Normally, for the quiescent operating condition of an Op Amp, the differential voltage needs to be Zero between the + and - input terminals.
That might have both resistors return to ground;
or both resistors return to +3V;
or one resistor returns to + 4V and the other resistor comes from the Op Amp output and the output swings to +4V.
The gain of the Op Amp is very high, so it only takes a very small difference from +4V to make the + and - inputs be essentially equal.
Then, the other factor is input offset specification.
Because Op Amps generally are not perfect, there can be a very small input offset voltage or current that causes the output not to be at zero volts when both the + and - inputs are zero volts with respect to each other.
Often, with a feedback resistor from output to the - input, the output will put out just the right small voltage above or below zero volts, so that the small input offset voltage is automatically compensated for.
That might have both resistors return to ground;
or both resistors return to +3V;
or one resistor returns to + 4V and the other resistor comes from the Op Amp output and the output swings to +4V.
The gain of the Op Amp is very high, so it only takes a very small difference from +4V to make the + and - inputs be essentially equal.
Then, the other factor is input offset specification.
Because Op Amps generally are not perfect, there can be a very small input offset voltage or current that causes the output not to be at zero volts when both the + and - inputs are zero volts with respect to each other.
Often, with a feedback resistor from output to the - input, the output will put out just the right small voltage above or below zero volts, so that the small input offset voltage is automatically compensated for.
Thanks @6A3sUMMER !
I would like to propose this variation I was simulating yesterday evening.
The feedback squashes the distortion when it becomes very high as pentodes easily do.
Then I've applied 23% UL (it will simplify the PSU as well), obtaining the following:
Rout is 1.35 Ohm that gives a DF of 2.9
0.5% THD at 1 Wrms and 2.9%THD at 10 Wrms
but mainly... it can be easily driven by a DAC without any driver stage.
On top of that, with this feedback configuration, it partially increases the bias current flowing thorugh the tube when the anode is going up, so in fact increasing the available swing on top. I can indeed increase the voltage of the working point and get more power because I increase the power dissipation where it is usually lower.
...so... what am I doing wrong?
I would like to propose this variation I was simulating yesterday evening.
The feedback squashes the distortion when it becomes very high as pentodes easily do.
Then I've applied 23% UL (it will simplify the PSU as well), obtaining the following:
Rout is 1.35 Ohm that gives a DF of 2.9
0.5% THD at 1 Wrms and 2.9%THD at 10 Wrms
but mainly... it can be easily driven by a DAC without any driver stage.
On top of that, with this feedback configuration, it partially increases the bias current flowing thorugh the tube when the anode is going up, so in fact increasing the available swing on top. I can indeed increase the voltage of the working point and get more power because I increase the power dissipation where it is usually lower.
...so... what am I doing wrong?
Last edited: