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Seemingly Simple Circuits: The TL431 Voltage Regulator, Part 3

In Seemingly Simple Circuits: The TL431 Voltage Regulator, Part 2, we looked at how voltage gain of the TL431 is measured. Now we plunge deeper in, to look at closed-loop output resistance, derive some formulas, then calculate transconductance and output resistance from manufacturer part data.

Output Resistance

The closed-loop incremental output resistance of the TL431 is specified using a test circuit similar to the gain-test circuit but with its divider removed so that νB = νO ; R1 = 0 Ω , R2 → ∞ . The changes to the gain-test circuit are

A slightly modified block diagram results. With the Gm amplifier nulled, the open-loop

νG is varied (νg ≠0 V), and νg and νo are measured. As in the gain-test circuit, the open-loop contribution of νg to νo is Τg x νg . The resistance-test circuit equations reduce to

The closed-loop output-port rout is found applying Ohm’s Law with RL sourcing io = ig , in series with νg ;

and can be calculated from measurement of νo and νg .

From circuit analysis, feedback reduces the open-loop voltage Τg x νg by 1/(1 + G ). The divider Τg is effectively

This is the Τg divider with RL and rout (cl) resistances. Equating the second and last expressions and solving,

The quasistatic output resistance of the TL431 is ro in parallel with the 1/Gm source which is dependent on νe and is also across νg = –νb = –νo ;

This result also follows directly from the substitution theorem : the voltage on which the current source of Gm depends is also across it, thereby reducing the dependent current source to a resistance, 1/Gm .

rout from Gm Currents

For the resistance-test circuit, the current of the Gm source can be decomposed into currents in ro and RL :

The fraction of Gm current in ro with νb = νo (as in the resistance-test circuit) is

Polarities of current and voltage can be somewhat confusing in this circuit. Because the Gm amplifier sinks current (into the amplifier) as positive, then for KCL to apply, positive current in ro flows upward to a negative νo . The same applies to RL ; νo (RL ) = –io x RL where –io and RL are non-negative.

Applying G = Gm x rout while solving for the closed-loop resistance of ro ,

The Gm current in RL is

and the closed-loop resistance of RL is

Thus the closed loop effectively reduces the resistance of the branches of the output node by G .

Combining the closed-loop resistances contributing Gm current,

As a circuit, Gm and rout form a dependent current source of resistance 1/Gm in parallel with rout . The same current flows in the parallel loop formed by the two resistances. Gm and rout are independent only if νo across them is 0 V. Without νG (from the test circuits) or a non-static VR , no voltage variation is expected at the output node; νo = 0 V, and rout and 1/ Gm are independent. The above equation implies that νo = 0 V and it also tells us that the 1/ Gm source is effectively equal in resistance to rout /G .

The closed-loop rout derived previously can be expressed in terms of this result as

Thus rout (cl) is reduced from the open-loop rout by the feedback factor, 1 + G , a result consistent with feedback theory; voltage outputs are generally reduced in resistance with a closed loop by feedback factor 1 + G x H .

To measure rout (cl), the output node must be driven with either an independent current source or a voltage source such as νg behind RL . Its contribution to νo causes the resulting current into the node to be that of the closed-loop rout . Substituting the closed-loop νo into the closed-loop io ,

from which

The closed-loop contribution of νg to νo is the amplifier closed-loop gain with Τg x νg applied to the output node.

Gm and ro from Test Data

Returning to the overall problem of finding Gm and ro , we now have two equations that can be solved for them simultaneously. The first is

and the second is

One of the TL431 specs gives rout (cl) = 0.2 Ω, though it varies considerably, from 0.13 Ω to 0.5 Ω, among manufacturers. The value of ro can be found by substituting the first into the second equation and solving;

Open-loop ro is hard to measure and must be calculated from closed-loop data. The input pin cannot simply be left open. Ideally, it could be connected to the internal VR , and then rout measured, though VR is not an available node. Consequently, in most circuit analysis, ro is considered negligible in its effects and is ignored.

In the output-resistance test circuit given by TI, RL = 1 kΩ behind a source shunted by 50 Ω. The source is not specified any further, though what is suggested by the circuit is a voltage source with 50 Ω output (as is typical of test equipment) terminated by the shunt 50 Ω shown in the circuit diagram. If this is correct, then the resistance in series with the 1 kΩ resistor is 50 Ω || 50 Ω = 25 Ω and the total RL = 1025 Ω.

Substituting G0 = 600, RLG = 230 Ω, rout (cl) = 0.2 Ω, and RL = 1025 Ω,

For rout >> 1/Gm , rout (cl) ≈ 1/Gm . Most circuit designs can make this approximation though some value of ro is needed when including the Rf path in the analysis. An accurate value of ro is difficult to determine, however. On the low end of the range, G0 = 400 and ro = 123 Ω, still much larger than 1/Gm . For G = 600, if rout (cl) ≈ 0.383, then ro → ∞ , with values within the specified range. The output BJT of the TL431 can be expected to have ro of 100 kΩ or more. What reduces ro is the shunt path of the TL431 power supply. As the amplifier changes supply current, it adds to the output current at pin 3 and reduces ro .

15 comments on “Seemingly Simple Circuits: The TL431 Voltage Regulator, Part 3

  1. rajeshkhanna11
    September 9, 2016

    Looks Awesome

  2. D Feucht
    September 9, 2016

    Rajesh,

    I am hoping that you independently derived and verified all the equations in the article! (I write, smiling)

    What in particular caught your attention?

    I like this circuit as article material because it presents a good example of a feedback problem that can be solved by carefully applying basic feedback theory.

  3. Thomas7
    October 3, 2016

    Thanks so very much for taking your time to create this very useful and informative site.

  4. TimTom12
    October 5, 2016

    Thanks for this post man! Love the way it is written. Also shared this page with some of my best friends and they are also excited!

  5. Thomas7
    October 6, 2016

    Richtige und erstaunliche Artikel, teilten mir sehr! Vielen Dank

  6. Thomas7
    October 6, 2016

    What in particular caught your attention?

  7. D Feucht
    October 7, 2016

    Bitte, schreiben Sie auf Englisch, wenn Sie wollen. Meine Deutsche ist nicht so gut.

  8. Steve Taranovich
    October 7, 2016

    Thanks Dennis—I can't speak German either, but we all would like to understand what the comments are about.

  9. Steve Taranovich
    October 11, 2016

    Bitte schön

  10. Benefiter
    November 17, 2016

    It's actually a cool and useful piece of information. I am glad that you shared this helpful information with us.

  11. D Feucht
    November 17, 2016

    “Gibt es eine neue Aktualisierung Ihrer großen Artikel? Ich bin für sie eifrig.”

    Ich verstehe halb. Bitte sehr, schreiben Sie auf Englisch, wenn Sie wollen.  Als meine Deutsche Sprache, deine Englisch braucht nicht sehr gut sein. Wir werden genug verstehen! Danke vielmals. Koennen Sie deine Frage auf Englisch sagen?

    (Meine Familie Name ist Schwabschen Deutsch aber meine Deutsche Sprache ist ungenugent! Mangelhaft! Es tut mir leid. Ich war ein Ami.)

  12. ClaireEllison
    November 26, 2016
    Excellent article plus its information and I positively bookmark to this site because here I always get an amazing knowledge as I expect.
  13. Benefiter
    November 28, 2016

    It's actually a cool and useful piece of information.

  14. ClaireEllison
    December 4, 2016
    Excellent article plus its information and I positively bookmark to this site because here I always get an amazing knowledge as I expect.
  15. jamescoom
    September 20, 2017

    Hi Dennis, I like this circuit as article material because it presents a good example of a feedback problem that can be solved by carefully applying basic feedback theory. Thanks for your kind sharing!

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