-10 to +4 balanced driver amp

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Sleeper

Well-known member
Joined
Jun 6, 2004
Messages
649
Location
Los Angeles
Hi all-

I know i've seen the amplifier I'm looking for, but I can't seem to find it tonight... I have something in mind, but I wouldn't mind a little steer in the right direction. That would be boss.

This would be for my mastering/monitoring console.
I want one of my input choices to be able to run a CD player in direct to my 2k balanced attenuator, so I need 12db of gain and a low to mid impedence output.

I have loads of 5532's and 5534's and I've got a +-15v supply running to the box for my meter buffers and phase metering.

If someone has a discrete idea that would be cool too.

Nothing too colorful is needed, just some plain old vanilla gain.

Thanks
Sleeper
 
Here ya go:

-10to4.gif


Decoupling caps at power supply pins can be .01-.1uF PP, or a 1uF 63V MPP from BC, which is actually nice and low-impedance at high frequencies. Put a 1000uF 50V Panasonic NHG series on the + and - rails too. The feedback caps could be 20pF.

For a 5532, you'll probably need to put a 100uF cap on each output, sad but true. If you use an OPA2604, you won't need them.

The slight extra gain in the top amp compensates for the loading of the 2k attenuator and the 100R output resistors, which are there to keep everything stable.

Peace,
Paul
 
[quote author="Svart"]Hey Paul, how do you calculate the gain from a specific feedback resistor?[/quote]

In the case of the above design, the top amplifier is non-inverting; the gain will be (R3/R2) + 1. In this case, (22600/20000) + 1, or (1.13) + 1, or 2.13. To get decibels multiply 20 * log(gain) ~= 6.6dB.

In the case of the bottom amp, that's an inverter, so the gain is (R6/R7) * -1. Since both resistors are the same value, the gain is simply -1 -- in other words, the same output level as input, but with polarity inverted. The gain in dB will be zero.

The total amplifier has a gain of 12.6dB. Where did the extra 6dB come from? It's there because there are two outputs in opposition to one another, so you're doubling the voltage between the two outputs, and that's equivalent to adding ~6dB.

Nominal -10dBV ~= -7.8 dBu. Adding this amplifier will boost the signal up to a nominal +4.8dBu; the attenuation of the 100R output resistors (R5 and R10) in conjunction with a 2k load brings that down to +4dBu.

A 5532 or OPA2604 can easily put out +18dBu when fed by +/- 15V supplies. Two of them, therefore, can put out +24dBu between them, or +23.2dB when the output attenuation is taken into account. So you have 19.2dB of headroom over nominal level, probably a bit more.

Peace,
Paul
 
Hi Paul, thanks also for a good explanation...
I was in the process of writing this up for svart, looking through my introductory ac/dc electronics book and I'm surprised to have gotten it right.
One part of the circuit made me a bit unsure of the actual math going on here...

I don't understand how this works.:? (but this is not unusual) :wink:

shouldn't the gain on the top amp be (R3+R4)/R2
because one of the 100 ohm output resistors R4 is inside the feedback loop?

and if so, why doesn't R9 raise the gain of the inverting amp above unity, and if the inverting amp isn't at unity, won't my CMRR go crazy.

Sleeper
 
[quote author="Sleeper"]shouldn't the gain on the top amp be (R3+R4)/R2
because one of the 100 ohm output resistors R4 is inside the feedback loop?[/quote]

Actually 1+[(R3+R4)/R2]. And yes, that's what the gain would be if you took the output from the output pin of the opamp. But you don't; you take it from a point down the voltage-divider ladder composed of the three resistors R4, R3 and R2. Which reduces the gain to the point where it works out to, within a close approximation,
1+(R3/R2).

Another way of looking at it: think of R4 as the output impedance of the IC. Open-loop it's 100R (roughly), but you're putting a whole heap of feedback around the amplifier, so its output Z gets reduced to a fraction of an ohm, so it doesn't affect the gain much at all.

The TL07x opamps actually have a resistor built onto the chip at the output -- I forget the value, maybe 75 ohms? Anyway, it's there, but the gain equation for a TL07x is still (for non-inverting) 1+(Rfeedback/Rin). (Pause to go look it up: 128R, plus a couple of 64R emitter resistors on the output devices.)

and if so, why doesn't R9 raise the gain of the inverting amp above unity, and if the inverting amp isn't at unity, won't my CMRR go crazy.

Same answer as above; the resistor is effectively nil because of the amp's feedback. The effective output Z of the circuit is 200 ohms, the sum of the two 100R resistors. Okay, yes, a tiny bit more than that, but not enough to matter much. If you really want to get fussy, add a 154k resistor from the - terminal of the bottom amp to ground. That gets the noise gain of both amplifiers to the same value, at least within the resistor tolerances. I suspect, though, that the tolerances in the 100R resistors will make a bigger difference.

Oh, while we're here, both a 5532 and an OPA2604 will drive your low (2.2k) impedance. But the 5532 will be operating way outside of Class-A, while the OPA2604 has a tricky output stage that keeps it effectively Class-A even into this difficult load. If this is a critical audio stage I'd go for the 2604.

Peace,
Paul
 
Thanks for the further explanation Paul. I'll be sure and match those 100Rs.
the 2604's are on the way and I'm just about thru with the artwork.

Thanks a million!

Sleepr
 

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