I've been working on various QUAD 405 amplifiers for 30 years. I've tried many variations, mostly without the results I was looking for. I just finished this one and it looks promising so far. The goal was to take the input OP AMP out of the signal path and to update the Class A amplifier. I haven't run this up to +/- 50V yet, just to +/- 40V and the square wave response is good with no overshoot. Thoughts and opinions on the circuit?
I don't know how bad it is, but R22 looks like an unnecessary noise source. I think the OP07 has base current compensation, making R22 useless.
Don't you get a 10 dB peak around 113 Hz from the DC loop with the present values? You can design this kind of DC loop to have an accurate second-order Butterworth response, but if my calculations are correct, yours has a Q just over 3.
Don't you get a 10 dB peak around 113 Hz from the DC loop with the present values? You can design this kind of DC loop to have an accurate second-order Butterworth response, but if my calculations are correct, yours has a Q just over 3.
I calculated the quality factor correctly in post #2, but forgot a square root when calculating the frequency. It should be 4.24 Hz rather than 113 Hz.
Getting back to the DC loop:
The circuit consisting of the amplifier and the op-amp with R28, C8 and R17 actually acts equivalent to an LR parallel network to ground in parallel with R29.
Calling the gain of the amplifier A, which in your case is ideally A = 1 + R10/R8:
LDCloop = R28 C8 R17/A
RDCloop = R17
Hence, the natural frequency is
fn = 1/(2 pi sqrt(LDCloop C2))
and the quality factor is
Q = sqrt(C2/LDCloop) * (R17 * R29)/(R17 + R29)
To get a Butterworth response, Q must be 0.5 sqrt(2). You get that when
R29 = 1/(sqrt(2 C2/LDCloop) - 1/R17)
Keeping everything else the same, you would have to reduce R29 to 2686.088913 ohm to get a maximally flat response. As that is awfully small, it may be more practical to reduce C2 and increase R28 to end up with a higher R29. For example:
C2 = 1 uF
R28 = 2.2 Mohm
R29 = 30176.89049 ohm
Getting back to the DC loop:
The circuit consisting of the amplifier and the op-amp with R28, C8 and R17 actually acts equivalent to an LR parallel network to ground in parallel with R29.
Calling the gain of the amplifier A, which in your case is ideally A = 1 + R10/R8:
LDCloop = R28 C8 R17/A
RDCloop = R17
Hence, the natural frequency is
fn = 1/(2 pi sqrt(LDCloop C2))
and the quality factor is
Q = sqrt(C2/LDCloop) * (R17 * R29)/(R17 + R29)
To get a Butterworth response, Q must be 0.5 sqrt(2). You get that when
R29 = 1/(sqrt(2 C2/LDCloop) - 1/R17)
Keeping everything else the same, you would have to reduce R29 to 2686.088913 ohm to get a maximally flat response. As that is awfully small, it may be more practical to reduce C2 and increase R28 to end up with a higher R29. For example:
C2 = 1 uF
R28 = 2.2 Mohm
R29 = 30176.89049 ohm
Last edited:
You are correct about the servo. I was in a hurry and carried it over from my initial testing on a clone board I was using. I've since built a second breadboard and will change it to the servo shown below.
With that large ratio of R12 to (R8 in parallel with R10), you have to look out that the DC control op-amp isn't driven into clipping when there is some DC or deep subsonic content applied at the amplifier input. The control range is about +/- 19.3 mV (the +/- 12.5 V output swing of the OP-07 attenuated by 1 + R12/(R8 in parallel with R10)). Up to +/- 15 mV may be needed to trim out the LSK389B offset, leaving +/- 4.3 mV for other offset sources, DC offset applied at the input and subsonics below the DC loop cut-off frequency (0.08551 Hz).
By the way, the type of DC servo loop in the opening post is actually my favourite. I've explained why in post #23 of this thread: https://www.diyaudio.com/community/threads/dc-servo-using-the-opa1656-cmos-opamp.388777/post-7082208