resistors for NOT gate (ohms law help)

Discussion in 'General Electronics Chat' started by Gazza2, Aug 25, 2013.

  1. Gazza2

    Thread Starter New Member

    Aug 25, 2013
    5
    0
    Hi everyone.

    I am a new member and beginner in electronics so please be gentle with me.

    I am really struggling to understand ohms law when it comes to finding what resistors I would need for a simple NOT gate to light a resistor.

    I know the equation I need is R = V/I but after reading a lot of stuff on here and from around other sites I am not sure what else I need.

    I have a 9v battery, 1 bc547c transistor and 1 5mm LED.

    My main problem is all the different terminology that I don't understand. If someone could explain or point me in the right direction as to where I would get the other figures for the formula I would be very greatful.

    Sorry for the long post and thanks in advance for any help.

    Thanks
    gareth
     
  2. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    light a resistor, that is a very strange description, I think probably you want to say "light up a LED"?

    a simple NOT gate - do you want to using a transistor to light up a LED, and control from B of bjt, and the transistor as a inverter?

    The function of inverter just like Seesaw, when one side is low then another side will be high.

    How about the voltage of 5mm LED, color, V and I ?
     
  3. Gazza2

    Thread Starter New Member

    Aug 25, 2013
    5
    0
    Hi scott.

    Yes that was a typo, it should have said LED.

    The LED is red and has a voltage of about 1.6v and the current is 10mA.

    This is what I am not understanding if the supply voltage is 9v do I take the 1.6v of the LED from the nine volts then divide by 10mA or am I looking at that all wrong.

    Also its the transistor base resistor that I am not sure how to calculate.

    I found the following page:

    http://tronixstuff.com/2010/06/07/electronic-components-the-transistor/

    which sort of helps but wont help me much if I don't understand how to read the datasheet of the transistor(working on that at the moment).

    Thanks for the reply
    Gareth
     
  4. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    You want to use the transistor as a switch, so you give the base 1/10th of the collector current. That's 1 ma in this case. 9V - .6V for the transistor is 8.4V
    8.4V/.001A = 8400 ohms. Use 8.2k or something close to that.

    As for the LED side, 9V - .3 volts for the transistor - 1.6V for the LED is 7.1 V that the resistor must use up.
    7.1V/.01A = 710 ohms
    Use something around 680 ohms or 750 ohms.
     
  5. absf

    Senior Member

    Dec 29, 2010
    1,493
    372
    A picture is better than a thousand words. Here's the simulation on proteus.

    Allen
     
  6. ScottWang

    Moderator

    Aug 23, 2012
    4,855
    767
    The circuit just the same as what I thought.

    For the life and brightness of LED, if using around 80% of 10mA is better.
     
  7. Gazza2

    Thread Starter New Member

    Aug 25, 2013
    5
    0
    Ok guys thanks for the replies.

    If I have this right you have to minus the voltage that each component will use from the supply voltage to get the actual voltage for the equation.

    The current is what the output load will use.

    The only question I have left is how would I find the voltage for the transistor on the datasheet ( in #12 answer it is .6v for the base and .3v for the collector).

    Thanks again all, you have been really helpful
    Gareth
     
  8. LDC3

    Active Member

    Apr 27, 2013
    920
    160
    The Vbe (base to emitter voltage drop) is very close to the voltage drop of a diode (0.7 V). This should be specified on the datasheet.
    When saturated, the Vce (collector to emitter voltage drop) of a BJT is about 0.3 V. This (or the resistance) might be specified on the datasheet.
     
  9. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    The voltage of the base to emitter junction and the saturation voltage from collector to emitter are often found as graphs on the datasheet. They are called Vbe(sat) and Vce(sat). They are on page 3 of this datasheet.

    We commonly throw around ".6 volts" for a base to emitter voltage and a little less than that for a collector to emitter saturation voltage because it's close enough, most of the time. As you can see, the final values in the simulation were exactly what I calculated while guessing at the transistor voltages. From this, I conclude that it is more important to remind you to include something for the voltage loss of the transistor than to get it exactly right.
     
    Last edited: Aug 26, 2013
  10. Gazza2

    Thread Starter New Member

    Aug 25, 2013
    5
    0
    Ok guys sorry for being such a dunce about this (I am going to sit in a corner on my own after this) but I can`t seem to make out where the figures come from. I am now getting frustrated because this should be relatively simple but even with all your explanations I cant seem to understand what matches what between the figures you guys have given me and the charts on the datasheet that #12 pointed me too.

    I have been staring at the posts, the picture of the circuit and the datasheet but still cant make head nor tail from it. I can understand the figures on the picture but I really want to understand how to find the figures for the transistor.

    Again extremely sorry for being a bit slow

    Thanks
    Gareth
     
  11. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    On the datasheet I included, page 3 top right. A graph of current and volts. The bottom line is labeled Vce(sat) at a gain of 10. Go horizontal to 10 ma and slightly up to see the line at about .07 volts. That's about an order of magnitude different from what the simulator said, and so close to zero that I think the graph is wrong.

    The top line on that graph is Vbe(sat) @ a gain of 10. Look on the bottom of the graph for 1 ma and run up to the line to see...about .66 volts. That seems reasonable. Not exactly the same as the simulator, but reasonable.

    I tend to throw rocks at simulators, but this time it looks like the graph is wrong. If you actually have the transistor, please measure it for us.
     
  12. Gazza2

    Thread Starter New Member

    Aug 25, 2013
    5
    0
    #12 you are an absolute god.

    Thanks for the simple explanation, I think it was because I was looking at it the wrong way round and trying to figure out where the 9v from the battery was.
    Once I have the circuit setup I will measure the voltages and post back to let you know.

    Again thanks for you patience, and thanks to everyone else who replied.

    Gareth
     
  13. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    I wish the women thought so?:cool:

    There are people here that would disagree?:(

    (We don't have an embarrassed smiley.)

    OK. Nine volts is not on the graph because it doesn't matter to the transistor. The transistor doesn't have nine volts when it's in saturation as a switch. Glad you figured that out.:) Now, go melt some solder.:D
     
Loading...