Post by: physics on April 24, 2010, 08:37:23 pm
i'm confused but someone told me it means the current is reversed for a diode or somethin and i dont get how there can be a -ve voltage etc etc.
Post by: physics on April 24, 2010, 08:45:27 pm
i used the vout formula where the v out is 3 and i get 668933ohms which is wrong :(
Post by: physics on April 24, 2010, 08:46:24 pm
Post by: physics on April 24, 2010, 09:03:49 pm
Post by: Chavi on April 24, 2010, 09:08:40 pm
what does a -ve voltage represent in a input voltage graph (sin) graph shape that goes up and down from +10 to -10V ? what does it meanNot 100% sure, but I think it has to do with Alternating current (which you then convert to DC by the RMS forumula if you need to)
i'm confused but someone told me it means the current is reversed for a diode or somethin and i dont get how there can be a -ve voltage etc etc.
When Vout is 6 V what is the resisitance of the LDR?Vout = (r2)/(r2 + r1) x Vin
i used the vout formula where the v out is 3 and i get 668933ohms which is wrong :(
So, 6 = r2/(r2 + 2000) x 9
Rearranging gives:
6r2 + 12000 = 9r2
so r2 = 4000 ohms
Can someone help me with what p and n type is? i have no idea.From my basic understanding in trying to decode Heinmann 12, a semiconductor is "doped" (mixed with another element) to create a diode. Depending on the numer of electrons in the added element, the semiconductor will either lack an electron (p type - for overall positive charge) or have an extra electron (n type - overall negative charge).
Post by: the.watchman on April 24, 2010, 09:10:07 pm
When Vout is 6 V what is the resisitance of the LDR?Vout = (r2)/(r2 + r1) x Vin
i used the vout formula where the v out is 3 and i get 668933ohms which is wrong :(
So, 6 = r2/(r2 + 2000) x 9
Rearranging gives:
6r2 + 12000 = 9r2
so r2 = 4000 ohms
Doesn't
That's what the diagram says...
Post by: physics on April 24, 2010, 09:13:37 pm
what does a -ve voltage represent in a input voltage graph (sin) graph shape that goes up and down from +10 to -10V ? what does it meanNot 100% sure, but I think it has to do with Alternating current (which you then convert to DC by the RMS forumula if you need to)
i'm confused but someone told me it means the current is reversed for a diode or somethin and i dont get how there can be a -ve voltage etc etc.When Vout is 6 V what is the resisitance of the LDR?Vout = (r2)/(r2 + r1) x Vin
i used the vout formula where the v out is 3 and i get 668933ohms which is wrong :(
So, 6 = r2/(r2 + 2000) x 9
Rearranging gives:
6r2 + 12000 = 9r2
so r2 = 4000 ohmsCan someone help me with what p and n type is? i have no idea.From my basic understanding in trying to decode Heinmann 12, a semiconductor is "doped" (mixed with another element) to create a diode. Depending on the numer of electrons in the added element, the semiconductor will either lack an electron (p type - for overall positive charge) or have an extra electron (n type - overall negative charge).
why for the voltage do u use 6 and not 3 cause i was trying to find the diode and not the resistor? i did
3=9 * (diode/ 2006800(total resistance)) = 668933ohms which is fail
sorry i'
m confused
Post by: Chavi on April 24, 2010, 09:20:38 pm
So the answer (if my calculations are correct) would be 4Mohm's for Vout
Post by: physics on April 24, 2010, 09:22:11 pm
Post by: Chavi on April 24, 2010, 09:25:30 pm
So in this case
Vout is 6 V
Vin = 9V
R2 = 2Mohm
R1 = unknown
Post by: physics on April 24, 2010, 09:31:26 pm
We know thati get it but the answer is ...wierd
So in this case
Vout is 6 V
Vin = 9V
R2 = 2Mohm
R1 = unknown
Post by: Blakhitman on April 24, 2010, 09:54:25 pm
and yeah the answer should be 4Mohms
Post by: the.watchman on April 24, 2010, 09:55:29 pm
R2 is in the numerator.
and yeah the answer should be 4Mohms
Yeah that's what I got (i did say that didn't i anna??? :P)
Post by: physics on April 24, 2010, 10:13:32 pm
noooooooo u said 4Mohms which is 4000000 and not 4000 and hte answer is 3700ohmsR2 is in the numerator.
and yeah the answer should be 4Mohms
Yeah that's what I got (i did say that didn't i anna??? :P)
Post by: Blakhitman on April 24, 2010, 10:27:29 pm
but then again, why is the answer 3.7 k ohms? o.O
Which textbook? and what chapter and question?
Post by: physics on April 24, 2010, 10:38:55 pm
Post by: superflya on April 24, 2010, 11:29:23 pm
The voltage across the photodiode? shouldnt it be 6V since its in reverse bias and no current hence no volatage can pass to the rresistor anyway :S but the answer is 3.5V
start by finding the voltage across the resistor using v=IR, plug in the values and ull get 2.5 v (careful with ur sig figs)
the voltage supplied is 6v so by subtracting the voltage across the resistor from the supply voltage, ull get the voltage across the photodiode. 6-2.5=3.5 v
Post by: Blakhitman on April 25, 2010, 10:19:19 am
heinemann physics book 5.2 Q6b
do you mean question 4?
Post by: Blakhitman on April 25, 2010, 10:34:35 am
here is the question,
4. When the LDR shown in the diagram is in darkness,
it has a resistance (RLDR) of 2 MΩ.
(http://img441.imageshack.us/img441/6809/asdasdfafa.jpg)
a) Calculate the total resistance of the series circuit
when the LDR is in darkness.
Simply add the two resistances, 2Mohms + 6.8kohms.
b) Calculate the voltage drop across R2 (Vout) when
the LDR is in darkness.
use the vout formula:
c)The LDR is now illuminated with a light source and
its resistance decreases. Determine the resistance
of the LDR if Vout now is 6 V.
Again use the Vout formula.
Post by: physics on April 25, 2010, 07:06:04 pm
Firstly, you copied the diagram wrong, lol.oh i know what i did wrong now :D thankssssssssssssssssss
here is the question,
4. When the LDR shown in the diagram is in darkness,
it has a resistance (RLDR) of 2 MΩ.
(http://img441.imageshack.us/img441/6809/asdasdfafa.jpg)
a) Calculate the total resistance of the series circuit
when the LDR is in darkness.
Simply add the two resistances, 2Mohms + 6.8kohms.
b) Calculate the voltage drop across R2 (Vout) when
the LDR is in darkness.
use the vout formula:
c)The LDR is now illuminated with a light source and
its resistance decreases. Determine the resistance
of the LDR if Vout now is 6 V.
Again use the Vout formula.or
Post by: Blakhitman on April 25, 2010, 07:20:34 pm
Post by: TyErd on May 02, 2010, 09:45:07 am
The voltage across the photodiode? shouldnt it be 6V since its in reverse bias and no current hence no volatage can pass to the rresistor anyway
That would be the case for a normal reverse biased diode but this is a photodiode and when they are reverse biased they conduct photocurrent.
Post by: kyzoo on May 02, 2010, 12:36:19 pm
Can someone help me with what p and n type is? i have no idea.
It's requires some Chemistry to fully explain but you do Chem so it doesn't matter.
~ Some semiconductors (such as diodes and transistors) are composed of a lattice of silicon or germanium atoms, which have four unpaired valence electrons and form a stable octet (outer shell of 8) of valence electrons by forming 4 covalent bonds with four other atoms in the lattice
~ At room temperature, there is negligible current flowing through the semiconductor, as the bonds which hold electrons in place in the lattice are too strong
~ However we can increase the conductivity of a semiconductor by doping it, which is to replace some of the silicon/germanium atoms in the lattice with other elements.
~ We create an n-type semiconductor by placing phosphorus atoms into the lattice (doping the semiconductor with phosphorus). Phosphorus has 5 valence electrons, hence when it forms 4 covalent bonds in the lattice, there will be 9 valence electrons when only 8 are required to form a stable octet. The one extra valence electron is bonded weakly to the atoms in the lattice, and hence it becomes a delocalised electron: It is able to flow freely through the semiconductor.
~ We create a p-type semiconductor by placing silicon atoms into the lattice (doping the semiconductor with silicon). Silicon has 3 valence electrons, hence when it forms 4 covalent bonds in the lattice, there will be 7 valence electrons when 8 are required to form a stable octet. The absence of the single electron required to form a stable octet produces a hole, which valence electrons from neighbouring atoms in the lattice are able to flow freely into. However when a hole in the valence shell of Atom A is filled by a valence electron of Atom B, the hole is then transferred to the valence shell of Atom B as Atom B is now lacking an electron. The hole then moves to Atom C, Atom D, and so on. In this way the hole moves through the lattice.
It's like a neat single-file line of people. If the person in front moves forward, there is now a gap between the 1st and 2nd person. If the 2nd person then moves forward to catch up to the 1st person, then the gap is then transferred to between the 2nd and 3rd person. By the repetition of this process, the gap is eventually able to find its way to the other end of the line.
~ When you connect an n-type semiconductor in a circuit, the electrons flow towards the positive terminal of the battery
~ When you connect an p-type semiconductor in a circuit, the electrons flow towards the positive terminal, hence the holes flow towards the negative terminal.