Diode Question

Hi, this is my diode: http://img201.imageshack.us/img201/2862/untitledmq1.png

Ok, I know Vcathode < Vanode in order for the LED to light up through the current sent through it.

Now, my question is, if the three Diodes are connected to 0v, Vanode > Vcathode, therefore the LED will light up. But since I havent' had circuit theory class yet, what will be the V_6 output (towards the right side of the diagram)?

The diodes will drop 0.7 v across it's terminals, so does that make V_6 = 4.3V?
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 Recognitions: Gold Member Science Advisor Staff Emeritus What do you mean, "when the diodes are connected to 0V?" I don't see any nodes labelled 0V in your diagram. Diodes drop about 0.7V when forward biased (i.e. when current flows in the direction of the arrow on their symbol), but drop an essentially unlimited voltage when reverse biased. - Warren

 Quote by chroot What do you mean, "when the diodes are connected to 0V?" I don't see any nodes labelled 0V in your diagram. Diodes drop about 0.7V when forward biased (i.e. when current flows in the direction of the arrow on their symbol), but drop an essentially unlimited voltage when reverse biased. - Warren
Ok, sorry. If anyone of the cathodes are connected to 0v*, what is going to happen to V_6.

Now when you say "drop 0.7v" - does that mean it will take away .7v from the +5v input from the top of the diagram, therefore releasing 4.3v to V_6?

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Diode Question

Rather than thinking of the diodes as "taking away voltage" or "releasing" voltage to their terminals, consider simply that a forward-biased diode looks like a 0.7V voltage source.

If you begin at the cathodes of the diodes, taken to be 0V, the voltage at the anodes of the diodes will be 0.7V. Thus, each diode has 0.7V across it.

- Warren

 Quote by chroot Rather than thinking of the diodes as "taking away voltage" or "releasing" voltage to their terminals, consider simply that a forward-biased diode looks like a 0.7V voltage source. If you begin at the cathodes of the diodes, taken to be 0V, the voltage at the anodes of the diodes will be 0.7V. Thus, each diode has 0.7V across it. - Warren
I see, that makes better sense now. Now, for V_6, is that going to be 0.7v because of the junction wth the diode across from it?
 Recognitions: Gold Member Science Advisor Staff Emeritus V_6 is the same node as all the anodes of all the diodes -- in other words, V_6 and the anodes are connected with nothing more than wire. Since the anodes are all at 0.7V, so is V_6. - Warren
 I see, that makes a lot of sense. Thanks! =D
 Another thing, you said that these anodes are "voltage sources" of .7v. what happens to the original 5v input? (i'm sorry if these are siple questions, i haven't had my circuits class so i'm kind of jumping into this naked)
 Recognitions: Gold Member Science Advisor Staff Emeritus No, I didn't say that the anodes were voltage sources. I said that a forward-biased diode always has a voltage at its anode about 0.7V greater than the voltage at its cathode. The voltage at one end of the resistor is 5V. The voltage at the other end is 0.7V. The voltage across that resistor is therefore 4.3V, and it carries 4.3V / 10 kohms = 0.43 mA of current. - Warren
 I see. that makes sense. Do you know of any programs that will do circuit design/analysis?
 Recognitions: Gold Member Science Advisor Staff Emeritus There are about a thousand of them, some costing $10,000 a seat. There are also numerous free simulators with fewer capabilities. A good, free SPICE simulator is made by Linear Technologies. It's called SwitcherCAD III. Its main purpose is the simulation of switched-capacitor circuits, but it'll do a fine job with anything you're going to come across in a basic circuits class. http://ltspice.linear.com/software/swcadiii.exe - Warren  Recognitions: Gold Member Homework Help Science Advisor Just a small point. The forward drop on an LED diode varies by color. For red it's about 2.2v not 0.7v. And if you wire them as your diagram then they will not all light up. Probably only one will.  Quote by NoTime Just a small point. The forward drop on an LED diode varies by color. For red it's about 2.2v not 0.7v. And if you wire them as your diagram then they will not all light up. Probably only one will. why will only one, and not all 3?  Quote by chroot There are about a thousand of them, some costing$10,000 a seat. There are also numerous free simulators with fewer capabilities. A good, free SPICE simulator is made by Linear Technologies. It's called SwitcherCAD III. Its main purpose is the simulation of switched-capacitor circuits, but it'll do a fine job with anything you're going to come across in a basic circuits class. http://ltspice.linear.com/software/swcadiii.exe - Warren

Ill give that shot when im on xp next (im on linux now).

Also, what are some other programs for a little more advanced circuit classes?

Have you heard of Quartus II? Opinions?
 Recognitions: Gold Member Homework Help Science Advisor It's all in that about. Depending on manufacturing tolerances the 2.2v (and the color) will vary slightly. So the one that has the lowest forward drop will turn on. The current flowing thru the 10k resistor will drop the voltage below the forward voltage of the other two leds. With 10k and 5v supply it's going to be really dim too. The 22 ohm average internal resistance of the led would be unlikely to overcome any difference in Vf. Note that this can also occur with the ordinary silicon diode with 0.7 drop in this arangement. No current would flow in two of them.
 Can any body tell me..the Basics of ZENAR DIODE....with the help of circuit diagram
 Can any body tell me the Basics of Zenar Diode...