Transistor Voltage Inverter

Discussion in 'General Electronics Chat' started by Timothy Elery, Jun 4, 2015.

  1. Timothy Elery

    Thread Starter New Member

    Jun 4, 2015
    I am a newbie and was hoping someone could assist me. Given the following schematic which takes a voltage of 12v and considers it off and a voltage of 2 volts and considers it on can someone tell me what changes would need to take place regarding capacitors and resistors (if any) to account for 20v instead of 12v? Meaning that when 20v is considered off and 2v is considered on.

  2. WBahn


    Mar 31, 2012
    I'm having a hard time figuring out what you are asking.

    When you say "takes a voltage", takes a voltage where? You have at the input of the right hand circuit a notation that it can be 2-12 Volts. Is that the point that "takes a voltage"?

    You have a +12V supply in the circuit? Are you changing that to 20V and making no other changes?

    This doesn't strike me as being a particularly good circuit. But maybe I'm wrong. Why not start from the beginning and describe what it is you are trying to achieve? What is the basic problem? You want a black box that has what as an input and what as an output and what relationship do you want between them?
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
    As far as I can see the 1k-10k-4.7k resistors form a voltage divider.
    12*4.7k/(1k+10k+4.7k)=3.59 volts

    20*4.7/(4.7+R)=3.59 volts
    R=21.48 kOhm

    I would probably keep the 10k resistor. And replace 1k resistor with 21.48k-10k=11.48 kOhm (or close to it) resistor.
  4. dl324

    Distinguished Member

    Mar 30, 2015
    The 4.7K resistor is across a base-emitter junction, so voltage can't be above about 0.7V.

    I can't make sense of the request. Why is there a 5V supply? How is Q2 connected? Does the notation near the output transistor mean output is either 12V or 2V? If so, how do you get 2V?? Is the dashed box supposed to provide 12V or 2V to the resistor network on the base of Q1?