'Lectrical Circuits 101

General Questions
#1

I want to learn about logic circuits, but my knowledge of the principles of electricity is next to nil. I do know that current, measured in amps, is like water flowing in a pipe. I know that the voltage is like a pump that propels the water. And I know the length of the wire, plus any loads on it, like lights or resistors, offers resistance measured in ohms.

Looking at elementary logic circuit diagrams, I find myself asking several questions.

(1) I notice that the circuit is often split two ways, with one side going to a capacitor and one side going through a resistor. What is this accomplishing? Does the capacitor delay the current on that side? On the resistor side, does the resistor diminish the current.

(2) Also, when a current is split two ways, is the situation analogous to a Y joint in a water pipe, e.g. each side gets half the current?

(3) When additional current is introduced into the system through another input, does that again behave like water, just as when the Missouri flows into the Mississippi, and increase the current in the system by that amount?

Sorry for these very basic questions, but my college physics was a LONG time ago!

Thanks

#2

You’ll have to show me an exampe of your logic circuit that uses capacitors. It sounds to me like you’re describing an integrator or differentiator circuit. The most common logic circuit is a gate, which performs very simple logic functions (see here).

When I used to teach logic gates, my students were already familiar with basic electricity so I could use equivalent circuits built from SPST (single pole, single throw) switches to demonstrate the operation of the gate, and derive the truth table. The following switch configuration demonstrates “AND GATE” operation (–X X-- is a switch in an open state and --X–X-- is a switch in a closed state):



    +5V----------X  X-------X  X---------- output (Y)
                  A=0        B=0                   Y=0

    +5V----------X  X-------X--X---------- output (Y)
                  A=0        B=1                   Y=0

    +5V----------X--X-------X  X---------- output (Y)
                  A=1        B=0                   Y=0

    +5V----------X--X-------X--X---------- output (Y)
                  A=1        B=1                   Y=1

To demontsrate more complex logic functions like inversion, we have to use transistors. Resistors are sometimes used as “pull-up” devices to generate voltage drop & prevent floating grounds where no clear logic level is present. This is really hard to explain in print.

Unless you need to do some kind of filtering, there is really no good reason to have capacitors in binary circuits because TTL logic uses steady-state dc levels of zero & 5 volts for the logical ones & zeros. A capacitor would distort the clean DC levels and cause them to look more like ramps or spikes, which digital circuits don’t particularly like.

Your questions 2 & 3 are really inapplicable in the context of logic circuits, because logic levels are determined by the presence or absense of voltage levels. Changes in current isn’t usually discussed because (although the current needs to be there to generage a voltage level), it’s the presence or absence of a voltage levels that determine what the logic value (1 or 0) is.

Can you show an example of your R-C logic circuit?

#3

If you understand how diodes can act as switches, here is a page that shows logic gates built from diodes. Nowhere can I find any linear logic circuits that use capacitors. Capacitors are used for filtering, shaping & dc blocking.

http://www.play-hookey.com/digital/electronics/dl_gates.html

#4

Thanks for the replies so far.

Attrayant, do you have Visio, and which version? I’ll copy out the circuit I’m looking at and email it (if you don’t have Visio, I think I can do it as a Word document).

#5

I’m looking at your link and it seems to give a pretty thorough explanation.

#6

I don’t know much about logic circuits per se, but I can answer these. Yes to both, except that in a current split, it’s not necessarily 50-50. The sum of the currents going out will be equal to the original current, though.

There are two basic laws for analyzing circuits, called Kirchoff’s Laws. First, whenever you have any sort of juncture (wires splitting or wires joining), the sum of the currents going in is equal to the sum of the currents going out. Secondly, if you take any closed loop in the circuit, the sum of the voltage drops around the loop is equal to zero: In other words, you have to get back to the same voltage you were at originally.

#7

Oops, sorry, I should have mentioned that the book I’m looking at is very old (pub. 1963), so maybe with the technology of that time they did use capacitors. At least I think so: the symbol looks something like this ( with current flowing rightwards):

-------|(---------

and in one of the examples, this symbol is annotated “680pf”.

#8

If you can screen capture the circuit (alt+printscreen, then paste into Word) I’d be happy to take a look at it, but wouldn’t be able to respond until Sunday morning. Others here might help you more quickly. I’ll be online today until about 9PM to check this thread but no access to email. :frowning:

In general you shouldn’t have to understand electricity to understand simple logic circuits… or even complicated ones, really. That’s the beauty of them, in a way. Apart from excessive fan-out (too many components hanging off of a single output) there isn’t much “electricity-wise” to think about.

I think your questions have been answered well enough without trying to delve into AC theory too much. But I will anyway. :slight_smile:

In general, however, capacitors stop DC from flowing and only pass AC. To AC they look almost like a resistor; to DC they look like an open circuit. The equivalent resistance, X, an AC wave sees at the capacitor is determined by its frequency, f, and the capacitance, C, in the formula:
X = 1/(C2pi*f)
(technically, then, as f-> 0, X -> infinity, which is why DC can see capacitors as open circuits; this is really not what happens, but good enough for government work)

When you have a parallel circuit it behaves as Chronos implied. If the resistance down each path is equal then they will receive the same current. If a single AC source is driving the circuit you describe, and if X is the same value as the resistance in the other leg, then each path here will also get the same amount of current.

For the formula to figure stuff out when the values aren’t equal, you must find what the equivalent resistence of a parallel circuit is. It will equal
1 / [(1/R[sub]1[/sub]) + (1/R[sub]2[/sub])]

If you have three paths then you can just keep adding terms in the denominator. If the frequency isn’t changing then you can substitute any “R[sub]n[/sub]” with “X” of a capacitor.

Was this too complicated? Any more questions?

#9

[sub]actually, you cannot “simply” replace X and R, but I didn’t want to get into that unless you were following the rest of it…[/sub]

#10

Yes I have Visio. And yes that’s a capacitor, although back in '63 (before my time) they called them condensers. A 680pF capacitor will only be used to pass or filter out quite high frequencies, which is why I’m wondering why they are appearing in your “digital/logic” circuits.

680pF is 680 picofarads, or 680 ×10[sup]-12[/sup] farads. You can calculate the reactance offered by such a capacitor at a particular frequency by using

X[sub]c[/sub] = (2[symbol]p[/symbol]fc)[sup]-1[/sup]

X[sub]c[/sub] is capacitive reactance in ohms, which is not the same as resistance but can be thought of as similar to resistance.
f is the frequency in hertz
c is the capacitance in farads

So a 680pF capacitor offers about 234 kohms of reactance to a 1KHz sine wave. Reactance decreases by a factor of ten as the frequency increases by the same factor, so a 10KHz sine wave would see only 23.4 kohms from the same capacitor. When you add resistors to a capacitive circuit, then you have to start talking about impedance (Z). On a right triangle, impedance is the hypotenuse, while resistance & reactance are the adjacent & opposite sides.

So a lot has to do with the frequency of the waveform, and things get really complicated when you start talking about square waves because of the extra harmonics involved.

As I mentioned before, distorted square waves don’t agree well with digitcl circuits, and there is a circuit called a Schmidt Trigger that is used to “square-off” or condition a mis-shapen waveform so they will be well received by digital circuits.

#11

A capacitor and resistor wired in series in a logic circuit is probably intended as a signal delay. When the signal on the input side of the R/C network changes, there will be a time delay before the signal on the output side of the R/C network changes. The time delay can be determined from the value of the resistor and capacitor.

Technically the output voltage starts changing the moment the input does, but in a digital logic circuit the voltage has to change past a certain threshold before the downstream gates will see it as a state change.

#12

You guys have given me some fantastic links. I have a lot to look at now.

#13

pF means Pico Farad (10^-12 F). Farad is the unit of capacitance, however typically Capacitors are in the micro Farad (mF or µF) to pico Farad (pF) range. Generally ( i’ve seen many exceptions) the kind of symbol you drew is for Electrolytic Capacitors which have their positive and negative terminals different (its like a rechargeable cell). Other, kinds of Capacitiors (Ceramic, Polyster) have no polarity and are denoted as --| |-- . But , I think your capacitor is actually a Ceramic/Polyster one, because you don’t get electrolytic capacitors in the pF range.
Now as to why use capacitors in your logic circuits? The answer, many teachers at school will not be able to tell you is simple. When you have logic switches going on and off in the circuit it amounts to a lot like having varying voltages/current thru different parts of the circuits (which is kind of like AC) which you need to filter off . These capacitors act as decouplers - and keep the “glitch” noise at bay. To sum - decoupling keeps ‘glitch’ noise under control that occurs as a result of rapidly-switching transistors in the logic gates, it prevents high-gain
circuit elements such as operational amplifiers (ok they’re analog) from self-oscillating (a bad thing!), and it keeps the power rail voltages cleaner overall.

#14

Yeah, andy_fl, no one doubts that capacitors are great for filtering switching noise. But you don’t put capacitors in series with the signal to accomplish that, they must go to ground.