LM10 application circuit - how does it work?

[Clear as mud]

In another forum thread, someone recently posted an article about Bob Widlar, and since he designed the LM10, the article included this picture (attached) of a circuit designed around the LM10.  It's from the TI application notes, figure 49 here . The figure includes the equation Vout = Vref x (R2/R1).  At first I was totally confused because I didn't see a reference voltage, and then I remembered that the LM10 includes an internal 0.2V reference.  When multiplied by 1000 (2M/2k resistor ratio), that gives the 200 V output.  So I can kind of see how it works now.  Let me know if this is correct:

When the voltage is less than 200V (or 204V), the transistors are in conducting mode, and Vout will track Vin, minus only a very small voltage drop across Q2.  The diodes provide enough voltage drop to keep the LM10 powered (pins 7 and 4).  The 0.2V reference on pin 1 is referenced to Vout, so when Vout is less than 200 volts, R1 drops less than 0.2 volts, keeping the non-inverting input positive with respect to the inverting input, and the transistors stay on.  As Vout rises above 200V, R1 will drop more than 0.2V, bringing the non-inverting input negative with respect to the inverting input, and the output pin 6 will go low and shut off the transistors.  D3 and D4 provide enough voltage drop to keep Q1 conducting just enough to keep the LM10 powered, while D1 and D2 provide enough drop to keep the base of Q2 low despite the small amount of current still coming through Q1.

I'm hoping that I am correct about the circuit operation, because if I am, it seems this would work quite well as a voltage limiter circuit.  As long as your voltage doesn't go above the set point, it seems very energy efficient, dissipating power only as the small voltage drop across Q2 and the quiescent current in other components.

If you really wanted your output to be accurate, the 2M resistor would have to have a much tighter tolerance, correct?  At 1% in the circuit shown, that's plus or minus 20k.  Given this inaccuracy, it seems like you could just throw in any old 2k resistor for R1 (instead of a ±1%), and get similar performance.

[741]

My minor contribution is to note that
   200.2*2020/2022 = 200.00198   //10% change in 2Meg, 2k accurate
   200.2*2000/2002.2 = 199.98      //10% change in 2k, 2Meg accurate

If the top resistor is some fraction, f of the lower one, the divider gives R/(R+fR) = R/R(1+f) = 1/(1+f)

If f is 0.001, the fraction is 1/1.001     = 0.999001 and then *200.2 gives 200

If f is 0.0005, the fraction is 1/1.0005 = 0.999500 and then *200.2 gives 200.1

[Clear as mud]

Both you and I looked at those resistors the wrong way.  It looks like a voltage divider, and without doing any math at all, I looked at it and thought inaccuracy in the 2M resistor would make a whole lot more difference to the output voltage than inaccuracy in the 2k one.  You did the math and showed the opposite.

But actually it does not operate as a voltage divider.  Because of the op-amp feedback, the junction between the resistors (connected to one op-amp input) is held at essentially the same potential as Vout (connected to the other op-amp input).  This means that while the op-amp is within its bounds of operation, the voltage across R1 is always 0.2V, the reference voltage.  So, the current through the two resistors is 0.2/R1, and consequently the voltage at the resistor junction (and thus, also Vout) is that current times R2.  So, the equation from the data sheet is indeed correct, and Vout = Vref x (R2/R1).

Then, since the equation for output voltage is not based on a voltage divider, but rather a simple proportion of resistance values, the inaccuracy in R1 would have approximately the same effect on the output voltage as a similar percentage of inaccuracy in R2.  So, it makes sense to specify a tight tolerance for both resistors.

[T3sl4co1l]

Yes, that is more or less correct.

Mind that, when you say "turns on" or off, this process happens gradually, over time.  Not digitally, and not instantaneously (nor simply delayed, either).

If it simply were on and off, it is plain to see that it can never be stable, it must oscillate.  As it happens, there is a happy medium between the two extremes (uh, taking something like purely proportional control -- like replacing the voltage regulator with a (very wasteful) resistor divider -- as the opposite extreme), where for the right degree of gradual change, it settles into a stable, quiet state.

There is, of course, a body of mathematics dedicated to the study and calculation of these processes -- dynamics.  While we can wave our hands and make a simplified, colloquial statement, we must be careful not to take something too literally, but we should strive to understand the underlying process.

Regarding the resistors -- I mean, it's still a resistor divider.  If we assume the REF pin has some voltage, and the +in pin draws no current, then its voltage must be the divider ratio times that input voltage, simple as that.

A practical circuit should of course have current limiting, and maybe foldback limiting at that, some transient protection, and maybe thermal protection, or an enable, or other add-ons as suits the end use.  The main reason those aren't shown is because they complicate a simple idea.  Appnotes have a careful line to follow, between showing an idea simple enough that any joe schmoe engineer can remember and understand it, and showing something that's complete enough to be usable as a drop-in circuit.  (And I suppose manufacturer FAEs are one bridge between those extremes... nice, if you can get a visit from them.)

So, I would caution against using the circuit verbatim, if you happen to need its function; but it's a fine core, and can be built upon without too much trouble.  Compare with other circuits like classic LM723 supplies and whatnot (which usually include current limiting, and often use emitter follower outputs just like this does).

Tim

[741]

...Because of the op-amp feedback, the junction between the resistors (connected to one op-amp input) is held at essentially the same potential as Vout...

I can't identify where there is any DC feedback (originating from pin 6) to the R1, R2 node on pin 3. Pin 3 is a non-inverting, so feedback has to be opposite polarity to Vout on pin 6. (The buffer amplifier has it's own feedback loop shorted so as to be a follower of course).

[Clear as mud]

When pin 6 goes low, it tends to turn the transistors off, making Vout lower and thus, pin 2 lower.  Pin 2 is the inverting input, so this has the effect of making the output pin 6 higher.  That is the negative feedback. The junction between resistors going to pin 3 sets the operation point.