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I am trying to create a voltage divider circuit to off offsset two BJT input voltage is for a push pull amplifier configuration, and I am having trouble choosing the proper resistor values for my circuit. My source voltage is 12V Vcc and I need to create something like below. I know the basic theory behind voltage divider but I’m having trouble making it. Also, the solution has unlimited solutions so what are the best range of resistors to use? I will be powering an 8 ohm speaker with my amplifier. Thank you kindly.

enter image description here

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  • \$\begingroup\$ You pick them based on current consumption/heat since there is always current flowing through the divider from power to ground. \$\endgroup\$ – Toor Feb 25 at 20:22
  • \$\begingroup\$ How much bias current does your circuit require? \$\endgroup\$ – Dwayne Reid Feb 25 at 20:25
  • \$\begingroup\$ You may wish to show your whole circuit. Do keep in mind that if your "12V" source is a car electrical system, you should plan on it ranging from 12V to 14V or so in "normal" use, and you should withstand excursions to +60V and -40V or so (Google "automotive load dump" -- you'll learn more than I know). If you're aiming for one diode drop across the transistors' BE junctions, then this circuit will be problematic. \$\endgroup\$ – TimWescott Feb 25 at 20:28
  • \$\begingroup\$ This kind of biasing is not usually done using resistors. The usual technique is to use diodes or diode-connected BJTs to get the 0.6V drops. \$\endgroup\$ – Elliot Alderson Feb 25 at 20:28
  • \$\begingroup\$ It is simple to calculate a divider if it is unloaded. As soon as you load the output (take current from them) the simple scheme collapses. To work around this you can pass a lot of current through the divider and take only a fraction out. But that is very inefficient. So we need to know a lot more about your circuit (To quote No5 : need input) \$\endgroup\$ – Oldfart Feb 25 at 20:32

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