 changes how much output signal current gets back to the - input node, and so the gain.

C5 and R7: C5 blocks DC voltage from leaving the opamp stage. R7? Not sure....

C5- yes. R7 forms the top of the diode "voltage divider" and limits current into the diodes. Actually, R7 is a form of clipping control as well. As we saw above, R7 limits the total current that the opamp can put into the diodes before hitting opamp clipping because the output voltage can't go high or low enough.

C6: Small cap to ground: what does this do?

It's part of the diode "voltage divider" but a frequency dependent part. R7/C6 forms a low pass network that keeps the signal at the diodes from being as large when the frequency is high. So the voltage at which the diodes will clip starts decreasing at the R7/C6 rolloff frequency and you get less distortion on treble.

Question 2: What does it sound like if we only clip off the top, or just the bottom? In other words, would it be of any use to play around with having only D1 or D2 in circuit?

In this case, only the side with the diode is clipped - at least until you run into opamp clipping on the non-diode polarity. What happens is that one side is heavily clipped, one is not. This will produce an octave effect, as you are introducing heavy non-symmetrical distortion.  The signal level increases because you now no longer have both sides limited to one diode drop, so the un-limited side can go to maybe 3.5V, and the total signal is about 4V peak to peak instead of 1.4V peak to peak.

Question 3: There was a posting a while back from a fellow who had created a clipping circuit that had a very odd shape. I think it was like devil horns, or some similar analogy—does anyone recall this?

Yes.

Here's another question. You know how R7 works with the diodes. What happens if you put a 1K resistor in series between R7 and the diodes, and take the signal output from the junction of R7 and this new resistor between R7 and the diodes?

[Fret Wire]

Question RG: R7/C6 form a lowpass of 15.9khz. One of the popular mods for an existing D+ is to lower R7's value for more output volume, 5k is popular. The lowpass filter is now at 31.8khz, if my math is right. C6 would have to be raised from .001 to .0022uf to get the lowpass back to 14.5khz if you want more volume, but the original rolloff.

But now R7 is halved and more current is going to the diodes (Ge 1N270's). Is this going to obliterate any roundness hitting the knee, or will it even be noticeable with Ge diodes? Would halving R7 be more noticeable with si or led diodes?

Come to think of it, but the popular mod for more volume when building a D+ from scratch, is to use a 100k volume pot (leaving R7 at 10k). I can't help but think the volume pot value also interacts with the values of R7/C6. Correct?

[John Lyons]

Just a friendly bump.

There were some un answered questions and it would be great to keep this going.
Yeah, it's an old thread...but it's a good one.

John

[JOHNO]

what a great read.bump

[R.G.]

Question RG: R7/C6 form a lowpass of 15.9khz. One of the popular mods for an existing D+ is to lower R7's value for more output volume, 5k is popular. The lowpass filter is now at 31.8khz, if my math is right. C6 would have to be raised from .001 to .0022uf to get the lowpass back to 14.5khz if you want more volume, but the original rolloff.

But now R7 is halved and more current is going to the diodes (Ge 1N270's). Is this going to obliterate any roundness hitting the knee, or will it even be noticeable with Ge diodes? Would halving R7 be more noticeable with si or led diodes?

There's no simple answer. Some of all of that at the same time. Sorry - Mother Nature says so. Some things She does not necessarily make easy for us.
Yes, if there is no other change, decreasing R7 to 5K kicks the highpass rolloff up. That may or may not be a factor, since it was already near the top of the audio range anyway. Us old guys have ruined our hearing listening to loud music and can't hear all that much over 10-12kHz for those of us that can still hear at all. Many younger guys too. There may or may not be much audible change in decreasing R7. The higher diode drive will be noticeable With Ge as a  little more volume, not much, but a harder sound. It is difficult to say whether the actual volume increases much or whether this is a psychoacoustic effect of the harder clipping.

If you sub in Si or LEDs, the sound will certainly change. But at least part of that is that you can't drive them as hard as you drove the Ges because there just isn't enough voltage from the opamp. Here's a secret of distortion: it's not so much what kind of diode you use for clipping, within reason, it's how far above its clipping level you drive it. Ge is kind of special because it's pre-clipping region is so small, comparable to its knee region. But after that, whether it's Si or LED, you can look at whether the incoming signal just barely crests the clipping region or is 10x the clipping voltage for how sharply it will clip.

Opamps do this too, but in a slightly different way. For an opamp banging against its power supply limits, the clipping is always sharp. That's because the signal that matters to the opamp is a virtual signal - it's the one inside the feedback loop that would exist if the feedback loop was opened. Opamps may have all kinds of odd distortions and nonlinearities inside the feedback loop where you can't get at them. In fact, that's WHY opamps exist - they are designed to make it easy to hide those distortions under the feedback. All amplifiers do this, but opamps are the logical conclusion, with such high gain that they can hide whatever internal signal exists.

Come to think of it, but the popular mod for more volume when building a D+ from scratch, is to use a 100k volume pot (leaving R7 at 10k). I can't help but think the volume pot value also interacts with the values of R7/C6. Correct?

Actually, not all that much, I think. The impedance of the R7/C6/diodes is well below 50K AND 100K over most of the audio band. The bigger pot lets any following cable capacitance have more effect, I guess. I would not look at the volume pot as a major source of tone change.

[Lenni]

Hey folks.

I'm new to the forum and the whole electrical engineering stuff and just try to teach myself all the basics.
Last week I started to read theses Electrical Engeneering Training Series which I found in this Thread.
I just got a very simple question to it: What is the difference of Ampere and current flow? Isn't that basically the same? I'm just confused because they always use both terms.

Thank you.

[R.G.]

Amperes are the unit of current flow.

An ampere is the flow of one coulomb per second.  Right - so what's this coulomb thing?

A coulomb is similar to a gallon of water. Water molecules are too tiny to count, so we simply say that one gallon is 23432832489782193478 molecules of water in a gallon, and then ignore the number from then on, only referring to gallons.

In the same way, a coulomb is the charge you get by accumulating 6.241 times ten to the eighteenth power electrons. And that many electrons flowing in one second is an ampere. I made up the number of water molecules in a gallon, but the number of electrons in a coulomb is correct to three decimal places - I looked it up.

Water is a commonly used analogy to electricity. Pressure is to water what voltage is to electricity. Rate of flow is to water what amperes is to electricity.

[Lenni]

Thanks R.G.

I already knew the coloumb thing but the water example is pretty nice.
Thank you for your answer. I just tried to avoid learning things wrong.

[Lenni]

And again I've got a little tiny questions which bothers me a lot.

My problem is in the formula for amperes I=E/R. For example we have a DC source with 1.5V and a 1ohm Resistor we would obviously have 1.5amps right?! But if we have for example a Resistor smaller than 1 or even no resistance the amps would obviously be higher than the Voltage? Is that actually possible? How could it be that more electrons are going from the - Terminal to the + Terminal in one second than the potential difference is? Did I get something wrong? What happens at 0ohms resistance or does the electrical engineer doesn't care about it because as soon as you have a conductor you have a resistance?

Thanks folks!

[pjwhite]

And again I've got a little tiny questions which bothers me a lot.

My problem is in the formula for amperes I=E/R. For example we have a DC source with 1.5V and a 1ohm Resistor we would obviously have 1.5amps right?! But if we have for example a Resistor smaller than 1 or even no resistance the amps would obviously be higher than the Voltage? Is that actually possible?

Sure it's possible.1.5 volts into a 0.1 ohm load would give you 15 amps, for example.  Not that you would get that much current out of your typical AA battery -- batteries have an internal resistance that must be added to the circuit load to get an accurate calculation.  The connecting wires in your circuit also have a resistance that must be accounted for.  We usually can ignore wire resistance in our calculations, as it is so small as to be insignificant in most cases.

How could it be that more electrons are going from the - Terminal to the + Terminal in one second than the potential difference is?

That's kind of like saying, "How can a car go 50 miles per hour when it only travelled 1 mile?"

Did I get something wrong? What happens at 0ohms resistance or does the electrical engineer doesn't care about it because as soon as you have a conductor you have a resistance?

Unitl you get into the realm of superconductors, all wires have some resistance.

[Lenni]

Ok I guess I should answer my question a little bit different.
For sure it makes sense that 1.5V with 0.1 Ohm makes 15amps BUT (here we going to your need car example that I totally don't understand) If the + terminal has 1.5 x 6.28x10^18 less electron than the - terminal why should suddenly 15 x 6.28x10^18 run over there? puh that scares me that those things already bringin me in trouble.
Is it probably just that ampere = coloumb per second thing that makes me feel so stupid?

[R.G.]

If the + terminal has 1.5 x 6.28x10^18 less electron than the - terminal why should suddenly 15 x 6.28x10^18 run over there?

Let's go back to water and hoses.

The water flows from the output of the pump through the hose because there is a pump pushing it. Electricity flows because there is a voltage pushing it - a battery or a generator or some such. There are not a spare coulomb or two of electrons sitting around at the negative terminal looking for something to do. The electron pump forces just enough of them to the negative side and away from the positive side to be X volts of electron pressure.

Electron pumps, like water pumps, can make the most pressure when there is least flow. So if you bottle up the pressure of a water pump, you get maximum pressure. Let some of the water out, and some of the work that was making pressure is now moving water, and the pressure drops as the flow comes up. You get maximum flow when the pump is just spraying into air, no restriction at all.

An electron pump holds the maximum pressure when there is no electron flow - no current. When you let some electrons flow, the pressure drops a bit because of losses inside the electron pump. If you short the + and - terminals together, the current flow is all that the pump can do.

An electron pump - battery or other voltage source - recirculates electrons. They flow out of the - side, through the circuit, and back into the + side. So even through many, many coulombs flow, it never runs out of electrons to pump.

So a (-) terminal doesn't have more electrons than the (+) terminal (Well, Ok, it does, but only enough to make that many volts). What it has is a pump behind it that will recirculate as many electrons as it takes. And matter contains many, many more electrons than the coulombs of electricity that flows, so you don't run out of electrons.

[Lenni]

First of all thank you very much. That is a really nice discribtion and makes it a bit easier to understand the whole thing. BUT! It doesnt' really answers my question (actually no wonder now where I look at how I asked the question ...) anyways. I ask it again in the hopefully right way
Why does a resistor between 0-1ohm increases the courrent flow and a resistor above 1 decreases it? Shouldn't a resistor always decrease the courrent flow?

Hope I'm not to anoying for you guys!

Thanks for your help!

[R.G.]

Why does a resistor between 0-1ohm increases the courrent flow and a resistor above 1 decreases it? Shouldn't a resistor always decrease the courrent flow?

Ah. OK, I think I see.

Your concept is incorrect. It is not true that resistors between 0 and 1 ohm increase current flow and above 1 ohm decreases it.

A resistor of 0 ohms is only possible with superconductors, which are not encountered in our kind of electronics anyway. So let's just say that a resistor of 0 ohms is impossible (for us, in practical use). In reality, even a copper bar a square meter in cross section has some resistance. It may be a micro-ohm, but there will be some resistance. In fact, let's just say that it IS a micro-ohm, a millionth of an ohm, and see what the math tells us.

Ohm's law tells us that the current which flows when there is a resistance of value R connected across a voltage V will be equal to the voltage V divided by the resistance R, or  I = V/R

It also says that the voltage V which happens when we force a current I through a resistance R is V = I * R.

In fact resistance itself is defined as the ratio of the voltage across a circuit element divided by the current through it, R = V/I. As you recognize, these are the same relationship, just solved for the three different elements. When you know any two of them, you can ALWAYS calculate the third one from the other two.

So a one ohm resistor is a component which causes one volt of difference from end to end when one ampere (that is, one coulomb per second) is flowing through it.  Our copper bar with its millionth of an ohm, is really one one-millionth of a volt per ampere. That is the same as one volt per million amperes. The exact amount of current flow does not matter - only the RATIO between volts and amps.

So with a copper bar of 1/1,000,000 ohm resistance, the current that flows is the voltage V across it divided by one one-millionth, or a million amperes per volt. Let's increase that to only one one-thousandth of an ohm. Now we get a voltage of one thousand amperes per volt. OK, go to one ohm.

The current that flows is one ampere per volt.  (it's still I = V/R, for the V and the R you have).

At one thousand ohms, the current is I = V/1000 or one one-thousandth of an ampere, which we call a milli-ampere.

At one milliion ohms, the current is I = V /1,000,000 = one micro-ampere.

There is no break point at 1 ohm. It's a smooth change, and you can always calculate it by I = V/R; or V = I*R; or R = V/I, which are all the same thing, algebraically.

It is possible you are confusing series and parallel circuits. What happens when we have two resistors? If I have a voltage V and a resistor R, then the current which flows when I put the resistor across the voltage is I = V/R.

But what if I now add a second resistor R2 in parallel with the voltage source and with R? The voltage across each resistor is the same because they both have their ends tied across V. So two currents flow. One is I = V/R and the other is I2=V/R2. So the total current coming out of V must be the sum of the two, I + I2. Adding more resistors in parallel increases the current flow because each resistor demands and gets its own current equal to the voltage divided by its resistance. Now, hold on for some math.

Let's define a third resistance Rp to be the resistance we get when we divide V by the sum of the two currents that come out of V with two resistors. That is, Rp = V/(I+I2). We know that I = V/R and I2 = V/R2, so we substitute that in.

Rp = V/(V/R+V/R2) There is a math trick that lets us write (V/R +V/R2) as V*(1/R+1/R2), so we can say
Rp = 1/(1/R+1/R2)
That looks clumsy, but it's the way I usually use the equation for parallel resistors. If R = V/I (that is, volts per amperes) defines a resistance, Y = I/V (that is, amperes per volt) defines a conductance, how well it conducts, not how well it resists. By adding resistors in parallel we are adding additional conductances, more paths to let current flow, and thereby make more total current flow.

In the only-two-resistors case, the algebra lets you come up with Rp= R*R2/(R+R2) which is how most people use it. The conductance view seems more intuitive to me, and it works for an indefinite number of resistors, not just two.

There is an other way to hook those two resistors up. If I connect them in series so the current must flow first through R then through R2, what happens. Well, we know that if all the current that flows must flow through both of them, then the current must be identical in both of them, right? So there is only one current I in this case, not a unique one in each resistor. The voltage in the two resistors must be
V1 = I*R1 and V2 = I*R2.

We know that since the total voltage V is all we have, that the two resistor voltages must add to be V, or V = V1 +V2. That means that V = I*R1 + I *R2. We can use the math trick that I*R1+I*R2 = I *(R1+R2) and get that V = I *(R1+R2).

From that we see that resistors in series add as ohms. So adding resistors in series with each other adds resistance and lowers the current that flows. Resistors in parallel add as conductances and let more current flow as you add more resistances.

But each separate resistor lets current flow that makes the voltage across it be equal to the current through it times the resistance. Always.

[Lenni]

Allright now I got it! Suddenly it seems more logical. 0 ohm would actually mean unlimited current flow right? For some reason I was thinking that the current flow can't be higher than the Voltage. For example a Batterie with 1.5V can't deliver 3amps because it only has 1.5V.

Anyways! Thanks so far but you just mentioned another problem that I have.

There is an other way to hook those two resistors up. If I connect them in series so the current must flow first through R then through R2, what happens. Well, we know that if all the current that flows must flow through both of them, then the current must be identical in both of them, right?

Why does that current has to flow threw both of them? Why doesn't the first resistor decreases the current a bit and the second one a bit more. I mean the current first flows threw the first one and than threw the second one right? But why is the effect of the two resistor coming right after the electrons leaving the source? What I don't understand is why can you measure ... lets say: 3amps ... all over a circuit with 18V, a 2ohm resistor and a 4ohm resistor?

[R.G.]

Because electricity must flow in closed loops.

An electron pump (like a battery) can produce 18V. If you put one resistor across it, some current I = 18/R flows. The current going into the resistor must be equal to the current coming out of the resistor because electricity must flow in closed loops. If we want the current I to be 3 amps, then we can calculate that R must be 6 ohms.

But the same result happens if we have one resistor of 6 ohms, three resistors of two ohms each in series, or six resistors of one ohm each.

The voltage across whatever string of resistors we put there will be the same - the 18V of our supposedly-perfect battery. Since current out has to equal current in to the battery (HAS to - otherwise the battery gets charged up with a static charge because it's exporting charge, right?), then the total current out is equal to the current in.

If we put a 1 ohm resistor across it, we get I = 18A. If we put a second 1 ohm resistor in series with the first oh, and the series combination across the battery, then the total voltage across two resistors is still the 18V. But since the resistors are identical, then there must be half the voltage across each one, right? So each one-ohm resistor has 9V across it, and we know that I = V/R = 9V/1ohm = 9A. And since what's going through the first resistor can't go anywhere else but through the second resistor, we get that the total current is just the same as the battery current of 9V.

Notice that we get that same current if we use one resistor of 2 ohms instead of two resistors of one ohm.

Now we add a third 1 ohm resistor in series. The voltage gets distributed across three identical resistors, so each one must have 1/3 of the total, or 6V across it; and the current in each one must be 6V/1ohm = 6A. That's the same result we'd get if we put one three-ohm resistor across the 18V battery.

Current always goes SOMEWHERE; Nature says it goes in closed loops except in the rare cases where it's piled up on something in the form of static electricity. In electronics, it flows in loops out of and back into voltage or current sources. Series resistors can't decrease the current a little more each step - they all have to work on the same current at the same time, because it's all they have to work with.

In your question, an 18V source has a 2 ohm and a 4 ohm resistor in series across it. From the little thought experiment above, we know I can change the 2 ohm to two one ohms in series and get the same result. I only did up to N=3 resistors, but the same applies for as many resistors as you'd like; so the 4 ohm can be replaced by four one-ohms in series. Then we have six identical one-ohms, each of which has the same identical current flowing through it, and 1/6 of the voltage across it. So each resistor gets 3V across it, (18/6) and the current through it must be 3v/1ohm = 3A. And it's the same for all six identical resistors.

[PerroGrande]

Hi Lenni,

Yep -- a truly zero-ohm object would imply infinite current as I = V/R which goes to infinity as R approaches zero.  Of course, even with superconductors, the capability of a source to provide current is limited due to its own internal resistances, etc.  Perfect current sources, along with frictionless surfaces and massless springs, can be found at your local Physics store... (or perhaps Acme, if you're a fan of the old Warner Brother's cartoons)   

When considering series circuits, the current flowing through each of the devices in the series is the same.  In a series circuit, there is only one path for the flow of current.  The movement of charge (i.e. the current) is the same everywhere throughout a series circuit.  Charge doesn't pile-up or accumulate, nor is it used up by the resistors.