Suggestion on a transistor circuit

Discussion in 'The Projects Forum' started by guilio_2010, Mar 20, 2013.

  1. guilio_2010

    Thread Starter New Member

    Mar 20, 2013
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    I was wondering if this would work as a visual indicator if a switch was opened for a PNP transistor at the base. Basically, as long as the switch is closed, then I don't want the led on, but as soon as the switch is opened, then I wanted to have an external battery supplying the power to indicate the switch is opened. There are more than 1 switch at various locations, but any one of those would flip a relay and I wanted to have 1 switch be the initial indicator for all switches. Appreciate the advice an help.

    Thanks,
    guilio
     
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  2. upand_at_them

    Active Member

    May 15, 2010
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    No, it won't. For a couple of reasons.

    The current through the LED will only flow if the transistor base current is flowing.

    But the LED won't light at all, because the voltage supply is too low.

    And why are you connecting 24V to the transistor base?

    You should spend some time Googling "transistor switch" before you post that mess.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    Let's chill out and at least recognize him for taking the time to put a schematic together and post it. That's far more than most first posters do.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    As pointed out, you do have some problems. You need a higher voltage in the right hand source. Assuming you need 10mA through the LED and that it drops about 2V, you probably want something on the order of 6V to go along with that 100Ω resistor.

    With a 24V supply on the left, if you connect that directly to the base you will most likely destroy the transistor because of excessive reverse base-emitter voltage (typically small signal transistors are rated for something on the order of 6V). Also, your LED won't light when the switch is open because there is no path to ground for the needed base current.

    You mentioned multiple switches and a nontrivial interaction. I don't know if I followed the description correctly. Could you throw together something that gets what you have in mind for, perhaps, three or four switches?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Oh, Guilio, what package did you use to draw your schematic in. It seems much cleaner than most I have seen. I'm looking for something with which I can easily produce electrical and logic schematics that are suitable for presentations.
     
  6. guilio_2010

    Thread Starter New Member

    Mar 20, 2013
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    upand at them - I apologize for not being and expert as circuit, but I am here to find help and/or advice on how to be better as well as provide advice to others if I have it. Here is my thought process on that circuit as to why I put it together that way based off an electronics class I had years ago. A transistor is a switch. If the base is higher than emitter, current flows through the collector and emitter. for a PNP, my understanding is it's the opposite effect. If base is lower than emitter, current flows which is the effect I'm looking for. Why 24V? because that is what I have to work with. Nothing more, nothing less and I expect to have to use a different componet but that is all I had for the circuit design.

    WBahn - I wanted to use a higher rated transistor. I just didn't have that option on the builder.

    For the right hand source, I was looking at a smaller ~3V lithium battery and using a different resistor. I'll figure that combination out and revise. I'm just looking as if the switch is open, the current flows through the Emitter to Collect to light up the LED.

    "Also, your LED won't light when the switch is open because there is no path to ground for the needed base current."

    This is where I'm not follwoing. If the base voltage is higher, then no current should flow. If the switch is opened, the base voltage is less then current should flow through the emtter and collector and to my understanding, the batter with LED and resisotor is a complete circuit. What would the base have to do with it?

    The website is:
    https://www.circuitlab.com/

    Thanks,
     
  7. guilio_2010

    Thread Starter New Member

    Mar 20, 2013
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    WBahn,

    for the switching, just picture 5 more switches in series with the one that is current there. They are just scattered in different locations.

    Thanks,
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Don't worry about that, but do expect to have to go back and forth a bit because you don't "speak our language" and so we have to do some give-and-take to get the right ideas communicated. We're game if you are.

    That's fine, but it doesn't change the fact that if you apply more reverse voltage than the transistor is rated for then you will let all the magic smoke out of it.

    Your current design will let the smoke out.

    Correct, but if it is too much higher you will destroy the transistor.

    On what is this claim based? If the switch it open then the voltage on the base is undefined. But more to the point, you cannot get any base current.

    A BJT (bipolar junction transistor), which includes NPN and PNP transistors, operate by amplifying the base current. The ratio of the collector current to the base current is called the β of the transistor. For most small signal transistors the β in the active region is a few hundred, but when used in saturation as a switch it can be quite a bit lower, like 10 to 20. So if you want 10mA of current in the LED, you need to allow for about 1mA of base current.

    A fairly simple way to do this is to tie a resistor between the base and ground. With the switch closed, the base will be pulled up and the PNP will be turned off. When a switch is opened, the resistor will let current flow from the emitter to the base to ground.

    Let's say that you had a 6V supply for the right hand side and that the diode has a 2V drop and you want 10mA in it. If you make the base pulldown resistor 1kΩ, then 1mA flowing in it will put the base at 1V and that will put the emitter at about 1.7V. Another 2V for the LED and you are up to 3.7V, leaving 2.3V across the current limiting resistor. To get that voltage with 10mA of current you need a 230Ω resistor -- the nearest standard size is 220Ω so let's go with that.

    Now, as it turns out, this circuit configuration prevents the transistor from going into saturation (unless a collector resistor is added). So what if β is 200? Well, not a lot. The base current would be on the order of 50μA which would put the base voltage at 50mV and the emitter voltage at about 0.75V. The LED brings that up to about 2.75V and so the 220Ω current limiting resistor has 3.25V across it resulting in a current of about 15mA, which is probably safe. But you might find that using a 330Ω would work better (assuming that 10mA is your target).

    Now, how about the little problem of blowing out the transistor? That's easy to take care off. You have a 1kΩ resistor between the base and ground. Put another resistor between the base and the first switch and then, when the switches are all closed, you have a voltage divider. If you are using a 6V source on the right hand side, we only need to get the base up to about 3.3V to turn it off (because of the 2V LED drop and the 0.7V base-emitter drop). To be safe, let's pull it to 6V, which means that there will be 6V across the base pulldown resistor and we need to drop the remaining 18V with that same current. To drop three times the current we just need three times the resistance. So use a 3kΩ (or, based on standard values, perhaps a 3.3kΩ) resistor.

    Thanks. I tend to like having whatever software I use installed on my machine, but maybe this will be good because I can tell my students to use it and they don't have much basis to claim they can't. As long as I can save my schematics locally.

    I'll check it out.
     
  9. guilio_2010

    Thread Starter New Member

    Mar 20, 2013
    4
    0
    Sorry for the delay.

    So there needs to be a current that goes from Emitter to base and emitter to collector for it to work. I'll revise and breadboard this and see what I get now.

    Thanks,
     
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