Simple NE5532 gain stage?

[capnspoony]

Is it ok to use a NE5532 chip for 20db of makeup gain as a simple inverting amplifier using the

Inverting amplifier:

Vout//R2 (R2 being the feedback resistor) = -Vin/R1 (R1 being the minus input resistor)

Vout/Vin = -R2/R1 = voltage gain.

Could I make R2 200K and then R1 10R and wala .. or is it not that simple?

how does voltage gain convert to db?

best,

Richie

 

[bigugly]

Okay, let's run some numbers.

Voltage gain has a logarithmic relationship to dB. The equation is: 20*log10(X)=dB. Where X = your voltage gain.

Vgain=R2/R1.
Using your figures: 200,000/10=20000 whoa, that's a lot.

You need a Vgain of 10 to get 20dB. So that could be: R2=10000, R1=1000 or R2=100000, R1=10000.

Also R1 will determine what your input Z is so depending on what is feeding your input you should keep it high(ish). I wouldn't go less than 1k on the inverting input if the source Z is low, like the output of an opamp. If it is higher I would suggest making R1 roughly 10 times the source Z. What exactly are you trying to do?

Hope that helped,

James

 

[capnspoony]

I have a step down transformer 2:1

on my input .. then a passive eq which goes another 12db

so I have around 24 db to make up

-richie

 

[pstamler]

Actually no; your 2:1 transformer (I assume that's the winding ratio, not the impedance ratio?) drops the voltage level by 6dB. Combine that with the 12dB you're losing elsewhere and the total loss is 18dB.

What kind of level do you want your next device to see? +4dBu nominal, maybe +24dBu max? A 5532 will drive +24dBu if you configure it as a balanced driver (using both halves). So if you use one half as a non-inverting amplifier (gain = 4x, or 12dB), drive one side of the output with that amplifier but also feed it into the other channel which is wired as an inverting amplifier with unity gain, you'll get 18dB of balanced makeup gain. (Where does the extra 6dB come from? The extra equal-but-opposite leg of the output.)

With a 5532, though, you'll need to look at issues of input bias current and DC. Possibly an OPA-2604 might be more appropriate; it has a FET input and essentially no input bias current.

Peace,
Paul

 

[capnspoony]

thanks paul

yes sorry it is 6db and it is the winding ration I was talking about.

what do you mean by gain = 4x? just four times the amount? Or does that x hold some value?

To wire an opamp with unity gain I simply have a wire from the output to the minus inpur, correct?

I'm not familiar with input bias current... Im just reaching my first birthday of DIY.

thanks
--richie

 

[pstamler]

[quote author="capnspoony"]yes sorry it is 6db and it is the winding ration I was talking about.

what do you mean by gain = 4x? just four times the amount? Or does that x hold some value?[/quote]

No, I'm talking about 4 times. For voltages, db gain = 20 x log (Vout / Vin). The log of 4 is about 0.6; 20 x 0.6 = 12dB. So a gain of 4x = +12dB.

To wire an opamp with unity gain I simply have a wire from the output to the minus inpur, correct?

Only for a non-inverting stage. Non-inverting stages have a gain of (Rfeedback / Rinput) +1. Rfeedback is the resistor from the - terminal to the output; Rinput is the resistor from the - terminal to ground. So if, for example, Rinput is 2.5k and Rfeedback is 7.5k, gain is (7.5 / 2.5) + 1, or 3 + 1, or 4. Which is 12dB, as described above.

For an inverting stage, the math is different. There, the gain is equal to -1 x (Rfeedback / Rinput), where Rinput is now the resistor going from the input to the - terminal. So if Rinput and Rfeedback are both 10k, gain is -1 x (10k / 10k) = -1. The - sign means the signal's polarity is inverted.

So if you put a non-inverting amplifier after your passive network and feed its output to pin 2 on your XLR output jack and also to the input of an inverting amplifier, whose output goes to pin 3 on your XLR output jack, you'll get an output signal on pin 2 which is 12dB greater than the input to that opamp, and a signal of equal level but opposite polarity on pin 3. The double signal (balanced) is the equivalent of 6dB additional gain, so your total is 18dB of makeup gain, which is what you need.

I'm not familiar with input bias current... Im just reaching my first birthday of DIY.

That explanation takes longer; I suggest finding a copy of Walter Jung's IC Op Amp Cookbook. It's out of print, I think, but a lot of public libraries have it.

Peace,
Paul

 

[capnspoony]

thanks for the help paul.

I'll be sure to look for the cookbook... wonder if ebay has it.

-richie :thumb:

 

[PRR]

> inverting amplifier

Why "inverting"? Mostly the ear hardly cares. Reading ahead, you have a transformer, you can always swap two leads and make the overall polarity come out right.

In general, use the non-inverting plan until you find a compelling reason otherwise. It usually can have higher input impedance and lower noise.

Mix-amps, including some active EQs, can be advantageously built inverting. Many other odd corners where inverting is best or simplest. And inverting avoids common-mode distortion, which is real, but went undiscovered for so long I wonder what the problem is for everyday chip-slapping.

> then a passive eq which goes another 12db

Many passive EQs need a load. My old Bogen did nothing unless I hung 600.00 (or 470) ohms on the end.

If the required load scales out to be a happy inverting amp, then it will be simpler than a non-inverting. You need 18dB make-up? That's near 1:10. If the EQ load is 600, then the opamp feedback is 6,000. Seems fine to me.

But FWIW, on my Bogen I used a non-inverting makeup amp. Habit, and also I could modify the EQ load independantly of the make-up gain. I had a 100K input resistor for initial testing, and then I shunted that with 2K, 1K, 470 until the undocumented EQ gave sufficient dips (it was dip-only). Then I could fiddle the gain resistors for overall gain without upsetting the EQ response. It could have been done in less parts, but the gig was only hours away. "Works NOW" was more important than "saved $0.12".