[tangent]

Quote:

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

I'm sorry I offended you. I'm not trying to be insulting.

To understand this breakdown between us, it is not important which of us is right. All that matters is that I understand this circuit differently than you do. If I had no concept at all of how it worked, I might just take your statements without question. The problem is, I do have a mental model of how it works, and that model makes sense to me. It may not be a correct model, but I haven't yet seen anything that makes me change it.

In this sort of situation it does me no good to take without question things you say that go against my understanding. That would crush my curiosity, stop my learning process, and turn me into a jeffreyj hero worshiper. Not that there's anything wrong with jeffreyj being someone's hero. It is simply unrelated to the practice of engineering, that's all.

Given that I understand this circuit differently, I must question things you say that go against my understanding. You can:

1. ...take my response as the unschooled piker telling the Real Engineer that he's a blithering idiot.

2. ...try to explain it a different way.

3. ...decide that it's too stressful to try to communicate with an unschooled piker.


If I thought you were an idiot, Jeffrey, I'd just stop replying. I'd have no incentive to continue. I'm still participating in this thread because I do respect your knowledge, and I hope to resolve this question.

I do not possess all of the mental tools or experience you have, but I'm willing and able to learn. That should be enough.



Still with me? Okay. Allow me to explain how I think this circuit works.

Let's begin with the Jung circuit, but with the diode removed. What we have now is basically a Sulzer variant with the pass transistor's collector connected directly to the raw supply, and a CCS which is just being a nuisance. (I'll get to that in a bit.) There are some different component values and we're using some improved component types, but basically it's a Sulzer regulator at this point, right?

Let's say R1 and R2 are 1K. The op-amp slews to put out 2Vref + Vbe. The drop across the pass transistor puts 2Vref at the output of the regulator, which balances the op-amp's inputs.

Now imagine some 120 Hz ripple on the raw supply. We'll assume that the PSRR of the op-amp is infinite, so the only thing that happens is the ripple gets through the pass transistor without much attenuation. This ripple appears on top of the 2Vref output voltage, so as the ripple voltage rises, the op-amp's output voltage goes down, cancelling the ripple. Vice versa when the ripple voltage starts dropping again.

The CCS, as I said, is just sitting there being a nuisance. The CCS current splits among the op-amp's output and the base of the transistor. The current going into the base of the pass transistor probably will make it saturate, as you said above, but Vce doesn't really matter here. All that's important is that the emitter remains one diode drop below the base voltage. The base of the pass transistor is at 2Vref + Vbe despite the current being driven into the op-amp's output. The fact that the op-amp is sinking current may affect its performance, but as long as the current level is within a reasonable range it doesn't make the op-amp deviate from maintaining 2Vref + Vbe at its output. Hence, the regulator's output remains 2Vref despite the current source's action.

I apologize for taking so long to get to this point, but I wanted to make certain that I understand everything, and that we agree on the basics. Now let's throw the proverbial monkey wrench into the works: the diode.

When the ripple voltage rises, the op-amp's output voltage starts dropping. The pass transistor's base is now one diode drop above the op-amp's output voltage. (Yes, I'm ignoring the 10 ohm resistor's drop.) The emitter is a diode drop below that, so now when the op-amp's output voltage goes to 2Vref, the regulator's output is also 2Vref, right?

If all this is correct, the system is still voltage-controlled. It's true that the op-amp will only sink current with the diode there, but when the op-amp's voltage moves, does it not "push" the diode's anode voltage around relative to ground? And in turn, would that not push the base of the pass transistor, which pushes on the emitter?

 

[tangent]

Okay, problem solved. Morsel and I had a long conversation, where we cleared up several problems. The big one are:

1. "Comparator" -- I realize that ideal comparators are the same thing as ideal op-amps. The only thing that gets a chip labelled one or the other is one of practical implementation issues. However, when I think of a comparator, I think open-loop and an output voltage that is always pegged to one or the other rail. I didn't think that was likely, because that would imply that the op-amp's output voltage was PWM square waves, where the pulse width governed how much correction was applied either way. That would be an interesting regulator, but I didn't think that was what was happening here. Hence, the error amp is not a "comparator" in the way I'd use the term. It boils down to a question of semantics.

2. I see that the CCS can drive the pass transistor into saturation. I also see that the op-amp can pull current away from the CCS. What I didn't see is where the I/V conversion occurred. For instance, if the regulator's input voltage is 14V and the regulator is configured to put out 10V, where is the 4V drop? If the output voltage of the regulator is directly governed by the op-amp, the regulator works like I said above. But if the regulator is current-controlled, where is the conversion from current to voltage? Well, it must be across the pass transistor, of course. But how did that voltage drop develop? I am now thinking that by pulling current away from the base of the pass transistor, the op-amp varies the Vce of the pass transistor, thus converting a current to a voltage. The Vce varies from Vsat (~0.2V) up to whatever value is necessary to regulate the output voltage. Is this correct?

 

[jeffreyj]

Quote:

Originally posted by tangent I'm sorry I offended you. I'm not trying to be insulting.

Don't worry, my ego isn't *that* delicate

- you didn't offend me so much as frustrate the hell out of me for a minute, there, and with good reason, I think.

I'm all for questioning the status quo and the powers that be - I think that's rather apparent - and I am not averse to being questioned, either, but, like anyone else, I'd prefer that the questioning be the result of reading what I wrote, then disagreeing with it... ahem.

And don't go calling yourself an unschooled pike, or me a Real Engineer - appellations like this only serve to dismiss anything the former says and engrave in stone anything the latter says - neither action being all that useful, as you seem to fully understand.


Quote:

Still with me? Okay. Allow me to explain how I think this circuit works. Let's begin with the Jung circuit, but with the diode removed. What we have now is basically a Sulzer variant with the pass transistor's collector connected directly to the raw supply, and a CCS which is just being a nuisance. (I'll get to that in a bit.) There are some different component values and we're using some improved component types, but basically it's a Sulzer regulator at this point, right?

So far, so good.

Quote:

Let's say R1 and R2 are 1K. The op-amp slews to put out 2Vref + Vbe. The drop across the pass transistor puts 2Vref at the output of the regulator, which balances the op-amp's inputs. Now imagine some 120 Hz ripple on the raw supply. We'll assume that the PSRR of the op-amp is infinite, so the only thing that happens is the ripple gets through the pass transistor without much attenuation. This ripple appears on top of the 2Vref output voltage, so as the ripple voltage rises, the op-amp's output voltage goes down, cancelling the ripple. Vice versa when the ripple voltage starts dropping again.

Yep.

Quote:

The CCS, as I said, is just sitting there being a nuisance. The CCS current splits among the op-amp's output and the base of the transistor. The current going into the base of the pass transistor probably will make it saturate, as you said above, but Vce doesn't really matter here. All that's important is that the emitter remains one diode drop below the base voltage. The base of the pass transistor is at 2Vref + Vbe despite the current being driven into the op-amp's output. The fact that the op-amp is sinking current may affect its performance, but as long as the current level is within a reasonable range it doesn't make the op-amp deviate from maintaining 2Vref + Vbe at its output. Hence, the regulator's output remains 2Vref despite the current source's action.

You are mostly right. The voltage at the base will be held constant as long as load at the pass transistor's emitter does not exceed Ib * hfe; what I think you are missing, though, is the behavior of the circuit when this current level is exceeded, and that is where the real difference between this one and a conventional series-pass regulator lies.

In a conventional regulator design, the error amp needs as much DC gain as possible to minimize the error between output voltage and reference; it truly does function as a comparator (practically open loop at DC, but with the AC response tailored to prevent oscillation). In the Jung circuit, the op-amp is not running open-loop because it is directly controlling the output voltage by doing whatever is necessary to oppose the effect of the current source: remember, without the op-amp present, the CCS will drive the pass transistor to saturation while without the CCS, the only thing the op-amp can do is short the pass transistor's base to ground. The CCS is the key ingredient in this circuit that lets the op-amp retain a very high bandwidth while still ensuring the output voltage is accurate (otherwise two mutually exclusive requirements in a conventional regulator design).

 

[aos]

So the current source' current is set so that if all of it is going to the pass transistor it would operate at (or below) is max rated power output? And as mentioned the output voltage / current are regulated by taking away current from the transistor base. And the diode makes sure that current source is the one determining maximum current through the transistor' base.

I do think that I/V occuring in the opamp output stage is a misconception. Opamp really changes its output voltage in reaction to its inputs. Net effect is change in the split of the current between transistor and opamp but it's indirect. Change of voltage on opamp's output will change the voltage differential between opamp's output and the transistor base which is what determines the current that is to flow to opamp. Isn't that why you really need the 10Ohm resistor?

 

[jeffreyj]

Quote:

Originally posted by aos So the current source' current is set so that if all of it is going to the pass transistor it would operate at (or below) is max rated power output?

If the designer is smart, then yes.

Quote:

And as mentioned the output voltage / current are regulated by taking away current from the transistor base. And the diode makes sure that current source is the one determining maximum current through the transistor' base.

Correct again.

Quote:

I do think that I/V occuring in the opamp output stage is a misconception. Opamp really changes its output voltage in reaction to its inputs. Net effect is change in the split of the current between transistor and opamp but it's indirect. Change of voltage on opamp's output will change the voltage differential between opamp's output and the transistor base which is what determines the current that is to flow to opamp. Isn't that why you really need the 10Ohm resistor?

Well, you're right in that calling it an I/V conversion is a misconception, because the opposite is occuring (Voltage at the inputs controlling the amount of Current sunk by the output). Let's assume that the desired output voltage is 10V and that the unregulated supply voltage is, oh, 100V (just for fun). If hfe is 100 and Ib is 10mA, then the output current will always want to be 1000mA. So far so good? Let's say, though, that the load resistance is 20R, not 10R; left to its own devices, the CCS will raise the voltage on the base of the transistor to 20.6V to force 1000mA to continue to flow, correct? But the op-amp will not rest until the voltage at both of its inputs is 10V. The way that it will cause this to happen is by sinking current away from the current source, until 5mA goes through it and 5mA goes to the base. The voltage at the base will always be 10.6V, but the current going to the base will vary between 0mA (no load on the emitter) and 10mA (full load on the emitter). Whatever is left over from 10mA - Ib is sunk to ground through the lower transistor of the op-amp's output stage.

Ergo, the op-amp is acting as a V/I converter. (Ok - strictly, a voltage-controlled current sink, but that's close enough in my book).

added text below:

I think the 10R resistor is there to isolate the junction capacitance of the diode when it is reverse biased by the op-amp during overload conditions. If the overloading was due to high-frequency current peaks being drawn, the op-ampwould vary it's output voltage sympathetically. the diode's junction capacitance would transfer the AC component of the voltage (even if it was always positive) to the base of the transistor as a valid control signal which could turn the whole regulator into an oscillator.

 

[aos]

I agree with all that you say, voltage at base is constant, but I think the opamp effects the change in the current by raising its own voltage which changes the differential from its output to the base. As you know the current through a resistor is always equal to voltage differential on its ends so that would change the current going to the opamp. Basically what I'm saying is that opamp is voltage-controller voltage source, not voltage-controller current source as you're saying. I.e. the change in opamp out current is a reaction to change of the opamp out voltage, not vice versa. I'm talking about changes once the whole thing has stabilized into some operating point (small signal if you will).

 

[tangent]

Quote:

Basically what I'm saying is that opamp is voltage-controller voltage source, not voltage-controller current source as you're saying

The output voltage of the op-amp will be equal to the output of the regulator, due cancellation of the Vf of the diode and the Vbe of the pass transistor. Therefore, one can think of the system in voltage terms.

What one cannot do is ignore the current sinking activity of the op-amp. It must at least sink current away from the base of the pass transistor in order to satisfy the current gain behavior of the pass transistor.

 

[aos]

Many circuits are typically analyzed by first caluculating the DC operating point and then applying voltage/current sources that provide signal under analysis, a signal that is much smaller than the voltage/current already in the circuit so that operating point doesn't move much (small signal models/analysis). So in quiescent state of this regulator you'll have opamp voltage roughly equal to output voltage. Then to figure out how the circuit works (how it regulates) you apply a small change in voltage or current somewhere in the circuit and analyze the impact. This is done so that you can utilize simple formulas for active elements (e.g. ic = ib * hfe) which are typically linear. If you didn't do this, you wouldn't be able to calculate much by hand - either because it'd get too complex to handle or it would be impossible to solve analytically (and have to be solved numerically). You get imprecise results of course (the rule was 10 or even 20% allowable deviation), but you'd get sufficiently good description of how the circuit works.

That's why I say that a (small) change in output voltage would translate into a (small) change in distribution of current between transistor and opamp, as I described above. At least that is my understanding of this circuit.

 

[jeffreyj]

Tangent - I think you've got it now, no?

aos - A current sink and a current source are very different things. The diode in series with base to op-amp output prevents the op-amp from ever acting as a current source. However, upon further reflection I have to agree that the op-amp will maintain a consistent voltage at it's output, disregarding the very small variation the 10R resistor will necessitate, for all load currents from 0 to (Ib * hfe).

Oh, and I don't do SPICE. It caused me no end of grief back in school and every time since then that I have said to myself, "surely it's gotten more accurate," I find that, no, it hasn't, and it still takes more time to simulate most functional cells in SPICE than it does to whip up a quick dead-bug prototype. (Yes, I realize there are plenty of engineers out there who somehow, someway, have managed to master SPICE and make it do their bidding, but I'm not one of them).

edit: inpexplicably referred to op-amp's output as "base"

 

[tangent]

Quote:

I think you've got it now, no?

Mostly. The only thing I'm worried about now are the complex implications of all this.

Here's one: The voltage drop across the pass transistor must be varying to accomplish voltage regulation, and this is accomplished by varying current to the base. Does this not imply that the regulator output current vs. dropout voltage is highly variable? And since hfe varies with Ic, it would be nonlinear variability.

The Jung article talks about the CCS giving the circuit LDO. I wonder if this is isn't because for high load currents, the op-amp pulls less current away from the CCS so the pass transistor is closer to saturation. The question is, does dropout not go up at lighter loads? That's backwards from the way the LM317 family behaves.

 

[jeffreyj]

Quote:

Originally posted by tangent Mostly. The only thing I'm worried about now are the complex implications of all this.

It's good that you are worrying, because you are right to do so. TANSTAAFL


Quote:

Here's one: The voltage drop across the pass transistor must be varying to accomplish voltage regulation, and this is accomplished by varying current to the base. Does this not imply that the regulator output current vs. dropout voltage is highly variable? And since hfe varies with Ic, it would be nonlinear variability.

Sort ofl the much higher bandwidth afforded by not needing to run the error amp open loop helps control the hfe non-linearity, but this non-lineariy does complicate the transfer function of the network so phase shift vs. frequency is a bit more difficult to determine (if calculable at all). As far as dropout voltage goes, see below:


Quote:

The Jung article talks about the CCS giving the circuit LDO. I wonder if this is isn't because for high load currents, the op-amp pulls less current away from the CCS so the pass transistor is closer to saturation. The question is, does dropout not go up at lighter loads? That's backwards from the way the LM317 family behaves.

No, dropout should stay at a little over 1.2V for light loads (I say "a little" because hfe is extremely non-linear in the saturation region); remember that while the equivalent resistance between C and E goes up with a lighter load, the voltage drop across said resistance is strictly a function of ouput voltage to input voltage. For heavy loads, the dropout voltage will increase, as with 78xx/3x7 regulators, because Vce[sat] will increase.

 

[aos]

Output current is really a function of load, not the regulator. It's the load that uses more or less current, changing the output voltage and forcing the regulator to change to compensate. So that current is the cause, not the effect. So load commands the current, regulators controls the voltage, and dropout can be all over the place but you don't really care about it.

jeffreyj - I see that you know that this is the way Spice works (calculates point, then uses small-signal models). But I was talking about by-hand analysis which was how I was taught to do it. I might experiment a bit with my Jung, doing something to that 10 Ohm resistor. In my interpretation, without it the circuit would not operate properly as the opamp's output impedance is probably too low to be enough. I wish I had the original article...

 

[jeffreyj]

Quote:

Originally posted by aos ... jeffreyj - I see that you know that this is the way Spice works (calculates point, then uses small-signal models). But I was talking about by-hand analysis which was how I was taught to do it.

Oops - I was sure you were talking about SPICE.


Quote:

I might experiment a bit with my Jung, doing something to that 10 Ohm resistor. In my interpretation, without it the circuit would not operate properly as the opamp's output impedance is probably too low to be enough. ...

Interesting supposition... would you explain it a bit more?

 

[aos]

I did realize while writing it that someone might recognize it as the way that Spice works (or at least it used to)... But indeed we used to calculate DC and then work on small-signal. You have to be careful, some of those transistor formulas only work in small-signal (and some further require transistor to be not saturated etc.). For example, when you assume Vbe = 0.65V and it's constant - like we do here - it's because IC = IC0 * e ^ (Vbe / VT) so essentially the voltage remains constant while transistor works as amplifier because very small changes in voltage yield large changes in collector current. So you'd assume that transistor is in certain state then just set up standard circuit equations assuming Vbe = 0.65V or whatever. hFE is a parameter of the small signal model if I remember well so if you use that, then you're definitely talking about analysing the small changes around operating point, not the initial DC point. That is fine if you go about it in a way where you'd say that load current went a bit up (or input voltage went down), let's see where it leads.

About 10 Ohm, as I am thinking that the change in current diverted from the pass transistor is because voltage between opamp and base changes, the change in that current is equal change in opamp voltage (as base voltage won't really change much) divided by resistance of resistor and the diode. Hm, since the diode already has non-negligible resistance perhaps the resistor isn't THAT critical. On the other hand, if I got it right, 1N4148 has only 0.22 Ohm resistance when forward biased - but is it forward biased? If they didn't exist you'd only have output resistance of the opamp - milliohms - to divide the change by so even a small change in voltage would yield big change in current. Perhaps it'd be too sensitive. I would imagine that the 14 page article would explain all this, I'd rather just read it but I don't think it's that easy to get a hold of Audio Amateur...

 

[tangent]

Quote:

without [the 10 ohm resistor] the circuit would not operate properly

Jung says it's for "stability". I don't recall any justification for this. I looked at the D44H datasheet, and it didn't give a spec for base capacitance. Maybe that's a bad sign. The resistor could be there to spoil the Q of the filter caused by that capacitance and the inductance of the trace and pins leading up to the base. A 4148 only has a few puffs of capacitance, but maybe the resistor is there to isolate that capacitance from at least the inductance of the various wires.

Quote:

I don't think it's that easy to get a hold of Audio Amateur...

http://www.audioxpress.com/magsdirx/audelex/aabilist.htm