[tangent]

There's been some demand for me to put the Sulzer and Jung regulator history that I traced out a few weeks ago up on my web site. Naturally I couldn't leave well enough alone so I added a bunch of material. It should serve as a good introduction to this type of regulator now. It doesn't replace the original articles by any means, but an experienced DIYer could probably make a credible regulator from just my article.

Here 'tis: http://tangentsoft.net/audio/opamp-linreg.html

Enjoy!

 

[jeffreyj]

Nice page; you've got a knack for narrating this stuff.

Of course, it wouldn't be like me if I didn't have a couple of quibbles to point out...

Both quotes are from the web page:

Quote:

The NE5534 was the hot new chip back in 1980, priced at about $11 in 2003 money. ... In 1974 when the LM317 was designed, the hot op-amps were all close variants of the μA741. Some look back on the NE5534 with disdain these days, but it's still miles ahead of the 741.

There are two topology differences between these two chips that I think are worth pointing out. The first is that the 5534 employs a second differential gain stage for driving the output, rather than a single-ended one. The second is that the output stage of the 741 is pretty much a straight Class-B common emitter so it is hampered by the very slow PNP transistor which was all the fab process could make back then; the quasi-totem pole only employs NPN's so the output stage of the 5534 is much faster (which leads to much lower distortion and a much more appealing slew rate capability).

These days, PNPs can be made nearly as fast as NPNs, so the superiority the discrete versions of the common emitter and complementary feedback pair output stages enjoy can be conferred upon ICs as well.


Quote:

In the Sulzer regulator, the pass transistor's base is driven directly by the op-amp. In the Jung regulator, the pass transistor's base is driven by a constant current source (Q2 and assocated parts) in parallel with the op-amp. Further, there's a diode inline with the output of the op-amp so that the op-amp doesn't try to drive current to the base of the pass transistor, since that job is already being handled by the current source. Since the op-amp doesn't need to drive a load, its slewing performance is maximized.

The pass transistor will allow an output current to flow of Ib * hfe, where Ib is the current from the constant current source. For loads that draw significantly less than the product of Ib and hfe, the transistor will merely saturate and therefore not be a regulator, but a current limited (asssuming Ib*hfe is below Ic[max]!!). The op-amp sinks current away from the pass transistor's base, which is how regulation is accomplished. The advantages of this design are that: the op-amp is in the inverting configuration, giving superior performance in all aspects save noise; only the NPN transistor in the op-amp output stage is used to modulate the pass transistor, which is where the speed improvement comes from.

 

[tangent]

Thanks for reading through it, Jeffrey.

Quote:

There are two topology differences between these two chips that I think are worth pointing out.

It's interesting to know, but the point of that paragraph is only to point out that the error amp in the Sulzer regulator is better technology than was available when the LM317 was designed.

The question you could answer for me, though, is whether my speculation that the LM317's error amp is 741-like is true or not. I'm not adept enough at following pure discrete circuits yet to make this judgement.

Quote:

The op-amp sinks current away from the pass transistor's base, which is how regulation is accomplished.

I thought some of that must be happening, but I thought it was a secondary effect.

The op-amp must be swinging a voltage to effect regulation, since a) an op-amp has voltages as inputs and it changes its output voltage to balance the inputs; and b) if it were current-controlled, there would have to be a resistance to drop it across to get voltage regulation. Maybe I'm missing something, but I see no I/V stage.

The purpose of the diode on the output of the op-amp isn't to make it a current sink per se, but rather so the op-amp doesn't try to push current into the base of the transistor, which it normally would since it's an NPN. It's a division of labor issue, rather than the op-amp's current sinking being balanced against the CCS's output.

The CCS will give about 10mA with a 1.6V LED, which an AD797 could easily sink if it wanted. I just don't see it's "motivation".

 

[aos]

Here's some pictures of the modified Jung (2000+ with pre-regulator stage) that I'm using for PPA and other stuff. I designed a board over a year ago and only recently made some, but I believe I used the published layout (diyaudio?) as the model.

I also think I've read somewhere that LM317 contains 741-like opamp. In fact if you read articles analyzing stability of most of these regulators (even modern ones), you'll notice that the performance of the opamp used is pretty low (usually they are very slow). E.g. their PSRR drops drastically immediately after the crucial 120Hz is filtered out.

By the way I'm very unhappy with the Sulzer-Borbely. It might be the way I made them but both that I have measure horribly when driving any amp (i.e. lots of ripple similar to that Creek/Meta comparison posted around here), and it looks bad on oscilloscope too. This Jung looks far better.

 

[aos]

Quote:

I thought some of that must be happening, but I thought it was a secondary effect. The op-amp must be swinging a voltage to effect regulation, since a) an op-amp has voltages as inputs and it changes its output voltage to balance the inputs;

I think you're right - current source would not set the output voltage, it'd be the opamp. In ideal world current source does not control directly the voltage accross it. So, we have a portion of the current going to the transistor and portion to the opamp. Now if opamp raises its output voltage in reaction to output voltage dropping for example, you have to re-arrange the currents to have base resistance times its current equal to the voltage at opamp out plus diode plus current through it and the 10 Ohm resistor. As opamp voltage went up, the current through it will have to be partially diverted to the transistor so that voltages accross both paths are again equal, so transistor would provide more current at the output and hopefully the output voltage would go up and achieve new equilibrium. At least this is how I think it works...

 

[JohnFerrier]

Tangent,

Thanks for taking the time to compile this. I've printed it and will read through it. I'm aware of a recent long thread discussing the Jung Super Regulator over at the DIYAudio forum. I wasn't going to post a link to it, but Mr Jung submitted his first post to that forum this morning!

http://www.diyaudio.com/forums/showt...147#post253147


JF

 

[aos]

Where can one buy Audio Electronics? Especially back issues?

 

[jeffreyj]

Quote:

Originally posted by tangent Thanks for reading through it, Jeffrey.

Your welcome

Quote:

... The question you could answer for me, though, is whether my speculation that the LM317's error amp is 741-like is true or not. I'm not adept enough at following pure discrete circuits yet to make this judgement.

The differential input amp circuit is actually a little less sophisticated than a 741's, but otherwise the error amplifier of the 317 is very similar to a 741.

Quote:

The op-amp must be swinging a voltage to effect regulation, since a) an op-amp has voltages as inputs and it changes its output voltage to balance the inputs; and b) if it were current-controlled, there would have to be a resistance to drop it across to get voltage regulation. Maybe I'm missing something, but I see no I/V stage.

I see your reasoning here, and it feels right, but it isn't...

Let's analyze the circuit a bit more in-depth so you'll see that, there is, indeed, a V/I stage hidden in plain sight.

R8/Z1 set the reference voltage to the error amp (+) input (comparator, really). The ratio of Vo divided by R2 and R1 feeds the opposing input. Let us imagine that the AD797 is powered from dual polarity supplies for a second. If the output voltage were to rise above the reference, what would the AD797 output look like? It would saturate to the negative rail (or close to it, in any case). Since the pass transistor is NPN, a negative voltage at the base will drive it into cutoff. Now, take away the negative rail and what's the best the AD797 can do when the Vo exceeds Vref? Saturate to ground (or close to it). This *sinks* current away from the current source; current cannot be sunk from the base of an NPN once the minority carriers have been swept away (exception: reverse bias breakdown of the base-emitter junction). Now, let's consider what the current source is doing in the circuit. As I mentioned before, it supplies a current which ensures saturation for all loads < Ib * hfe. The typical power transistor in the TO-220 package (which is about all I can recall of the D44 transistor, right now) has a DC gain (hfe) of ~75-100. The current source supplies 10mA, so the maximum output current is 0.75-1A. Without the op-amp, the circuit would merely be a poorly performing current source (because beta changes in response to just about every conceivable variable), but the op-amp is there, and if the two voltages at its inputs aren't exactly the same it will staturate either to the positive rail (output voltage is too low) or ground (output voltage is too high). Now, the only reasons Vo would be too low are: a) the input voltage is too low; or b) too heavy a load has been connected. If it's reason 'a' there's not a lot the op-amp can do to change things, but if it's reason 'b', the op-amp would source additional current to the pass transistor's base if it weren't for the diode in series! That's why the diode is there, my friend, to prevent the op-amp from gleefully destroying the pass transistor in its single-minded quest to keep Vin(+) and Vin(-) equal.

The Voltage to Current converter I said was hidden in the circuit is the lower output transistor in the AD797 that can do nothing more than sink current to ground; a higher voltage difference at the inputs will result in more current being sunk by the output.

Et voila.

edited: spelling

 

[tangent]

Quote:

what would the AD797 output look like? It would saturate to the negative rail

I started skimming once I read this, because I think you're missing the feedback path. The op-amp is not acting as a comparator (i.e. it never pegs the output to the rails in normal operation) it's acting as a differential amplifier. (But not a diff-amp -- diff input, single-ended output.)

Am I missing something? Should I go back and re-read what you wrote in more detail?

 

[jeffreyj]

Quote:

Originally posted by tangent I started skimming once I read this, because I think you're missing the feedback path. The op-amp is not acting as a comparator (i.e. it never pegs the output to the rails in normal operation) it's acting as a differential amplifier. (But not a diff-amp -- diff input, single-ended output.) Am I missing something? Should I go back and re-read what you wrote in more detail?

What do you think, you bonehead!

Tangent sez: "hey, should I ignore what you wrote or should I actually *read* it before I trash it six ways to sunday??"

Seriously, though... firstly, in normal operation, yes, the op-amp will not saturate to either rail; secondly, I intentionally ignored the feedback path to make the illustration simpler. Since you seem to be a real glutton for punishment though, why don't you tell me what the gain of the circuit is at, say, 1Hz, 10Hz and 100Hz.

Oops... I don't see any values specified for R2 and R1... Make some up and get back to me on it

Really, this will be good for you, and perhaps it will teach you to Respect Mah Authoritah when it comes to power supply design!

 

[tangent]

The op-amp isn't operating open-loop. I think it's reasonable for me to suspect that everything developed upon that premise is bogus.

Quote:

I don't see any values specified for R2 and R1... Make some up and get back to me on it

Yes, R1 and R2 have no values, on purpose. That's the voltage adjustment for the circuit.

We have a 6.9V reference, so if we make both of them 1K, we should get just under 14V out. g=2 at DC, and it drops quickly to g=1. I believe the corner frequency would be 1/2*pi*R2*C1, so it'd be 3dB down at about 1Hz. At 10Hz, an additional 6dB down, at 100Hz 6 more dB, and so forth.

 

[tangent]

No wait, I'm mixing up my decades and my octaves. It's -20dB at 10Hz, and -40dB at 100Hz.

 

[tangent]

...but that's just the response of the filter. I simulated it, and you get this shelving type curve. It starts off at +6dB below 1Hz, curves down to +3dB by 1.8Hz (I figured it to be 1.3Hz...wonder why the discrepancy?) and then it starts flattening out to 0dB, as I said.

Now that's just a simple op-amp sim. What's happening in the context of this circuit is that any noise is going to be sent through at g=1. Only DC and very low frequency gets through with significant gain.

EDIT: Thinking more on it, what's dropping at 20dB per decade isn't the gain of the amplifier, but the feedback Z.

 

[jeffreyj]

Ok tangent - skip the forest and just look at the individual trees...

You said:
Quote:

The op-amp isn't operating open-loop. I think it's reasonable for me to suspect that everything developed upon that premise is bogus.

Thanks, tangent. I appreciate the vote of confidence... no, really.

No crap R2 and R1 set the output voltage; had you bothered to read my explanation rather than dismissing it as bogus because of your own errroneous interpretation of the side comment (comparator, really), then you'd see I'd already wrote just that.

I then wrote to make up some values for R2 and R1 so that you *could* figure out the gain, and I was hoping that by doing so you'd realize that it is highly dependent on what the output voltage setting is, and that it drops to unity at a very low frequency.

But back to the main issue... I put that side comment in to help emphasize the role of the error amplifier in a regulator circuit. It compares the output voltage to a reference and reacts accordingly. I analyzed the circuit from two extremes so it would be easier for someone to follow who has stated twice in the last couple of days that he's not so good at circuit analysis.

The operation of the circuit with Vout very close to Vref is considerably more complicated than when there is a significant error between the two, especially since the error amp gain is so low at DC. For example, a disturbance in the output voltage is corrected asymptotically with a time constant inversely proportional to DC gain (higher DC gain = Vout settles to within 1% of Vref faster). And that's just the beginning.

Another detail is that the function of the capacitor across the feedback resistor is to not only roll the gain off at high frequencies, but to couple transient changes in the output voltage directly to the (-) input.

Now, will you please go take a look at my explanation? After all, I did take the time to write it in the interest of helping you; if I wanted to be pompous and only write it to prove how brilliant I think I am I would have stuffed it full of equations and skipped the explanation.

 

[aos]

Jeffreyj is right in as that opamp is functioning as an error amplifier, i.e. the circuit actually takes advantage of a very high gain of any opamp to minimize error between reference and measured output voltage. Most opamps in voltage regulators are used this way, and not as a classic negative feedback amplifiers with low gain. Using opamp in a very high gain configuration is also probably the explanation why regulators don't work well past a few kHz as the performance deteriorates because you're not trading enough gain for bandwidth.