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

[Lenni]

Sorry R.G. but I still don't get it.
I already "accepted" that two 1ohm resistors are actually the same than one 2ohm resistor but your description just confuses me even more.
Puh I don't know how else to answer the question.

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.

All that I can say to that is just: Why? I read it 10 times now but I don't understand it. 18/2 = 9 yeah that's easy but that resistors suddenly decreasing the voltage but not just drives me crazy. I'll think about it a bit more but I would appreciate if you could try it again.

Thanks so far!

[PerroGrande]

Hi Lenni,

Seems like double-speak, doesn't it?  On one hand, we say that resistors limit the flow of current and at the same time, we're saying that current through each device in a series circuit is the same... 

huh??

Well, let me try to take a stab at this...

Let's say we have a battery that says "100 Volts" on it...  This means that there is a potential difference of 100 volts between the side marked "+" and the side marked "-".   If I build an incredibly exciting circuit with this battery and a 100 Ohm resistor, something is going to happen...

Turns out that a long time ago, George Ohm figured out an "easy" explanation which we still call "Ohm's Law." 

Our circuit above isn't very exciting, but it qualifies as a circuit.  It contains a complete path (from + to -) and there is a potential difference.  With these criteria met, an electrical current can flow. 

Current is a measurement of how many charged particles fly past a point in a second.  We measure this with an Ammeter, and the unit of measurement is the Ampere (1 Coulomb of charge in 1 second). 

Sticking this ammeter in our little circuit  would show a measurement of 1 ampere.  (I = V/R = 100 Volts / 100 Ohms = 1 Ampere).

Okay -- so we've got a circuit that does nothing, right?

Wrong!

We have 1 ampere of electrical current flowing through a resistor.  It *is* in fact, doing something.  Yeah -- its draining our battery, but in addition to that, it is converting one form of energy to another -- in this case, electrical energy is being converted to HEAT by the resistor.  How much heat energy?  Well, we've come to learn that P (power) = IV, which for our little circuit is P = 1 Ampere x 100 volts = 100 Watts.  Yikes!  That resistor is getting HOT!  (Ever try to touch a 100Watt light bulb...)

So our "useless" circuit *is* in fact doing something...

But is the resistor "limiting" the current?   In the famous words of Sarah Palin, "ya betcha it is."

As you noted, if I = V/R, and R gets very small, I has to get very big...   If I reduced my resistor to 10 ohms, I would now have 10 Amperes of current.  Drop the resistance to 1 ohm, and I now have 100 Amperes... Drop it to .1 ohms, and I've got 1000 Amperes (welding, anyone?)...  All this, of course, assumes that our crazy 100 Volt battery can produce this much current...

So yeah -- the resistor IS limiting the current... 

Furthermore, it doesn't matter where in our circuit I connect the ammeter.  If I put it between the + side of the battery and the "top side" of the resistor it reads 1 Amp.  If I put it between the - side of the battery and the "bottom side" of the resistor, **it still reads 1 Ampere.** 

** The same current the flows into the resistor flows out of it. **

Now -- George Ohm got the crazy idea of hooking two 50 Ohm resistors in series with one another.  Lo and behold, the ammeter read exactly the same thing as it did with 100 Ohms (1 Amp)... If you were to try it today, 100+ years after George Ohm tried it, it would read 1 Ampere... 

So, from where our ammeter sits, there is nothing different about two series 50 Ohm resistors than one 100 Ohm one.  And just like before, it doesn't matter where the ammeter is placed, we still show the same 1 ampere.

In fact, George "O" got all crazy and tried connecting the ammeter in series between the two 50 ohm resistors...  Still... it read 1 Ampere!!

The reason is that the same current that "comes out of the plus side of the battery" returns to the negative side.  Current doesn't leave the series circuit part way through, and no external device is adding extra current somewhere in the middle.  We merely have our resistors and our battery. 

So -- for our series circuit, the total resistance that the battery "sees" and the voltage of the battery determines the current that will flow through the entire circuit.  The "total" current is limited by this resistance (for if it was truly zero, the current would be infinite).  If the resistances are in series, the current *has* to be the same through each of them (we're not taking any current away and using it anywhere else, and nothing external is providing us with current).

[Lenni]

Hm ok that all makes sense. Especially the part of "What leaves the batterie, goes back in the batterie". Where else should the current go right?!
I guess my actual problem is that I don't understand how the batterie "knows" how much current to send out?! because the Resistor drops the voltage? But how does it drop the voltage?

I draw a little circuit for the whole thing and how it would be logical for me. I know that it obviously has to be wrong since we know that what comes out, comes back in. I guess my question is: How does a resistor do its job and on what? Is it reducing the voltage and so less current is flowing? That was my idea on this example. On the left side of the resistor are many electrons (the little red dots) and on the other side are less so there is obviously a potential difference and on the right the whole voltage is dropped. How does that potential difference really look like? Probably we can figure out my problem with solving this wrong thinking.

[lazerphea]

Hi all!
I'm new to this great board, and I started lurking this thread a couple of weeks ago. Thanks to all the infos i found here I managed to build (over my breadboard ) the Booster, the OpAmp buffer and the Distortion+ with satisfying results; I then moved on to the Fuzz, even thought I don't like the effect very much, I thought that could be a nice exercise to practice a bit.
Now the problems: the circuit outputs a huge signal that saturates as soon as I lightly touch my guitar's strings...
The only mods I did to the original circuit have been the substitution of the 150k and 18k resistors (R3 and R2) with some resistors in series, because I didn't have the right values at the time.
I then started to debug the circuit, and I found out one strange thing: when I measure the 330k Resistor (R1) out of the circuit, my multimeter reads the correct value, but when I put it in place, it drops to 121K and I don't know why, because I take the measure exactly at its pins... could this have something to do with the fact that it's a 0.5V resistor and not a 0.4V one?
Another thing: I measure 2.60V at the Q1 base and 2V at the emitter... can someone help me please?

[lazerphea]

Here I am again!
I thought it could be useful to show exactly how I wired all the stuff up Here's the pic:



(if the image doesn't open here's the direct link: http://img150.imageshack.us/my.php?image=hemmofuzz2rv6.png)

Note: the pinout of the two 2n3904 seem to be E-C-B, front facing the flat side, according to my multimeter... 

[PerroGrande]

Hm ok that all makes sense. Especially the part of "What leaves the batterie, goes back in the batterie". Where else should the current go right?!
I guess my actual problem is that I don't understand how the batterie "knows" how much current to send out?! because the Resistor drops the voltage? But how does it drop the voltage?

I draw a little circuit for the whole thing and how it would be logical for me. I know that it obviously has to be wrong since we know that what comes out, comes back in. I guess my question is: How does a resistor do its job and on what? Is it reducing the voltage and so less current is flowing? That was my idea on this example. On the left side of the resistor are many electrons (the little red dots) and on the other side are less so there is obviously a potential difference and on the right the whole voltage is dropped. How does that potential difference really look like? Probably we can figure out my problem with solving this wrong thinking.

Lenni,

One of the problems with your picture is that you show electrons getting "lost" in the resistor.  This isn't happening.  All the electrons that leave the battery return to it.  Otherwise, our ammeters would show different currents at the positive and negative sides of the circuit.  This doesn't happen, so all our electrons are getting back to their home base.

What you're not seeing/showing in that picture is the work that the resistor forces the electrons to do.  The chemical composition of the resistor causes a conversion of some of the electrical energy into heat. 

Another thing the drawing does not show is that the electrons are being "pushed" by a pressure -- we call it voltage.  The resistor, by forcing some work to be done to make heat, causes a pressure drop -- or, as we see it, a voltage drop.  A common analogy is that the charged particles are forced through a "narrow" area, and while the same number emerge (current is constant), some the "push" behind them is gone (i.e. voltage drop).  Older textbooks referred to voltage as "Electromotive force", and you'll occasionally see Ohm's law represented as E = IR (which is how I learned it, so I'm dating myself again).  Resistors cause some of the electromotive force to get used up doing things inside the resistor (heating up, among others).  So, for your picture, the same number of dots emerge, they just have a little less "push" behind them.  The resistor ate some of the "push" -- not any of the dots.

[PerroGrande]

The battery doesn't really know how much current to send out...

A battery will have certain characteristics that are largely governed by its physical makeup and chemical composition.  Depending on these factors, it will have a typical voltage (which, for purposes of illustration we will say is a "fixed" value -- although in reality it is not).  Different chemistry will lead to different "fully charged" voltages between the positive and negative sides of a cell.  Various other factors, including plate area, determine the battery's ability to provide current to a circuit for a period of time.  AAA, AA, C, and D size batteries all show a terminal voltage difference of ~1.5 volts, but their ability to produce a given current for a given period of time varies (larger battery generally = more current for longer time).

The battery attempts to provide the current that the circuit is asking for -- and Ohm's law tells us this figure.  The circuit's total "effective" resistance to ground ultimately determines how much current it requires.  This is why you'll see designers saying something like, "this circuit draws 20mA of current at 12 Volts".  The 12V battery may be able to provide 2,000,000,000 Amperes (a *really* freaking large battery), but our example circuit will draw 20mA from this battery or from a much smaller lantern battery or series-connected AAA batteries.

[PerroGrande]

Hey Lazer,

Can you link the schematic you're talking about?

[R.G.]

Even better, please post your question and schematic in a separate topic so it's got an appropriate title and won't either get overlooked at the bottom of this one, or be lost in the archives under a wrong topic.

[lazerphea]

Perro, R.G: I'll create a new thread with the schematic included! Thanks!

[Lenni]

Ok That the resistor isn't eating electrons is logical (how I already posted with the schematic, I know that this way is wrong. I just don't get why ). You said that the resistors just taking some of the "push" means Voltage drop right? But how can there be a voltage drop or potential drop if there is still the same amount of electrons on both sides? Isn't the difference of the amounts of electrons on two sides actually what makes the potential difference?

So resistors are basically lowering the "speed" or "pressure" of the electrons. That's something that I can easily understand, but what does the Voltage drop actually means than? Back to my little schematic up there: electrons coming to the first resistor, lowering the speed to 50% (Voltage drop of 50V from 100V makes 50%, tada that boy can calculate) and than takes the rest of the 50% => 0% = no speed. Do they speed up again because the battery suddenly "pulls" on them or is the speed idea just bullshit?

[Lenni]

I was just continuing with the NEETS and suddenly I saw this expression ( 50μA ) without that it had been explained. I was pretty confused what that actually should mean? what is that &mu standing for? Does is mean &mu per Ampere?
I would really appreciate if somebody could tell me what this stands for.

Thank you!

[DWBH]

In order to give this thread a continuation...

... what does the marked resistor do? Some sort of 'negative feedback'? 

[R.G.]

It does multiple things.
(1) it provides the steady state base bias current.
(2) it stabilizes the DC bias point thus provided by providing negative feedback from the collector. If the collector goes down due to something like the transistor getting hotter and conducting/leaking more, the voltage across the resistor goes down, and so the base current goes down, tending to raise the collector.
(3) it provides AC negative feedback: If an input signal goes up, the collector goes down; the collector going down reduces the current supplied to the base, lessening the effect of the input signal.

This is the so-called voltage feedback transistor bias circuit. It was one of the earlier versions of transistor biasing.

[ballooneater]

I was just continuing with the NEETS and suddenly I saw this expression ( 50μA ) without that it had been explained. I was pretty confused what that actually should mean? what is that &mu standing for? Does is mean &mu per Ampere?
I would really appreciate if somebody could tell me what this stands for.

Thank you!

mu (µ) is a greek symbol for "micro", meaning to multiply by 10^-6. It's similar to mega, milli, centi, and other such prefixes, most of which can be found here.

In this case, there would be 50*10^-6 (0.00005) amperes

[Lenni]

oh ok. Yeah with a little brainwork I actually should have figured that. But what does the & symbol in there means? Is that just a mistake and a sign which shouldn't be there or is it just me?!

[ballooneater]

oh ok. Yeah with a little brainwork I actually should have figured that. But what does the & symbol in there means? Is that just a mistake and a sign which shouldn't be there or is it just me?!

It's a mistake.  In HTML, special characters are often proceeded by the ampersand (&) sign.  The person who wrote this must've typed it wrong.

[kurtlives]

It does multiple things.
(1) it provides the steady state base bias current.
(2) it stabilizes the DC bias point thus provided by providing negative feedback from the collector. If the collector goes down due to something like the transistor getting hotter and conducting/leaking more, the voltage across the resistor goes down, and so the base current goes down, tending to raise the collector.
(3) it provides AC negative feedback: If an input signal goes up, the collector goes down; the collector going down reduces the current supplied to the base, lessening the effect of the input signal.

This is the so-called voltage feedback transistor bias circuit. It was one of the earlier versions of transistor biasing.

Isn't that form of biasing a bit unstable RG? Doesn't it vary quite a bit with temperature (even though it is Si). If one put a resistor from base to ground that would be a different method of biasing. Then the biasing and transistor would be more stable. I think the gain would drop a lot though.

That correct? More or less...

[R.G.]

Isn't that form of biasing a bit unstable RG? Doesn't it vary quite a bit with temperature (even though it is Si). If one put a resistor from base to ground that would be a different method of biasing. Then the biasing and transistor would be more stable. I think the gain would drop a lot though.
That correct? More or less...

All forms of biasing vary with temperature, just some less than others.

The first transistor bias was simply a resistor to V+. Ugly, drifty, especially with the germanium devices they used. Next was the voltage feedback form. The four-resistor "stabilized bias" form was one of the last.

We spent about two years one week in circuits class deriving Stability Factors, those being the sensitivity of the bias point to the temp drift of Vbe and Icbo. It's possible (eventually, especially if you have a grade point average to uphold!) to come up with equations based on circuit analysis that show the change of Ic and Vce with changes in Vbe for any resistor network which biases.

The voltage feedback setup gets more stable as the voltage across the feedback resistor increases and as the voltage across Re increases. It's possible to have it be quite stable, or to be downright skittery by the choices of values.

The principle problem with the voltage feedback setup is that it has a low input impedance.

[theGlitch]

bump.

this thread should not have been left alone for over 120days. It's got far too much valuable information.