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I am trying to calculate the power dissipated in my buffer circuit in order to measure the efficiency of the circuit I have built. The circuit can be seen below, and you can also see the DC values of my voltages and currents. During the process I was in doubt how to calculate the power dissipated, since I have three sources involved in the process.

The thought I was trying to accomplish to calculate the power:

  1. Calculate the power in each element of the circuit individually. For the resistors I would do R.I^2 while for the transistors I would calculate the current passing between collector and emitter multiplied by the voltage from collector to emitter.

Is this way of solving correct? I am not so sure about the method.

enter image description here

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  • \$\begingroup\$ Seems mostly correct. For BJT's, you should also add Vbe * Ib. Most of the time Vbe * Ib will be much smaller than Ic * Vce but might as well be thorough. \$\endgroup\$
    – mkeith
    Jun 2 at 22:10
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    \$\begingroup\$ The most common way to measure efficiency is to measure power in and power out. Power in would be battery voltage * battery current. Power out would be V load * I load. Since the signal input is a sine wave, it would be best to average over N periods where N is an integer. \$\endgroup\$
    – mkeith
    Jun 2 at 22:13
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    \$\begingroup\$ Why not just get the sum of products of each DC supply's voltages and currents? This should work well if the signal is small relative to the supply. \$\endgroup\$
    – Transistor
    Jun 2 at 22:13
  • \$\begingroup\$ @mkeith Ok, so the way I talked is correct, calculating the power dissipated in each element of the circuit. So I could do Pout= (3* 101.10^-3) + 5 (50.6 *10^-3) + 5. (50.3 *10^-3)? Is it correct? For example, in my circuit I use a negative source equal to -3V, would the power related to this source be summed in the same way as I did in the calculation above? \$\endgroup\$
    – Breno Amin
    Jun 2 at 22:36
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    \$\begingroup\$ Pout is the power delivered to the load. So it would be Iload^2 * Rload. To measure the efficiency as an amplifier, you will need to do transient analysis including a full sine wave. Then average the power across one full period of the sine wave. Pin is the sum of power from each of the three supplies. Pout is the power delivered to the load. If you use this approach, you can ignore all the transistors and other passive components (except the load resistor). \$\endgroup\$
    – mkeith
    Jun 3 at 5:31
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It looks to me like you have 4 power sources. Add up the average powers for those and you have Pin You have one load. Work out the power in that and you have Pout
Eff= Pout/Pin

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  • \$\begingroup\$ So I could do Pdissipated= (3* 101.10^-3) + 5 (50.6 *10^-3) + 5. (50.3 *10^-3)? Is it correct? For example, in my circuit I use a negative source equal to -3V, would the power related to this source be summed in the same way as I did in the calculation above? By the Way, Pout (RL) is equal to 80 mW \$\endgroup\$
    – Breno Amin
    Jun 3 at 1:15
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    \$\begingroup\$ (3* 101.10^-3) + 5 (50.6 *10^-3) + 5. (50.3 *10^-3) <BR/> Is the P<sub>in</sub>. If you count the load power as dissipated then that is the dissipated power. Most people exclude the power in the load from dissipated because the thermal management of the driver is being considered. \$\endgroup\$
    – Ken Smith
    Jun 4 at 13:37

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