Connect with us

Calculating resistance and Base current

Discussion in 'Electronics Homework Help' started by Daniel280599, Mar 6, 2015.

Scroll to continue with content
  1. Daniel280599

    Daniel280599

    4
    0
    Feb 2, 2015
    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.
     
  2. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    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
     
  3. Colin Mitchell

    Colin Mitchell

    1,417
    313
    Aug 31, 2014
    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.
     
  4. Merlin3189

    Merlin3189

    250
    69
    Aug 4, 2011
    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.
     
  5. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    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.
     
  6. Ratch

    Ratch

    1,092
    334
    Mar 10, 2013
    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
     
  7. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Do you suppose we could get a negative LDR from the usual supplier of gyrators and interocitors?
     
  8. Ratch

    Ratch

    1,092
    334
    Mar 10, 2013
    You will have to ask Exeter about that.

    Ratch
     
  9. Merlin3189

    Merlin3189

    250
    69
    Aug 4, 2011
    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.)
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-