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What happens when different voltages meet?

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HerfGun

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voltagediagram.JPG

The voltages and currents are approximate.

I'm trying to design a circuit to use in my car. The IC is the instrument cluster and the switch is a limit switch. I need to use that limit switch to signal my circuit. It would simplify things alot if I didn't use a diode between the IC and the tap.

I can't quite wrap my head around what's happening when two different voltages and currents meet like this.

The first crude little diagram with the LED works, and nothing has caught fire yet, but I'm wondering if I'm asking for trouble or if there is a safer way. I don't care if the circuit dies but the instrument cluster is very pricey.

The second two diagrams were some other possibilities I thought I might explore.

Thanks,
 
WHat do you mean by when two voltages meet? If you are talking about Voltage X on one side of a component and Voltage Y on the other side (and a component can simply be a piece of wire), what will happen is that the voltage difference of X-Y will appear across that component the same way voltages appear across resistors. Enough current will flow from X to Y (or Y to X depending on which is higher) to produce that voltage drop across the component.

If it was say a piece of wire (which is really just a very low value resistor), a MASSIVE amount of current will flow through it since V=IR says that with a low R you need a very high I to get a certain V. Same thing with a resistor, except the current will probably be lower since the resistance is higher (though it could still be too high and overheat). If it's a diode an infinite amount of current will flow in theory, because in theory the voltage drop across a diode stays the same regardless of the current flowing through it.

In this case, that component appears to be and LED and resistor (and gate-source or base-emitter of a transistor depending on the circuit).

So the voltages don't really meet, as you say they do. It's more it "dies down" along the way across whatever happens to be laying in between. How it does down depends on the behaviour of the components sitting between the two voltage differences and will affect the current as well.

Currents, however, do meet with each other and merge. REmember: Current flows, but voltage does not. I repeat, voltage *DOES NOT* flow.
 
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Just in simple words:

the stronger one will be the winner. :D
 
If the output of your IC is 11V and you short it to ground with your switch then the IC might overheat or instantly blow up.
Its output current might be only 10ma when its output is 11V but the current could be much higher when its output is shorted to ground.

If the supply is 12V and the output of the IC is 11V then the resulting 1V across the LED will not light it anyway.
 
voltagediagram2.JPG

Obviously I'm a complete beginner, and I'm trying to use the "water analogy" to picture what's happening. When the switch is closed everything seems like it should be fine. I just don't know what happens when the switch is open.

This picture my explain better what I'm wanting to do.

Fig. A is how the car is normally wired. The micro switch tells the instrument cluster if the trunk is open or closed. When I put a meter on it, it's +11V and 10mA is flowing.

Fig. B is how I want to wire my circuit. I want to tap that wire so my circuit also knows when the trunk is open or closed.

The question is: How do I make sure my circuit doesn't damage the Instrument cluster?

Thank you for taking the time to try to explain this to me! :D
 
I assume you are saying that when the switch is open you detect 11V on the pin. ANd when the switch is closed you detect 0V with 10mA flowing. Is this correct? Assuming it is...

1. If the 11V on the pin is the directly connected to a power source, then when the microswitch closes it will produce a overcurrent short-circuit condition. THis is because of the ~0 resistance in the wire. But obviously, this is not the case since you say 10mA is flowing- Not a number that can be considered "infinite" for all intents and purposes of the circuit.

2. Inside the IC, the pin is connected to 11V through a resistor. The resistor is a large enough value to limit the amount of current flow from 11V to 0V when the microswitch closes and connects the pin to ground. This is called a pull-up resistor because it ties the line to +V and "pulls" the voltage up to +V in the absence of a voltage directly applied to the line.

WHat is a pull-up resistor?
Pull-up resistor@Everything2.com

This is how pull-up resistors work (pull-down resistors that connect the line to ground also work in a similar way): When the switch is open, zero current flows through the resistor. Zero current in a resistor means zero voltage drop in it, which means the voltage on both ends of the resistor is the same (pin voltage is same as source voltage). When the switch closes and connects the pin to ground, the pin becomes 0V. A 11V difference (11V-0V) appears across the internal "pull-up" resistor and enough current will flow through the resistor from 11V to 0V to allow the voltages to be different so that they don't "meet" which would produce a short-circuit.


In your words, what this resistor does is prevent two voltages from "meeting" head on. This would cause a short-circuit which is basically too low a resistance between two different voltages. THe result is that enough current will try to flow through the resistor to produce the voltage difference at both ends to accomodate their "meeting". But if a resistance is too low, the current has to be too high to produce the same voltage drop across the resistor. THis same resistor should be enough to protect the insides of your IC from meeting head-on with the LED's 12V if something goes wrong. After all, it is keeping 11V from meeting with 0V. It should be able to keep 12V from meeting with 11V, which is a much smaller difference.

Example: 12V meets 11V through a resistance (ranging from zero to infinite ohms). Enough current will flow through the resistor to produce a voltage of 12V-11V = 1V across it to accomodate their meeting. The amount of current (I) that must flows through a resistance R to produce a voltage across it of V is

I=V/R

If V remains constant (1V in this case), you can see that as R gets smaller, a higher I is required to produce the same V. At some point, R gets low enough relative to V that an I that is too high for the circuit to physically handle is required. The worst case is when R = 0, so I must be infinite.

Remember: Current flows. Voltage does not.

Water analogy:
Current is easy. It's the amount of electrons/water flowing past a certain point.

Voltage is a bit harder. It is more like the energy contained in the electron that determines how hard it can push to overcome an obstacle. The more voltage/energy it has, the harder it can push. But the more it pushes, the weaker it gets. THe closest I can get to this is water pressure which doesn't flow, but this is far from an exact analogy since the electron's nano world does't exactly line up with our macro world. When two pressures meet directly, the lower pressure cancels out some of the higher pressure, then the surplus pressure causes the flow to move in the direction of the higher pressure. The greater the pressure difference, the less it is cancelled out and the more will flow. Same thing, it's like a pushing "tug of war" match. You don't want too much flow since that produces overcurrent/short-circuit. Resistors and other things "tire out" water the stronger side so they can't stampede in uncontrolled numbers over top of of the weaker side. For water, it's like a hydro-electric turbine. Higher pressure water at the entrance gets reduced to the lower pressure at the exit. The energy resulting from the difference in pressures was used inside the turbine was converted to other forms of energy like heat or light. For electrons, this is like a resistor that produces heat or an LED that produces light.
 
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