Someone good at electronics? Resistors getting less effective?

[xtremetheme]

Ok, im not good at electronic calculations and i dont know why this happens.

I have a 5V source. I apply a cable to it with 1M ohm resistor.
After the resistor the V is 2.5.
I add another resistor, after that the V is 1.7.
I add another one and now have ~ 1.2at the end.
So i have like 3 million ohm resistance and i still cant block the 5V out.
And if i add more 1Ms i guess they will be less effective every time, but why??

[Hell-Fire]

Part of it is that the resistors are more than likely not exactlly 1M each...there is generally a 10% "play" in them where they can be high or low.

No matter what you do, there is always gonna be voltage across the circuit unless the current or resistance drops to zero. If the resistance drops to zero, you have a short circuit.

Everytime you increase the resistance, you decrease the current in the loop. So unless the current becomes zero, there is always going to be some voltage in the loop. As the resistance gets larger and larger, the voltage will start to drop to milli-volts and so on and so on.

[enzoR]

its the ohms law

V=IR

its not linear or how do u call it?.

as the volts decrease the resistance changes, well not quite. its really hard for me to explain, but its correct. their aint anything wrong.

what do u mean you wanna block the 5v out. turn 5v into 0??

[phobos]

Easy to explain:

I think you connect the resistor in series to the power supply. Than you measured the voltage "behind" the resistor with your multimeter wich has an internal resistor of round about 1MOhm. So the Voltage is devided 50:50. If you put another resistor in series it will devided in 1/3 on each resistor wich will give you 3.3V at the 2MOhms and 1.7V at your multimeters internal resistor.

And what the hell do you mean with "i still cant block the 5V out" ?

And enzoR:

The function V=I*R is absolutelylinear because the resistance is constant!
The resistance does NOT change if you decrease the voltage, only the current will decrease.

[Jasonxxx]

R = (Change in Voltage(E)) / Current(I)

or (Change in Voltage(E)) = R * I

E * I = Power(watts)

now if you put a resistor in series before a load you will drop the voltage, but up the current so you will still have the same power, just less voltage... The catch with powering electronics is that the above is "ohms law" and it is linear... the resistor is constant and the current is the one that will change when it needs more power.. So the "change in voltage will be greater" i.e. the voltage drop will be greater..

Example: say you are using the 3.3v line

.3v(change in voltage) / 6amps = .05ohms

now you have 3.0v after the .05ohm resistor for supply to the load

or

.05ohms * 6amps = .3v(change in voltage) i.e. voltage drop

now if the pc for example is loaded and the current draw is greater i.e. the current goes up, then you have this:

.05ohms * 9amps = .45v(change in voltage) this just dropped the voltage even more..

now you have 2.85v after the resistor for supply to the load

Now with semiconductor "electronic" components ohms laws is not linear anymore...

I have been looking into this, and came across pot trimmers.. they can divide the voltage i.e. a series parallel circuit....

Does someone know much about voltage dividing? If you do this, does the voltage stay constant? or do we need a mosfet to accomplish this?

:toast:

[Ragnarok]

resistive voltage dividers are useless... unless u know the exact impedance of the load...

btw, if u wanna block out that voltage, just open circuit it :D

[slavik]

Resistance isn't exactly constant ;)

there's the resistivity of the material and the amount of material u are using ... then temperature comes into play ...

[phobos]

Originally posted by slavik
Resistance isn't exactly constant ;)

there's the resistivity of the material and the amount of material u are using ... then temperature comes into play ...

...but you can disregard in many cases ;)

...and if you want to devide a voltage you better go for a little circuit with a z-diode and a transistor. Or get a linear voltage regulator like a LM78XX where XX is your desired voltage. Just put two capacitors parallel to it on the input and output side, let's say 10µF and 100nF to block interferences and you're done.

[Jasonxxx]

thanks guys... I will try it out..:D