Trouble with the Elliott "Ultra-Simple Microphone Preamplifier" (Project 122)

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

RiffRalf

Member
Joined
Dec 29, 2015
Messages
21
Location
Sweden
I'm new here ( as a poster but not as a reader  ;) ) and looking for some help I guess.

I've been trying to finish the ESPs NE5532 preamp for some time now. I built it twice with exactly the same result: no signal out and low voltages on pin 1-7.

I modified the schematic to have a XLR female as output instead of the 10 pin ribbon connector. Pin 9 ("-12")  is connected to ground and "out" is connected to pin 2 of the female XLR (pin 3 & 1 to ground). Other that I followed the schematic very carefully.

NE5532:
1. 1.30v
2. 0.83v
3. 1.23v
4. 0v
5. 0.25v
6. 0.84
7. 1.30
8. 18.30v

http://sound.westhost.com/project122.htm
p122-f1.gif

NE5532_0.JPG


 
JohnRoberts said:
You measure 0V on pin 4, that your schematic shows connected to -12V. Fix that.

JR

I may be a bit slow...  what should I do? Shouldn't the "-12" be connected to ground? :-[
 
RiffRalf said:
JohnRoberts said:
You measure 0V on pin 4, that your schematic shows connected to -12V. Fix that.

JR

I may be a bit slow...  what should I do? Shouldn't the "-12" be connected to ground? :-[
Ground is 0V, if -12V was supposed to be 0V it wouldn't be called -12V.

With the input of the op amp sitting at 0V, and AC signal swinging above and below 0V, the op amp needs to swing negative to follow the input signal.

pin 4 needs to be -12V so the op amp can pull negative to cleanly pass the positive and negative swings of the AC signal.

JR
 
Brian Roth said:
These types of opamp circuits require a bipolar power supply.  IOW, a positive and negative DC voltage referenced to "ground".

Bri

And here I thought I was a good boy for building a rectifier bridge for my 12ac adapter.  ::)
 
JohnRoberts said:
RiffRalf said:
JohnRoberts said:
You measure 0V on pin 4, that your schematic shows connected to -12V. Fix that.

JR

I may be a bit slow...  what should I do? Shouldn't the "-12" be connected to ground? :-[
Ground is 0V, if -12V was supposed to be 0V it wouldn't be called -12V.

With the input of the op amp sitting at 0V, and AC signal swinging above and below 0V, the op amp needs to swing negative to follow the input signal.

pin 4 needs to be -12V so the op amp can pull negative to cleanly pass the positive and negative swings of the AC signal.

JR

Turns out none of my multimeters (analog and digital) are able to display + or - just the voltages and the adapter do not tell if positive/negative is on tip or barrel.  Already tried with a led and resistor but I get light both ways, of course.
 
RiffRalf said:
Turns out none of my multimeters (analog and digital) are able to display + or - just the voltages and the adapter do not tell if positive/negative is on tip or barrel.  Already tried with a led and resistor but I get light both ways, of course.

That's not right. If it is a dc supply then the LED should light only one way. If it lights both ways it implies the output is AC. Is there nothing written on the adapter that tells you what its output should be?

Cheers

Ian
 
Hi,

There's an Elliott page with a simple bipolar power supply, too, here: http://sound.westhost.com/project05.htm.  Your circuit will work OK with +/-15V, like in the project on that page.

You could also get two regulated DC wall-warts and stack them, or use two 9V batteries (two batteries will be the easiest way to do it).

If you want to modify the circuit, you could use a single supply and bias the opamps to halfway up instead of to ground.

You will likely want a blocking capacitor on the output, too (you certainly will if you run it single-supply).

With a multimeter if you put the black probe to ground it should tell you whether the red one is more positive or more negative.
 
ruffrecords said:
RiffRalf said:
Turns out none of my multimeters (analog and digital) are able to display + or - just the voltages and the adapter do not tell if positive/negative is on tip or barrel.  Already tried with a led and resistor but I get light both ways, of course.

That's not right. If it is a dc supply then the LED should light only one way. If it lights both ways it implies the output is AC. Is there nothing written on the adapter that tells you what its output should be?

Cheers

Ian

Yes, it is indeed a AC adapter, as I wrote in a post earlier. Unfortunately it does not tell what is the positive or negative side of the barrel and tip.
 
RiffRalf said:
ruffrecords said:
RiffRalf said:
Turns out none of my multimeters (analog and digital) are able to display + or - just the voltages and the adapter do not tell if positive/negative is on tip or barrel.  Already tried with a led and resistor but I get light both ways, of course.

That's not right. If it is a dc supply then the LED should light only one way. If it lights both ways it implies the output is AC. Is there nothing written on the adapter that tells you what its output should be?

Cheers

Ian

Yes, it is indeed a AC adapter, as I wrote in a post earlier. Unfortunately it does not tell what is the positive or negative side of the barrel and tip.
AC is both, doesn't have a + and -...  You need to rectify the AC to make it + and - DC..

Do some more study time on power supplies...

JR
 
Sorry for all the stupid questions, I'm very new to this AC-businesses. DC is usually my home turf.

dfuruta said:
Hi,

There's an Elliott page with a simple bipolar power supply, too, here: http://sound.westhost.com/project05.htm.  Your circuit will work OK with +/-15V, like in the project on that page.

You could also get two regulated DC wall-warts and stack them, or use two 9V batteries (two batteries will be the easiest way to do it).

If you want to modify the circuit, you could use a single supply and bias the opamps to halfway up instead of to ground.

I don't have the required caps and regulators to build that supply and time is of the essence, I'm afraid.

Stacking two DCs is a way to go.  Did not know you could do that, thanks!

dfuruta said:
You will likely want a blocking capacitor on the output, too (you certainly will if you run it single-supply).

Should I add an electrolytic or non polarized cap?

dfuruta said:
With a multimeter if you put the black probe to ground it should tell you whether the red one is more positive or more negative.

Aha! I connected the black probe to one side of the tip/barrel and the red one to the opposite, that is probably why I only got positive voltage.


 
Connected an 12acv adapter and got this + a loud hum in my speakers. No hot ICs, at least not yet.

Black probe on ground (not black wire off the acv input) and the red probe on the pins.

NE5532:
1. 4.5acv
2. 4.1acv
3. 4.3 acv
4. -7.1 acv (guessing it is negative since my new multimeter apparently cannot not show + or - when measuring in V~ )
5. 4.45 acv
6. 4.48 acv
7. 4.5  acv
8. 7.1  acv
 
Black probe on black wire off the acv input and the red probe on the pins.

NE5532:
1. 1.2 acv
2. 1.6 acv
3. 1.7  acv
4. -14.3 acv switched black probe to red on volt input
5.  1.6 acv
6. 1.6 acv
7. 1.6  acv
8. 14.3  acv
 
RiffRalf said:
Connected an 12acv adapter and got this + a loud hum in my speakers. No hot ICs, at least not yet.

You can't just connect a 12 VAC adapter and expect it to work. The op-amp wants a DC supply. So you need to rectify the AC, which gives you DC, and then you probably want to regulate the rectifier output.

Since you have an AC output wall-wart which has only two wires, to get a bipolar DC output you will need a voltage doubler (see http://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html, it's the first diagram, the "full-wave voltage multiplier").

diode50.gif

The DC output of the doubler will be double the RMS AC input voltage. That doubler output is between the two diode outputs.

Look at that schematic again. One pin of the transformer secondary (one of your wall-wart's barrel connections) goes to both diodes, the other goes to a midpoint (where the caps meet). That midpoint is declared to be 0 V, so if you measure between that point and each of the two diode outputs you'll see your expected DC voltage. You want to measure the voltages at the doubler outputs on the DC meter, not AC.

Also remember that your wall-wart is specified to have an RMS AC output voltage. So a 12 VAC RMS input will give you about 16.9 VDC at each of the two doubler outputs.  You can use the standard LM317 and LM337 regulators to bring those down to solid -12 VDC and -12 VDC outputs, with the AC ripple attenuated.
 
Andy Peters said:
RiffRalf said:
Connected an 12acv adapter and got this + a loud hum in my speakers. No hot ICs, at least not yet.

You can't just connect a 12 VAC adapter and expect it to work. The op-amp wants a DC supply. So you need to rectify the AC, which gives you DC, and then you probably want to regulate the rectifier output.

Since you have an AC output wall-wart which has only two wires, to get a bipolar DC output you will need a voltage doubler (see http://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html, it's the first diagram, the "full-wave voltage multiplier").

diode50.gif

The DC output of the doubler will be double the RMS AC input voltage. That doubler output is between the two diode outputs.

Look at that schematic again. One pin of the transformer secondary (one of your wall-wart's barrel connections) goes to both diodes, the other goes to a midpoint (where the caps meet). That midpoint is declared to be 0 V, so if you measure between that point and each of the two diode outputs you'll see your expected DC voltage. You want to measure the voltages at the doubler outputs on the DC meter, not AC.

Also remember that your wall-wart is specified to have an RMS AC output voltage. So a 12 VAC RMS input will give you about 16.9 VDC at each of the two doubler outputs.  You can use the standard LM317 and LM337 regulators to bring those down to solid -12 VDC and -12 VDC outputs, with the AC ripple attenuated.

Thank you so much for the explanation Andy, very helpful! Last time I felt like such a beginner as I do now was when I started out with electronics. I got a lot to learn.

So you are telling me that if I take adapter and hook one of the wires up in between C1 and C2 and the other one in between D1 and D2 I'll have +16.9 and -16.9 VDC at the ends of the Electrolytics that would serve this circuit? I would like to skip regulators and keep the voltage at 16.9 to increase the headroom of the op amp, unless of course the regs are really necessary?
 
RiffRalf said:
So you are telling me that if I take adapter and hook one of the wires up in between C1 and C2 and the other one in between D1 and D2 I'll have +16.9 and -16.9 VDC at the ends of the Electrolytics that would serve this circuit? I would like to skip regulators and keep the voltage at 16.9 to increase the headroom of the op amp, unless of course regs are really necessary?

With no load, you will get the ± 16 V. But once you add a load, you will get ripple on your outputs, which can be significant. Adding the regulators will do much to minimize the ripple.

Remember that for voltage regulators, you must mind the drop-out, which is the difference between the input voltage to the regulator and its output which must be maintained in order for the regulator to work. For LM317-type regulators, that's usually 2 V for light loads and 3 V for heavy loads (Read The Fine Data Sheet). 

A further complication is what happens when you have low mains input to your wall wart? If the thing is spec'ed for 12 VAC output with 120 VAC mains input, what happens when the mains drops to 108 V?

-a
 
If time is of the essence forget about it for now, grab two VDC wallwarts (should be easy enough to find lying around?), and stack 'em up.
 
You are trying to build a quality mike preamp. Why dont you build a proper power supply instead of farting about with an a.c. plugpack? Its never going to give you a good result. There are loads of circuits, and boards available, to build a proper supply.
 
Back
Top