0
\$\begingroup\$

I want to to add about 20x gain to a microphone I have. I tested the following circuit with satisfactory results:

NE5532 circuit

I was going to build a PCB for it, but then I realized I had no NE5532 around in SMD packaging. I found some LME49990, and reading the datasheet I found the following circuit:

LME49990

So I modified it a little to get a balanced output. I just removed the second part that converts balanced to single ended:

my version

Two questions: 1) Will this work? 2) why is R5 shared between both amps instead of using a resistor to GND like a traditional noninverting amplifier?

My rationale behind the two opamp version is that I have less components in the audio path, so, less chance of adding noise.

Bonus question: do I need to worry about DC here? Should I add coupling capacitors at the inputs?

\$\endgroup\$
  • \$\begingroup\$ Why have you got 4 kHz low pass filters in each input line (final circuit?) \$\endgroup\$ – Andy aka Jan 9 '17 at 15:07
  • \$\begingroup\$ Why does your first circuit have 20 dB of gain when clearly, from the component values it's more like 54 dB? \$\endgroup\$ – Andy aka Jan 9 '17 at 15:13
  • \$\begingroup\$ @Andyaka the 4kHz LPF is a mistake, should be .0047 for 40kHz (for no particular reason), or even .0082uF for around 20khz. in fact, I'll decide when I build the actual PCB if I will populate those or not. The first circuit takes mic level to line level. I only need to boost the mic level to within mic-level voltage, because the recorder I have doesn't support line-level balanced inputs. Only mic. \$\endgroup\$ – hjf Jan 9 '17 at 15:22
1
\$\begingroup\$

Circuit 1 you say is fine except you don't have SMD op-amps - this is what you said: -

I tested the following circuit with satisfactory results

So, I would go with replacing the NE5532 with 2x LME49990 SMD devices. In fact the performance requirements for the final two op-amps are quite weak - they are just unity gain amplifiers to provide a differential output. Have you got any dual SMD op-amps that could be used?

My rationale behind the two opamp version is that I have less components in the audio path, so, less chance of adding noise.

No, once you have achieved the gain from the first stage noises do not add linearly any more because 100 uV noise + 10 uV noise does not equal 90 uV it equals 100.5 uV noise.

Bonus question: do I need to worry about DC here? Should I add coupling capacitors at the inputs?

Yes you do and you should do what you stated - add coupling capacitors. The microphone may be phantom powered or have some dc voltage across it so you don't want to compromise on this. Keep the 3300 ohm resistors to 0 volts - add series capacitors on the mic input connections to the left.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I should have clarified: I dont have 5532 or 5534 in SMD. I have LME49990 and OPA2134. For the test circuits I used TL072 for the first stage and it was much noisier than NE5532 (hiss was noticeable).But, if the second stage is not as important, could I replace the 5534 in the original circuit for a LME49990 and the inverter/follower with TL072? \$\endgroup\$ – hjf Jan 9 '17 at 15:43
  • \$\begingroup\$ You said "I used TL072 for the first stage and it was much noisier than NE5532 " - do you mean noisier than NE5534? \$\endgroup\$ – Andy aka Jan 9 '17 at 15:49
  • \$\begingroup\$ Yes, sorry, noisier than a 5534. As you can guess I'm not good at analog design. So I read up on the differences between these opamps and in some forum discussion someone explained this is due to TL072's JFET design vs 553x bipolar, and how, depending on the situation, one can be noisier than the other. \$\endgroup\$ – hjf Jan 9 '17 at 15:52
  • \$\begingroup\$ Yeah the 5534 is substantially better. The numbers are in the data sheets - it's called equivalent input noise voltage and is listed at nV per sqrt(Hz) which might be a tad mind-boggling but just compare numbers. TL072 is 18 and 5534 is 3.5. The 5532 aint too bad at 5. \$\endgroup\$ – Andy aka Jan 9 '17 at 15:56
  • 1
    \$\begingroup\$ Because you have done some testing on it and have established a level of performance that presumably is acceptable AND proven it's functionality. This means it has a head start on anything else. \$\endgroup\$ – Andy aka Jan 9 '17 at 17:21

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.