ohhh right, I think i'm with it now... it literally does mean a short of the LED then, not a short of the circuit.

thanks

What do they mean by that? how's an LED become "short circuited"  did someone twist the LED? and short it? or  are they talking about shorting the circuit elsewhere?

It is called a thought experiment, it is meant to test your understanding of what is going on. It does not mean that this did happen it is asking you what would be the consequences if it did happen.

Any component can fail, they can fail either open circuit ( like a fuse ) or fail short circuit, you have to know what will happen under each of the two circumstances.

I still don't understand what it means.

If i put a piece of wire between the anode and the cathode? is that a short? (because even though the LED would go out, the rest of the circuit would be fine) or
do they mean a short at the +/- terminals ?

Basically i'm not "stuck"  just a little confused as to what they're asking.

12. Construct the above Series LED Circuit on a physical breadboard with your calculated resistance value and ensure the LED lights up. Make sure you connect the anode (the longer lead) towards the positive side of the power supply.
`
13. Measure all circuit values and record in Table 2.

table 2, I already completed the table, it's attached as an image.

14.  Troubleshooting Exercise:

PREDICTIONS

(i) Assume your LED became ‘short circuited’ (ie. it’s resistance became zero). Determine values for the following and explain…
circuit current voltage across the LED
(*Insert your answers in Table 3)


(ii) Now assume your LED became ‘open circuit’ (ie. it’s resistance became infinite). Determine values for the following and explain … circuit current
voltage across the LED

"Assume your LED became ‘short circuited’ (ie. it’s resistance became zero). Determine values for the following and explain…
circuit current voltage across the LED "

What do they mean by that? how's an LED become "short circuited"  did someone twist the LED? and short it? or  are they talking about shorting the circuit elsewhere?

i use tip31 transistors with a small heatsink and pwm out to control the brightness (analogWrite) of my high power LED's...

trying to decide how i'd make a power supply that works from low voltage dc rather than rely on an ac 240 supply.  i have over 10 regulators coming only + regulators.

my atx connection to the motherboard requir+12  -12v   +5 -5v   3v so on, so it's going to be tricky then without some neg voltage regulators.

but virtual ground and what crossroads said is a useful and good starting point to learn from cheers (oh the car battery is not in the car but by me... )

That depends on the resistor you're using to the tip120.... is the tip120 a transistor? base collector emitter? vs a fet?

a fet has a high impedance, and only takes a small amount of voltage to switch on, where as a tip120 relies on current and without a resistor you could be damaging your pin out on the arduino. with a fet you'd need a pull down or the small capacitance keeps the led's on...

If i took a voltage regulator, or infact 2  12v or 5v positive regs.

then swapped 1 regulator output terminals (-/+), is that not -5 and +5 ?  I'm thinking of running a computer from a 12v car battery using a load of switching positive regulators....  and to get all the negative rails reversing the outputs.....

It's great that you're adding a voltage divider to make the voltage levels safe enough to be read in by an analog port!

but, I still don't know why?... Power is Amps * Volts (Watts) you're reading Volts, so it's not so much a meter which measures current, but potential voltage at any given point?

ac in to your transformer... out pops a low voltage ac sinewave... at most you'd simply filter the signal and then measure the low voltage ac side with a voltage divider....

the voltage regulators and the rectifier simply convert it to dc current, which is a step too far?

5v-----[R1 100ohm]-----[R2 1000ohm]-------gnd
........................Vout

So the key to triggering off the transistor is get a voltage out when it goes *dark* or when it goes *light*, so let's say we want it to go dark


LDR 10ohms = super bright
LDR 100,000ohms = Dark
LDR 1000ohms = Normal Light

5v-----[R1 2000 ohm Light Dependent Resistor]--*VOUT---[R2 1000ohm]-------gnd
So, here we have 2000 ohm and 1000ohm for your total resistance.

3000ohm RT correct.

Amps = 5/3000
= 0.00166 amps

So at that point in time, Vout = 0.00166 * 2000 = 3.333333333333333v   where as. a high light value reading
would show something like, 1 ohm for the LDR.

TR = 1001ohms
TC = 5/1001
=0.004995004995005amps

So Vout = 0.0049v

So when it's dark, Arduino will read a bright, it will be off, and when it's dark it will be on, re-arrange the resistors
to make it do the other way round... look up ohms law (larger the value eg more ohms,  the greater the voltage drop across it.)

http://www.ebay.com.au/itm/400483508568?ssPageName=STRK:MEWNX:IT&_trksid=p3984.m1439.l2649

I came across this guy selling them at $1.00 AUD which has had a bit of a crash, that makes it something like $0.95! USD a piece and he's giving
them out on free shipping!!!!

So i bought 8 of them, I'm always coming across new needs, LED driver (as a constant voltage for 10+watt LED's) keeping a float on a lead acid batteries from swiching ac power supplies (19v switching chargers of a couple of amps at 19v into one of these little guys floats a car charge with little effort, or powers a small DC powered device)....

Until he realizes exchange rates for the aussie dollar are dropping he's soon going to be out of pocket, i'd be quick.

[D]

the LDR is a light diode - now the key word "Resistor"

The D in "LDR" is for Dependent, not diode.

oh man, you got me!....  when's the trial?

the LDR is a light diode - now the key word "Resistor"

two resistors forms a voltage divider...

Your simulation - the 7805 model is imperfect - so just try to increase the battery to 35V for example and you get better results. My simulation shows the 7805 internal current is 5.27mA, therefore the output voltage is higher. But it copies what we had written before:

Code:

I=5/120+5.27mA = 46.9mA
V(180)=46.9mA*180=8.43V
Vout=5+8.43=13.43V

Again - you must consider the internal current of the 7805 as well (ie 4mA) to get precise results..

I'm with it now, cheers.

I Created this "12v" output circuit... and the simulated results show 10v not 12v!

(4 new replies have been posted!, i'll read them first)