0
\$\begingroup\$

I'm experimenting with a mic preamp based on NE5532. I'm using the following schematic with some modifications:

enter image description here

The modifications are:

  • the positive supply is directly connected to the power supply(no noise filtering for the power, but I'm using a chemical battery)
  • the 4.7uF cap is missing.
  • NE5532 instead of LM833

The part that is marked with a green rectangle is only for creating a 4.5V reference with low output impedance.

Voltages at the points marked with red circles, with almost no signal coming in from the electret mike:

  • Battery is 8.5V
  • 1 4.3V (5% metal film resistors, within error)
  • 2 4.27V
  • 3 4.33V
  • 4 1.4V

So it seems that the opamp cannot make its inputs equal. There is too much difference, I guess that cannot be an opamp offset error. Probably it cannot go lower than 1.4V at its output.

If I disconnect the 10uF capacitor from the +input, then I measure these:

  • 1 4.33V
  • 2 4.27V
  • 3 4.33V
  • 4 4.59V

First I thought that the 10uF capacitor is leaking. But it is not...

Here is the input vs. output signal when the electret microphone is receiving a 440Hz sinusodial sound wave:

enter image description here

It is clear that the output signal is clipped, and it is because the opamp cannot go below 1.4V. What I don't get is that WHY it needs to go below 1.44V to make its inputs equal? Does it have to do something with the 4.7uF cap being removed?

UPDATE: The cap was not simply removed but replaced with a piece of wire. Sorry for being ambiguous.

UPDATE2: Replaced 100 Ohm resistor with 1K, so the overall gain is reduced to about 10dB. The I put back the cap, but changed to 10uF. The corner frequecy changed from 339Hz to 16Hz, as it was suggested. Now the output at point #4 is also at 4.3V, and the distortion is not visible:

enter image description here

Conclusion: never forget to eliminate DC amplification when don't need it

\$\endgroup\$
  • 1
    \$\begingroup\$ So - based on a comment to an answer below, "the 4.7uF cap is missing" and "the 4.7uF cap being removed" translate into "I replaced the 4.7uF cap with a piece of wire"?? If so then that explains everything, and you should have been clearer up front, since what you wrote in your question implies that there's nothing there, not a piece of wire. \$\endgroup\$ – brhans Jun 10 '18 at 19:01
  • \$\begingroup\$ You are 100% right, I'm sorry for the bad description \$\endgroup\$ – nagylzs Jun 10 '18 at 19:31
1
\$\begingroup\$

Yep, the 4.7uF cap makes the gain at DC unity so that small offsets don't matter so much, as it is you have 40dB of gain (seems excessive for a powered capsule) right the way to dc, so it only takes a few mV of offset to drive the thing into saturation. A bipolar opamp with 100k on one leg and 100R on the other will typically have more offset then that just due to the bias currents. 4.7uF BTW is a bit small if going for that much gain, you will loose all the low frequency response.

I also don't really like that 10uf in the output of the mid rail generator, it hurts stability, at least put a 50 ohm or so build out resistor before it.

\$\endgroup\$
  • \$\begingroup\$ I think you meant 40dB gain= 10k/100R gain \$\endgroup\$ – Sunnyskyguy EE75 Jun 10 '18 at 17:18
  • \$\begingroup\$ I see, so by replacing that cap,with solid wire, I have changed DC unity gain into 40dB, and any input offset is multiplied by 100 on the output. \$\endgroup\$ – nagylzs Jun 10 '18 at 18:33
  • \$\begingroup\$ Urk, yea, not sure where I got an extra 20dB from, fixing it.... \$\endgroup\$ – Dan Mills Jun 10 '18 at 20:35
2
\$\begingroup\$

Post Mortem ( after answer accepted)

The real problem with this question is the user copied a design without understanding what every part does and then realize not every design you find is perfect. ( user needs to read more App notes until understood )

Lessons to Learn

Perfection in Engineering IMHO means meets or exceeds all specs or measureable requirements (cost, performance, quality metrics) but if you dont have any or don't know what is important, Warning! this is not Engineering rather a casual hobby interest. The best designs simply meet the best specs. Often when pressured for schedules we must accept this criteria even if we wanted to do more and so we think it's only 90% complete. So learn to make hard decisions on specs.

The Mic has an active JFET open drain so the pullup resistor of 10k determines AC voltage gain and DC operating point and if matched perfectly will be mid-scale on Op Amp output of full output swing or Vbat/2.

This makes the Vbat/2 buffer circuit in the green square completely irrelevant* along with 100k resistor. The Gain ratio is AC coupled by 4.7uF in series with 100 Ohms and drops -3dB at 340 Hz and ought to be 5 to 10x bigger for better bass response.

Why*?

Since the the 4.7uf feedback cap (or bigger!) blocks DC but needs low impedance due to 100R it can now go to ground and no longer needs the 100K bias R to V+/2. With matched input series resistors of 10K from FET pullup to Vin+ and 10k Feedback to Vin- the input bias currents produces the same voltage drop thus cancelling but since the DC gain is 1+1 we should expect the output to be roughly V+/2 within 10% or less.

It is also important to verify leaky caps.

Removing the 4.7uF proved that but then when you saw the half wave you doubted it. Consider that during the input zero crossing, the output ought to be at the same as the DC voltage 4.5 or so.

final

To determine AC gain. define impedance and minimum used voltage input and available output swing. Then define load impedance and choose high Z headphones (e.g. 300 Ohm Sennheiser ) for lower power or 60 Ohm Buds but if wanting to use a 9V battery , 8 Ohms will drain the battery fast Most likely. 40dB gain is too high. Examine any high quality audio amp specs and consider at least a dozen critical specs in your next design.

\$\endgroup\$
  • \$\begingroup\$ You are right, this is just a hobby for me. I only do this because I'm interested in electronics. I was experimenting for a while before I asked this question here. Didn't know any other place to ask, and I don't have a teacher at hand. It was a good lesson to learn. But I guess this is how we all learn: first read some books, then try things out and learn from mistakes. \$\endgroup\$ – nagylzs Jun 11 '18 at 18:01
  • \$\begingroup\$ Yes and the more of all the above, the better. What I did in 1975 before I graduated was read every DEsign Electronics magazine 1page submitted design and the Theory of Operation. This was much more helpful for me to start designing in my 1st job in R&D. You can do same since they are all on archive.org \$\endgroup\$ – Sunnyskyguy EE75 Jun 11 '18 at 18:13
  • \$\begingroup\$ I would be careful trying to use the mic as part of a potential divider for main amp bias, Idss is (Like bipolar Beta) usually horribly badly specified. Better practise is to bias the mic separately from the gain stage, makes things a lot more predictable. \$\endgroup\$ – Dan Mills Jun 12 '18 at 16:28
1
\$\begingroup\$

Battery is 8.5V

The minimum recommended supply for the NE5532 is 10 volts. I'm not ruling out other problems but this would be a show stopper for me.

If you want my advice show a circuit diagram that depicts exactly what you are trying out and don't expect folk to make a mental picture based on the incorrect diagram and a bunch of notes.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.