Post by: evaporade on May 02, 2009, 10:19:06 pm
Post by: TonyHem on May 02, 2009, 10:54:46 pm
Post by: /0 on May 02, 2009, 11:32:59 pm
Post by: Damo17 on May 02, 2009, 11:58:48 pm
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit.
Post by: TrueTears on May 03, 2009, 12:00:38 am
I thinkwas saying the same as that with tonyhem on msn.is 6V and the rest are 0V, but not sure.
But I'm not 100% sure...
Post by: /0 on May 03, 2009, 12:36:21 am
I would say 6v across v3 as well and 0v across v1 and v2.
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit.
Ah that's why
My teacher was explaining to the class that there was no current in the circuit. But how would this account for the voltage drop across the reverse diode?
So there must be current in the circuit after all.
Post by: TrueTears on May 03, 2009, 12:37:26 am
in Jarcaranda it says the circuit has very LITTLE current, so I assume there is still current, but just very very small...I would say 6v across v3 as well and 0v across v1 and v2.
My explanation:
The reversed bias diode would have an extremely high resistance. It would let a very small amount of current through, micro amps, thus it would take all the voltage being put into the circuit.
Ah that's why
My teacher was explaining to the class that there was no current in the circuit. But how would this account for the voltage drop across the reverse diode?
So there must be current in the circuit after all.
Post by: Damo17 on May 03, 2009, 01:01:18 am
If a reverse-biasing voltage is applied across the P-N junction, the depletion region expands, further resisting any current through it. Key word here is 'resisting', it does not completely stop all current just makes it harder for current to flow as the depletion region gets larger.
Post by: naved_s9994 on May 03, 2009, 04:23:38 pm
Post by: Mao on May 03, 2009, 10:51:26 pm
Post by: kurrymuncher on May 03, 2009, 10:53:27 pm
Post by: Mao on May 03, 2009, 11:02:28 pm
Post by: TonyHem on May 03, 2009, 11:36:00 pm
Post by: TrueTears on May 04, 2009, 03:38:57 pm
(http://img387.imageshack.us/img387/6089/vcenotes1.gif)I've confirmed.
V1 is 0
V2 is 0
V3 is 6V
basically the reverse bias allows no current in the circuit (There might be some, but let's assume it is a perfect diode and blocks out all current)
using Ohms law on the resistor we have V = IR when I is 0 there is 0 V
for the forward bias diode we can not use ohms law as it does not have a linear relationship.
Looking at the graph of a diode, we are only dealing with the positive half of the graph (Since forward bias implies positive and reverse bias implies negative)
when I is 0 we see that V = 0 Hence V2 = 0
However at V3 we see its a reverse bias diode, again looking back at diode graph, we are only dealing with the negative half. Since the battery is 6V and when I = 0, the voltage can be anything for the reverse diode, so the potential difference is 6 V.
Post by: Mao on May 04, 2009, 07:33:47 pm
A diode has a forward voltage as well as a reverse voltage (usually excluded as it almost always lead to permanent damage to it). I'm not sure if 6V is enough to kill the diode in reverse bias, but if it is, there will be some interesting complications.
Post by: Damo17 on May 04, 2009, 08:13:58 pm
TT: the only way to confirm something like this is to construct the circuit and experimentally test it out.
A diode has a forward voltage as well as a reverse voltage (usually excluded as it almost always lead to permanent damage to it). I'm not sure if 6V is enough to kill the diode in reverse bias, but if it is, there will be some interesting complications.
The max voltage a reversed bias diode can take before braking down is usually 50v+. But this depends on the doping of the p-n junction. Take for instance a zener diode which has a heavily doped p-n junction, it allows electrons to tunnel from the valence band of the p-type material to the conduction band of the n-type material, such that the reverse voltage is kept to a known value.
Post by: bigtick on May 04, 2009, 08:43:40 pm
http://itute.com/board/viewtopic.php?f=6&t=607&p=1228#p1228