# How does the feedback work in this linear power supply?

How does the feedback loop in this circuit work?

Image source

I am unfamiliar with this configuration. It seems to have positive feedback.

• Note that the output stage is inverting. Thus: you need to swap the (+) & (-) inputs on the error amplifier, exactly as shown in the schematic. Commented Oct 13, 2019 at 4:49
• Thanks man.But I still don't get how the 5v reference work.It seems there will be a constant 5v across this IC lm336 but not across the resistor??
– User
Commented Oct 13, 2019 at 5:27
• @aswincvenu everything is referenced to the positive voltage (+) Commented Oct 13, 2019 at 6:56
• It would help to provide some context, such as what the circuit is intended to do. It appears to be a linear power supply, but some additional details would be helpful. Also, you should state why you think it may have positive feedback. Commented Oct 13, 2019 at 7:25
• The op amp only compares the voltages at its inputs. It doesn't care what those voltages are referenced to. Commented Oct 13, 2019 at 16:44

It may be easier to see if you break things up into functional groups. Here are the first few of these:

simulate this circuit – Schematic created using CircuitLab

Starting from the left, you have an unregulated DC supply providing $$\V_-\$$ and $$\V_+\$$ rails, followed by a (specialized) $$\5.0\:\text{V}\$$ zener/resistor pair that yields a continual voltage that is $$\5.0\:\text{V}\$$ below the $$\V_+\$$ rail. (Keep in mind, now, that this is a reference against $$\V_+\$$ and NOT against either the chassis ground or the $$\V_-\$$ rail.) You also have a simple, adjustable to some degree, resistor divider. But in this case it spans from $$\V_+\$$ to the chassis ground and NOT the $$\V_-\$$ rail. So this samples a portion of the span from $$\V_+\$$ to chassis ground.

What's important to now notice, though, is that the following opamp will attempt to adjust its output in order to keep its two inputs (the zener reference output and the divider output) the same. So this means that you need to look at the resistor divider network "upside down" so to speak. In short, the difference from the $$\V_+\$$ rail to the $$\V_+\: \text{sample}\$$ will also always (supposedly, if everything else is arranged correctly) be $$\5.0\:\text{V}\$$. Since there should always be a $$\5.0\:\text{V}\$$ drop across $$\R_2+P_1\$$, it follows that:

$$V_+=5\:\text{V}\cdot \left(1 + \frac{R_3}{R_2+P_1}\right)=5\:\text{V}\cdot \left(1 + \frac{6.8\:\text{k}\Omega}{3.3\:\text{k}\Omega+P_1}\right)$$

In short, $$\P_1\$$ actually does adjust the output value. And, given the above values you should find that it can adjust it such that $$\12.9\:\text{V} \le V_+\le 15.3\:\text{V}\$$, roughly speaking. (Slight offset errors in the opamp, allowable variations in the resistors and potentiometer values, etc. You may want to work out the actual variations possible in this situation using whatever % variations you would like to select for the resistors.)

The next part of the circuit is something like this:

simulate this circuit

Here, all that's being done (ignore the LIMIT for a moment) is that the opamp output servos around, but of course cannot really supply a lot of base drive current. It doesn't have much current compliance. So a Darlington is added to boost that up so that the current compliance of the CONTROL output of the opamp can handle the necessary base drive for the low-side drive of the power BJTs in the next section (not yet shown.)

There is another important detail. $$\R_5\$$ is vital as a kind of "pull-down" to "help" the output of the 741 get closer to its $$\V_-\$$ rail in the case where the power supply isn't experiencing a loaded condition (it's unloaded or lightly loaded.) If there's no real load, then the power BJT group won't have any collector current to speak of and the power BJT group base voltages will be very close to $$\V_-\$$. The Darlington emitter follower's emitter voltage will therefore also be close to $$\V_-\$$. And therefore the Darlington's base voltage may only be perhaps $$\1.4\:\text{V}\$$ (or maybe less) above $$\V_-\$$. This is probably too close for comfort to allow the Darlington base to be directly driven by the opamp output. Using the $$\R_4\$$/$$\R_5\$$ divider resolves any question about this.

So the main purpose of the above section is to boost the current compliance of the 741 opamp while also permitting it to handle the no-load case. The LIMIT feedback control is there to handle the case where the output current rises (in one of the power BJTs, anyway) above some chosen limit. When that happens, $$\Q_1\$$ will sink current, causing an additional voltage drop across $$\R_4\$$ and therefore driving down the base voltage of the Darlington follower, which then drives the Darlington's following emitter downward, too (and thereby limiting the load current.)

The remaining section is just a current-sharing arrangement for BJTs:

simulate this circuit

The emitter resistors help with "current sharing" between the BJTs, as discrete BJTs can vary quite a bit in their saturation current and therefore in the required $$\V_\text{BE}\$$ for some given collector current. By adding emitter resistors, if one of the BJTs attempts to "hog" more of the load current than others, its emitter resistor will be forced to drop more voltage and this "pinches" its $$\V_\text{BE}\$$ by pushing upward on its emitter (it's base will be at the same voltage as all of the other BJTs in the power-sharing arrangement.) So these resistors act as local NFB to help force the BJTs to approximately share the load current. Nothing is perfect about this, but it gets the job done "close enough for government work," so to speak.

The LIMIT output will rise upward above $$\V_-\$$ as more current is carried by $$\Q_3\$$. (There's no particular reason why $$\Q_3\$$ is chosen. One could just as well have picked this off from one of the other emitter resistors.) If this rises sufficiently above $$\V_-\$$, then $$\Q_1\$$ will "turn on" sufficiently to force a larger voltage drop across $$\R_4\$$ and that will pull down the base of the Darlington, which will pull down on the shared bases of the output power BJTs, lowering the voltage drop across $$\R_7\$$ and limiting its current. (By implication, this will also limit the current in the other three power output BJTs, too.) Again, no particular "precision" here. But you can guess that $$\Q_1\$$ will start to become active when there is about $$\600\:\text{mV}\$$ across $$\R_7\$$ (or about $$\6\:\text{A}\$$.) So the limit should be about $$\24\:\text{A}\$$, or so. The web page says $$\20\:\text{A}\$$. And I've no reason to disagree. It works out close enough, I think.

Regarding the question of the (+) and (-) inputs of the opamp? Assume for a moment that the $$\V_+\$$ suddenly rises upward (relative to chassis ground.) This will pull upward directly on the REF output of the zener section and apply 100% of the change to the (-) input. But the divider (potentiometer) section will only apply a percentage, less than 100%, of the change to the (+) input. So does that help you see why it is arranged the way it is?

• Thanks man helped a lot.
– User
Commented Oct 14, 2019 at 1:03
• Also is it because both zener and divider are tied up at same point?
– User
Commented Oct 14, 2019 at 1:07
• @aswincvenu Is what because of what? I'm not sure what you are asking. Is this regarding the last paragraph I wrote? Or something else?
– jonk
Commented Oct 14, 2019 at 1:53

I see where your confusion is, and I think what you're missing is that the bottom rail there is not labelled ground anywhere, and the non-inverting input is chassis-ground referenced. The inverting input has feedback from the output via R4-R5-R1.

The opamp "+"-input gets a fraction of the output voltage which is the positive input voltage on one end and the negative output on the other.

When the opamp "+"-input increases (pin 3), the output voltage (pin 6) also increases. Therefore, the base voltage of Q3 to Q6 also increase. An increase in the base foltage of these "followers", increases their emitter and collector currents, which is an increase in the load current. That will increase the current in capacitor C2 and/or the load which is attached to the output. An increase in current there increases the output voltage by lowering $$\V_{out-}\$$.

The feedback at pin 3 is the result of $$\V_{out+} - a.(V_{out+}-V_{out-}) = (1-a).V_{out+} + a.V_{out-}\$$.

We can consider that $$\V_{out+}\$$ remains constant (thanks to C1), but that $$\V_{out-}\$$ changes.

We saw above that when the opamp "+" input changes, that $$\V_{out-}\$$ is lower. Therefore $$\a.V_{out-}\$$ becomes smaller and the feedback voltage at pin 3 decreases. This proves that the feedback is negative: we started out with an increase of this positive input, and the feedback is a decrease of the positive input.

We could argue that there is a second feedback loop through R11. Why? Because when the output on pin 6 increase, the current in R11 also increases and this current is essentially provided by C1 (and D1). This extra current tends to lower $$\V_{out+}\$$ and therefore also provides negative feedback as the positive opamp input decreases. This second feedback loop probably has a negligeable effect.

R5 does not contribute to normal load voltage regulation, but it is esssentially there to set and limit the bias voltage for Q2. R6, Q1 and R4 provide an overcurrent protection. If the current is "too high", Q1 becomes active and pull the input of Q2 down. That in turn drops the conductivity Q2 to Q6, which lowers the output current. It is in fact another feedback loop that acts as a current limiter. The opamp U2 will try to increase the output voltage but Q2 will counteract.

One other thing that is interesting in this circuit is that it uses a NPN resistor (darlington) in the loop and that this does not increase the open loop gain. But that was not part of the question and left as an exercise ;-).