Calculating resistance and Base current

[Daniel280599]

I'm stuck on a homework question and was wondering if any of you guys could help me out. Here is a link to a picture of the circuit I'm dealing with http://imgur.com/5WNf1Aw. The 2 questions I need help with are "Calculate the resistance of the LDR when Vout is 0.4v" and "Calculate the base current (LB) when Vout is 3.2v and Rb is 1.5kΩ" If anyone could please give me help and advice on how to do it I would appreciate it very much.

 

[BobK]

For the first question. Notice that there will be no more then 0.4V on the base, so it will not conduct. What you are left with is a simple voltage divider. You should be able to compute the resistance of the LDR from that.

For the second question. You know the voltage on one side of the 1.5K base resistor. What is the voltage on the other side? Once you know these two voltages, how can you determine the current?

Bob

 

[Colin Mitchell]

Your teacher doesn't know what he is talking about. Vout can never reach 3.2v. He has made a BIG mistake and I can see how he has made it.

 

[Merlin3189]

It all depends on the LDR and I agree, I can't find one with a low enough On resistance.
But if this is a theoretical exercise, it doesn't seem unimaginable to have an LDR with the required range.

 

[Laplace]

It all depends on the LDR

Not sure why the ON resistance of the LDR is an issue. For Vout to be at a high value the OFF resistance of the LDR must be high. If we assume the OFF resistance is infinity, then Vout is given by the voltage divider of the 4.3K bias resistor and the 1.5K base resistor. Vout = (5-0.6)(1.5/(1.5+4.3))+0.6 = 1.74V so Vout can never reach 3.2V no matter what LDR is used.

 

[Ratch]

Yes, the current in the 4k3 resistor is less than the current in the 1k5 resistor. That means current has to be supplied by the LDR. Mathematically, that means the LDR has to have a negative value.

Ratch

 

[Laplace]

...the LDR has to have a negative value.

Do you suppose we could get a negative LDR from the usual supplier of gyrators and interocitors?

 

[Ratch]

Do you suppose we could get a negative LDR from the usual supplier of gyrators and interocitors?

You will have to ask Exeter about that.

Ratch

 

[Merlin3189]

Not sure why the ON resistance of the LDR is an issue. For Vout to be at a high value the OFF resistance of the LDR must be high. If we assume the OFF resistance is infinity, then Vout is given by the voltage divider of the 4.3K bias resistor and the 1.5K base resistor. Vout = (5-0.6)(1.5/(1.5+4.3))+0.6 = 1.74V so Vout can never reach 3.2V no matter what LDR is used.

Sorry, I was assuming Vout was the voltage on the collector! (I couldn't read the diagram - it looked like omega 2t or something. I've been spending too much time on Bhuvanish's questions! And why is Vout on the input?)
BTW I still can't work out what Collin was hinting at.

Ok, so Vout isn't 3.2V but 1.74 V and can create about 0.75mA base current, which should energise the relay.

My worry was that the ON resistance was too high at say 2k, because it would not pull the base low enough to cut off the transistor. At 2k, the voltage divider of 4k3 and 2k gives Vout 1.58V ,if no base current, but that's enough to draw base current (I make it about 0.34mA , giving 35+mA collector current) That may not be enough to pull in the relay, but it may be too high to let it drop out.. (If it were a good BC109, Ic would be well over 100mA and the relay would be always on.)
Looking through Farnell I couldn't find an LDR lower than 2k (at 10 lux, however much that is, but it's what they quote On resistance at.)