Here's a sim close to open loop with the drive on the lower MOSFET
I was interested to see if there was symmetry from driving the lower side only and there does appear to be. BTW I have matched the area of the driver MOSFETs Kp's for equal gain.
I was also interested to see the available loop gain and it's rolloff frequency. I added 100k resistors across the ideal current sources otherwise there is no LF end gain limiting. (Unexpectedly, the driver MOSFETs drain conductance's do not appear to contribute to the rolloff).
The plot of I(G1) tells us there is about 20dB of feedback at play so the open loop gain can be estimated to be 40dB+20dB or 1000. And the open loop folloff is 1kHz/10 or 100Hz. So the linearity in this simulation looks reasonable below 100Hz, eg at 1Hz shown below (left) and then 100Hz (right)
So given sufficient negative feedback the linearity can be made acceptable. And the open loop linearity appears very similar to other autobias circuits I have looked at that give non-switchoff behaviour in Class-AB. With 20dB feedback here the THD is around 1% so 40dB more feedback would bring it down to about 0.01% and the HF rolloff should be acceptable up 20kHz.
I also show temperature of the power MOSFET's stepped with a 10 deg C change. The idle current increases from about 100mA to 150mA warmed. So thermal feedback from the power transistors will be needed (somehow).
Cheers, IanH
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I had started to think about the various places that emitter resistors could be used to modify the transistion, there's quite a few possibilities...
So thanks for that.
And thanks to you for that.
Part of the benefit of the Monticelli circuit is that it is supposed to drive the outputs symmetrically, even when driven on one side.
You have demonstrated this, your one sided drive ends up pretty symmetrical, is there any improvement if you connect the drive currrent source symmetrically to both sides?
And thanks for the FET version too.
What LTSpice model do you have for the subthreshold conduction?
I never doubted that thermal feedback would be needed for the power transistors.🙂
I think it will need no more than the usual "bias transistors as close as possible on the heatsink to the outputs" used in conventional audio power amplifiers.
Best wishes
David
What I find disturbing about this topology is its frequency behaviour when driven through its dedicated nodes:
The closed-loop bandwidth barely reaches 150kHz, yet the gain is low, there are no capacitors, and the 2N2222/2907 are not particularly slow. Quite puzzling
The closed-loop bandwidth barely reaches 150kHz, yet the gain is low, there are no capacitors, and the 2N2222/2907 are not particularly slow. Quite puzzling
The closed-loop bandwidth is about 400 kHz in the simulation. With a 1500 ohm feedback resistor, 250 pF of collector-base capacitance of the 40 output transistors together could cause that. That's 6.25 pF per transistor. I've neglected various current divisions and all other capacitances, when you don't, the result will be lower.
The OPA1611 has some local capacitive feedback around the OPS and this results in a pretty low open loop output impedance as the frequency increases.
So it makes sense.
One of my uses is as an amp for very efficient horn loaded drivers, so I am particularly concerned about noise.
So I planned to keep the feedback resistors quite low just to reduce Johnson noise, nice if there's other benefits.
What I don't yet understand is the capacitor between the level shifters in the 1st patent picture, supposedly a feedfoward for stability.
Need to check that once I am back at my sim PC.
Best wishes
David
Driving the circuit via the bases of and Q4 and Q8 looks more advantageous, as it benefits from their follower action. Is there a drawback though?
Cancellation between the signal currents from Q2 and Q1. You drive the bias setting rather than the output current.
I feel like the purpose is to improve the slew rate. For single-sided drive, either rise or fall can be poor. Isn't that an improvement?
Maybe, but so what?
It still works as intended, non-switching and always-on bias, and the linearity seems greatly improved; I say "seems", because the circuit is now inverting, and the loop gain is greatly increased, all of which obviously contribute, but the core linearity is certainly not seriously degraded, otherwise these structural improvements would be mostly nullified
I didn't know you also wanted to drive Q10 and Q6. If you had only driven Q4 and Q8, you would have got partial cancellation. Anyway, to me, it looks like an inelegant way to add a gain stage to the loop.
Leaving the other transistors dangling over nowhere didn't seem like a sensible option, which is why I kept the whole topology.
The gain increase is not that high in the end: ~4, and the core linearity doesn't look much better:
Yet, the fact is that the overall linearity is greatly improved; a cancellation mechanism somewhere maybe?
The gain increase is not that high in the end: ~4, and the core linearity doesn't look much better:
Yet, the fact is that the overall linearity is greatly improved; a cancellation mechanism somewhere maybe?
No, the drawing is somewhat convoluted because I made quick and dirty fixes to address the inverting configuration, but there is no positive feedback
What I thought you intended to do when you mentioned driving Q4 and Q8, was to drive Q4 and Q8 and add the appropriate current sources to bias all other transistors normally. In that case, you would essentially have driven Q1 and Q2 in common mode.
This is the simulation I did. All the transistors are default models (BF=100, Is=1e-16) because I am using the circuit from when I simulated this circuit before.
One for the original drivepoint
and one for the Elvee drivepoint.
Now I don't think wrong. Please check this.
Blue is Ic of Q3, Green is Ic of Q4, Red is the current through the load.
The dotted line is the original driving point, and the solid line is the Elvee driving point.
It's hard to see, so I enlarged it.
The gain (transconductance) is significantly lower, but the polarity is still the same.
Therefore, I would think that change the feedback from the emitter (CNF) to the base would be positive feedback.
Didn't latch up even with positive feedback because the gain was extremely low? I thought.
However, the simulation result of Elvee's circuit is indeed an inverting amplifier with a gain of exactly G=-3. It's a riddle. 😵