voltage level shifter

Discussion in 'General Electronics Chat' started by anhnha, Jan 20, 2016.

  1. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    I am stuck at design a circuit with the specs as below.
    Input voltage: VDD
    Output voltage: Vout
    VDD is also the ONLY power supply for the circuit.
    When VDD = 1V, Vout = 0V
    When VDD = 2V, Vout = 2V
    Any idea how this can be done without using IC?


    Specs.png
     
  2. EM.

    New Member

    Nov 13, 2015
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    If you can show the circuit it who'd be easier to try to figure out

    What is going on with the circuit?

    But I believe that the 1V is actually less than 0.7V and wen the current

    coms in, it goes throw a diode, and the diode doesn't let throw current

    at all as long the voltage is at least more than 0.7V

    you might see on the tester 1V if the tester is not 100% accrete.
     
  3. Bordodynov

    Active Member

    May 20, 2015
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    188
    Insofar powerful must be a signal. The Question it is necessary to assign more concretely. But if VDD=1.1V or VDD=0.9V that then?
     
  4. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    774
    47
    This is the circuit I need to come up with not ready yet.
    In these cases, it can be considered as 1V.
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,789
    945
    Draw a simple transistor circuit - common emitter configuration.
    Put a suitable resistor in the emitter leg to ground that keeps current to a safe level for the transistor you use.
    Oh, Use NPN. Connect Vdd straight to collector.

    You need to prevent your base voltage from turning on the transistor when Vdd is 1 volt but let enough voltage through when Vdd is 2 volts to forward bias the base.

    Do you know how much voltage 2 diodes in series drop when forward biased?
     
  6. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Is this what you meant?
    What you describe is common collector or emitter follower not common emitter.
    Common Collector.PNG
    The answer depends on the type of diodes.
    For silicon diodes, the typical forward voltage is 0.7 volts, so two diodes in series will drop about 1.4V.
    For germanium diodes, the forward voltage is only 0.3 volts, so two diodes in series will drop about 0.6V.
     
  7. Lestraveled

    Well-Known Member

    May 19, 2014
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    It will take one PNP transistor, one diode and two resistors (or one resistor depending on the load.) I'll give you a hint; the emitter is connected to the input. Can you figure it out??
     
  8. hp1729

    Well-Known Member

    Nov 23, 2015
    1,952
    219
    A simple voltage comparator? LM393?
    At what specific point do you want to recognize the 2 V level?
     
  9. Lestraveled

    Well-Known Member

    May 19, 2014
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    How about this

    [​IMG]
    The transistor will be off until the input voltage (left) rises above 1.4 volts.
     
    AnalogKid and Brownout like this.
  10. WBahn

    Moderator

    Mar 31, 2012
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    You haven't given enough information. You've said what you want Vout to be at just two particular values of Vdd. What about all of the other values? What if Vdd is 0.5 V, or 1.5 V, or 1.9 V, or 2.2 V?

    Instead of a plot of Vdd and Vout as a function of time, you need to plot Vout as a function of Vdd.
     
  11. anhnha

    Thread Starter Active Member

    Apr 19, 2012
    774
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    Not sure...
    Level shifter.PNG


    I just want to try a simple solution.
    Also, I am not sure if the comparator will work at 1V VDD.
    I will check it now.
     
  12. Lestraveled

    Well-Known Member

    May 19, 2014
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    See post #9
     
  13. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    I am designing a digital 2 to 1 multiplexer.
    Vout is will be the voltage for select pin for the multiplexer.
    VDD will have only two values 1V and 2V with some tolerance for example 1%.
    With VDD = 1V, I want the select pin will be 0V and when VDD = 2V, the select pin is 2V.

    Yes, I will check if it works.
     
  14. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    The simulation result is relatively good :D
    Schematic 1.PNG
    Waveform1.PNG
     
  15. Lestraveled

    Well-Known Member

    May 19, 2014
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    You can probably adjust the values of R1 and R2 downwards and get some improvement.
     
  16. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Thanks. I think that is already very good result for my purpose.
     
  17. dl324

    Distinguished Member

    Mar 30, 2015
    3,243
    620
    OP asked for a solution that didn't use an IC...
     
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