Simple circuit question

RonM

I am not sure if this forum is for questions like this so forgive me if it is not. I am new to the Stamp and to electronics as well.
Making my way through the WAM book, I am still confused on the point of current flow with respect to the boards and use of a battery.
Page 44 in WAM shows current flowing from the batteries negative terminal thru an LED then thru a resistor to the batteries positive terminal.
I wonder if someone can straighten me out on this. I'm looking for clarity.

1. When using a battery (DC), current flows from negative terminal thru the circuit to the positive terminal. True?
2. Given the pic on page 44, why is the resistor after the LED in the circuit and not before - the arrows show flow from - to + battery terminals?
3. If the current is flowing from the batteries - terminal thru the LED, why is it flowing through the LED cathode and not the anode?
4. This is another confusing point. The WAM refers to Vdd as the same as the batteries + (albeit regulated) terminal and Vss as the same as the batteries - terminal. Given this,
why is the flow of current from Vdd to Vss in a circuit using the parallax boards?


As you can imagine, I'm learning this on my own.


I'm looking in the printed WAM but the online version is the same.

Thanks,
Ron
·

Mike Green

Ron,
1) The issue of which way does the current flow is actually more complicated than you think. Electrons flow out of the negative terminal of the battery and into the positive terminal, but historically current is considered to flow from the positive terminal to the negative terminal. Go figure!

2) Current flows through the LED, then the resistor. The circuit functions the same when the resistor is "closer" to the battery. Th

RonM

Thanks Mike for the explanations and hint on wikipedia. Funny how you can roll forward and then get hung up on a single point. I'm not one to stroll forward through the training manuals without understanding what I am doing.

Thanks again

allanlane5

Classically, "Current" has been the positive flow of "charges" from a Plus voltage to a Lesser voltage. By definition.

In the circuitry we deal with using the BS2, Vdd is +5, Vss is Ground (zero volts), and "positive current" flows from +5 to Ground. The Diode symbol is a 'little arrow' pointing in the direction of "positive" current flow.

Now, for the Physics of the situation, what's really happening is the 'negative' terminal has a surplus of electrons, which are 'flowing' through the wire from negative to positive. This is a 'negative' charge, flowing in the 'negative' direction -- which is the same as a 'positive' charge flowing in the 'positive' direction.

Bottom line -- the math and the symbols become MUCH easier if you simply assume positive charges flowing in a positive direction (from plus to ground). You can be aware the Physics of the situation is different, but that's how this stuff is defined in the Voltage Law.

V = IR. Voltage == Current * Resistance. A +5 voltage will result in a +0.5 Amp of "positive" current through a 10 Ohm resistor.

dennisw

Amazing! Having been trained in electronics in the military and worked in telecommunications for 38 years, I was taught and always used the theory that current ( i.e. electrons) flow from negative to positive and holes flow from positive to negative. But, either way you look at it, the circuits still work the same way. When trying to understand electronics, KISS, don't make it too difficult.

dennisw