Circuit Question

[PerroGrande]

Most of the time, I can figure circuits out without too many problems.  However, occasionally my brain refuses to find fourth gear and this is one of those cases.  The following circuit appears all over the place in Boss designs, notably the BD-2.



From a macro level, I think I know what is going on here.  The configuration looks a lot like a long-tailed pair differential amp.  The output of the differential amp is amplified by the PNP Bi-Polar transistor (the 100pF cap limits its high-frequency response by providing more negative feedback as the frequency increases).  The output is taken from the collector of this transistor.  However, a portion of this signal is also taken into the "other" input of the differential amp.  Several voltage dividers are involved, including a potentiometer making this an adjustable gain stage.   The amplified signal (from the PNP) is *in-phase* with the original signal (flipped once by the FET, again by the PNP), so when it is "added" to the differential amp, it reduces the overall gain. 

There are several design aspects that I don't quite understand, however, and would like to ask/discuss.

1) Why choose this circuit instead of an op-amp or multi-stage amplifier? 

2) I can *not* wrap my brain around the DC biasing of this thing. 

3) The linearity of this design looks decent on the simulator.  I would expect that adding emitter degeneration would drastically improve the stability of the circuit, yet the designers did not do so -- and the results look good!

4) The "first" (Q1) FET is pulled up to a 4V VRef.  Is this done because the FET is acting more like a variable resistor (i.e. linearizing)?

Hopefully I'm not being too dense here..  I haven't had *that* much to drink this holiday!

[Sir H C]

Not being too dense.  It is a discrete op-amp.  Why use this instead of a packaged op-amp?  I can not really say, maybe they wanted the lower gain this offers, the ability to put the pole (the cap across the bipolar) where you want it, or some other effect this offers as opposed to a packaged op-amp.

The dc biasing is the R6 to 4 volts.  That sets the input to 4 volts, and then the op-amp function forces the output to near 4 volts (based on the gain of the circuit.

Emitter degen would help some with the DC offset of the output but you would lose some gain in the process unless you put a cap there and that would be one more part and cut some of the potential swing of the thing.

The signal taken back into the other side of the diff pair is the negative feedback.  What you see is what would be in most FET input op-amps with the exception that the tail and the bipolar load would most likely be active devices. 

[DDD]

Yes, there is just a "typical" Op Amp arrangement on the schematic diagram.
I think B@ss use this configuration often because of the following:
1. kind of "more soft" saturation relatively to the IC OA
2. there is less "junk" like integrated current sources, semiconductor resistors and capacitors, e.t.c. that usually can be found on the OA chip. Hence - less unwilling artifacts and mud in the sound.
3. specific discrete FET mojo :-)   

[PerroGrande]

No question that I see the building blocks of an op-amp -- at least the input stages (minus the active loads that were mentioned).  This circuit *will* swing very close to the rails before clipping -- but not any closer than a rail-to-rail op-amp will do. 

Back to the biasing --

I see the 4v operating point, which in turn will cause the source(s) to climb to a point above 4V.  What still eludes me is how the PNP transistor is kept within its operating point so well.  Obviously, some action at DC is doing this, but I just don't see it yet.  Sorry if I'm being dense here (it happens with age, they say, along with memory loss and something else that I can't recall at the moment) -- but I like to understand stuff dead cold.

[Sir H C]

The PNP is kept in check by the negative feedback going to the diff pair.  The opamp wants to keep the output at the same voltage as the input (with no signal), so this works to get the voltage across R2 to the right value to keep that bipolar happy.  Now this voltage varies dramatically with temperature, so you end up having that diff pair skewed all over the place at cold or hot.

the output can swing close to the rails with high gain, at unity gain, you can not get that close or the tail current source/resistor will crap out, nor too high based on the JFET pinchoff point.

[R.G.]

1) Why choose this circuit instead of an op-amp or multi-stage amplifier?

That's a tough question. Could just be cheaper, given the price of transistors, but that's not a big issue. There are some clear differences between discrete and monolithic opamps. The discrete one may actually have better characteristics for some audio situations, since it will usually have lower open loop gain and therefore need a less drastic compensation pole. It may have better large signal swing, there's a whole bunch of edge conditions that might be the choice.

Monolithics are far better at DC accuracy (not needed here) and gain accuracy (again not needed).

Maybe the designer was bored with opamps. 

In any case, it is just an opamp, if a non-standard one.

2) I can *not* wrap my brain around the DC biasing of this thing.

It's an opamp. The entire insides of the opamp do not matter as long as there is enough gain to drive the output to a point where the (-) input will balance the (+) input. The open loop gain is at a guess about 4000 to 10,000, so moving the output from 0V to +9V (assuming that it could go that far) would only take an input imbalance of 9/4000 to 9/10,000 volts, 900uV to 2.25mV. So with one input held at 4V by the bias resistor, the output only has to move until it gets the other input at 4V +/- a few millivolts. The difference in the Vgs and mu of the JFETs on the input will be far larger.

Feedback is amazingly powerful at setting fixed operating points.

3) The linearity of this design looks decent on the simulator.  I would expect that adding emitter degeneration would drastically improve the stability of the circuit, yet the designers did not do so -- and the results look good!

Actually, internal emitter degeneration might make it LESS linear by reducing open loop gain faster than it increases internal linearity. This is something that you really have to delve into opamps and feedback a bit to understand. The linearity of the internal gain is almost a moot point. If there is enough gain, feedback will reduce the internal nonlinearity by the feedback factor. So for an open loop gain of, say, 5000, and an internal open loop distortion of maybe 5%, using this at a gain of one as a follower would reduce the internal distortion to 0.001% by using up all the feedback to correct the distortion. It is by no means clear that linearizing one stage would reduce the internal distortion faster or at the same rate as it decreases open loop gain.

4) The "first" (Q1) FET is pulled up to a 4V VRef.  Is this done because the FET is acting more like a variable resistor (i.e. linearizing)?

No, it's done to put the input in the center of the input common mode range. For this to work as an opamp, the input devices have to be able to swing linearly. What follows is not a particularly well structured walk through the circuit, hitting numbers and things as I think of them. I'd do better if I was writing up an article; but I thought you might like to see what I think about as I go.

The BF245A has Vgsactive = -0.4 to -2.2V, Idss of 2 to 6.5ma, and mu = 3 to 6.5mmho.
First the gain: 3 to 6.5mmho is 3 to 6.5 ma/V.
With a gate at 4V, the device's source will be sitting at 4.4 to 6.2V. We'll assume the JFETs are balanced for the moment; so the current in the 4.7K source resistor is 4.4 to 6.2V/4.7K = 936uA to 1.3ma;
Since Idss is bigger than that, either JFET can, if forced, eat the whole current for the pair. So the noninverting input JFET can suck as much as 936uA through the bipolar's base if its gate is pulled up.
The gate only needs to pull up higher than the other gate by enough for about 1ma to flow; and that's 1/3 of a volt (from the mu value).
On the other end, the 2N5087 has a current gain of at least 250 at 1ma from its datasheet. So for it to move the output voltage from 0V to 9V (0ma to 4ma), it only needs its base to change from 0ma to 4ma/250 = 15uA. The input JFET only needs to move 0.015ma/ 3ma/V = 5mV to change its drain current enough to do that.
So a differential gate voltage of only 5mV can drive the output from fully on to fully off. Open loop gain looks like it's in the ballpark of 9V/5mV = a little under 2000. Good enough for discrete opamps.

With one gate at 4V, the other JFET gate only has to move 5mV higher or lower than the input JFET's gate to run the output from fully on to fully off. Because of the direction each gate moves the output (i.e. + or -) the inverting feedback gate will find a voltage somewhere within 5mV of the input gate; we've just shown that the output CAN do this, and since the direction of feedback is set up to allow it, it will happen.

The only problem with this analysis is that the JFETs are not identical unless selected. So we could have one JFET with the Vgsactive of 0.4V and the other at 2.2V. Likewise the mu of unselected devices will vary 2:1. A 2:1 mismatch on the mu will only make the possible differential 10mV instead of 5mV. But the Vgsactive will add directly to the input offset. So for completely unmatched JFETs, you could see the thing balance with the + and - inputs as much as 2.2-0.4 = 1.8V apart plus the 5-10mV of gain offset.

No biggie. There's enough room in either direction for a 1.8V offset. So it will balance, even with mismatched JFETs. Bucket-matched JFETs would cut that down a lot.

Yep, it will act like an opamp.

I think B@ss use this configuration often because of the following:
1. kind of "more soft" saturation relatively to the IC OA

That's possible. The sharpness of saturation corners is only increased by order of 2000, instead of 100000. Hard to tell.

2. there is less "junk" like integrated current sources, semiconductor resistors and capacitors, e.t.c. that usually can be found on the OA chip. Hence - less unwilling artifacts and mud in the sound.

Maybe. That "junk" gives you things like more gain accuracy, less offset, and such. I have yet to see anyone who can tell an opamp from non opamps AS LONG AS THE OPAMPS NEVER CLIP. Clipping behavior, where feedback cannot hide the inside operation of the opamp from the outside world, does make a difference. And I'm pretty sure that Boss the company never cared much about that difference. Individual designers, maybe. Demonizing the internal opamp junk is a favorite of the hifi tweakos, but it's oversold, just like oxygen free copper wire.

3. specific discrete FET mojo :-)   

And then they didn't even bother to get their mojo back in money by advertising "New! Opamp free circuits! Hand picked field mojo, fresh from the bins to you!"