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 *ν _{B} * =

*ν*;

_{O}*R*= 0 Ω ,

_{1}*R*→ ∞ . The changes to the gain-test circuit are

_{2}A slightly modified block diagram results. With the *G _{m} * amplifier nulled, the open-loop

*ν _{G} * is varied (

*ν*≠0 V), and

_{g}*ν*and

_{g}*ν*are measured. As in the gain-test circuit, the open-loop contribution of

_{o}*ν*to

_{g}*ν*is

_{o}*Τ*x

_{g}*ν*. The resistance-test circuit equations reduce to

_{g}The closed-loop output-port *r _{out} * is found applying Ohm’s Law with

*R*sourcing

_{L}*i*=

_{o}*i*, in series with

_{g}*ν*;

_{g}and can be calculated from measurement of *ν _{o} * and

*ν*.

_{g}From circuit analysis, feedback reduces the open-loop voltage *Τ _{g} * x

*ν*by 1/(1 +

_{g}*G*). The divider

*Τ*is effectively

_{g}This is the *Τ _{g} * divider with

*R*and

_{L}*r*(cl) resistances. Equating the second and last expressions and solving,

_{out}The quasistatic output resistance of the TL431 is *r _{o} * in parallel with the 1/

*G*source which is dependent on

_{m}*ν*and is also across

_{e}*ν*= –

_{g}*ν*= –

_{b}*ν*;

_{o}This result also follows directly from the *substitution theorem* : the voltage on which the current source of *G _{m} * depends is also across it, thereby reducing the dependent current source to a resistance, 1/

*G*.

_{m}*r _{out} * from

*G*Currents

_{m}For the resistance-test circuit, the current of the *G _{m} * source can be decomposed into currents in

*r*and

_{o}*R*:

_{L}The fraction of *G _{m} * current in

*r*with

_{o}*ν*=

_{b}*ν*(as in the resistance-test circuit) is

_{o}Polarities of current and voltage can be somewhat confusing in this circuit. Because the *G _{m} * amplifier

*sinks*current (into the amplifier) as positive, then for KCL to apply, positive current in

*r*flows upward to a negative

_{o}*ν*. The same applies to

_{o}*RL*;

*ν*(

_{o}*R*) = –

_{L}*i*x

_{o}*R*where –

_{L}*i*and

_{o}*R*are non-negative.

_{L}Applying *G* = *G _{m} * x

*r*while solving for the closed-loop resistance of

_{out}*r*,

_{o}The *G _{m} * current in

*R*is

_{L}and the closed-loop resistance of *R _{L} * is

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

Combining the closed-loop resistances contributing *G _{m} * current,

As a circuit, *G _{m} * and

*r*form a dependent current source of resistance 1/

_{out}*G*in parallel with

_{m}*r*. The same current flows in the parallel loop formed by the two resistances.

_{out}*G*and

_{m}*r*are independent only if

_{out}*ν*across them is 0 V. Without

_{o}*ν*(from the test circuits) or a non-static

_{G}*V*, no voltage variation is expected at the output node;

_{R}*ν*= 0 V, and

_{o}*r*and 1/

_{out}*G*are independent. The above equation implies that

_{m}*ν*= 0 V and it also tells us that the 1/

_{o}*G*source is effectively equal in resistance to

_{m}*r*/

_{out}*G*.

The closed-loop *r _{out} * derived previously can be expressed in terms of this result as

Thus *r _{out} * (cl) is reduced from the open-loop

*r*by the feedback factor, 1 +

_{out}*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 *r _{out} * (cl), the output node must be driven with either an independent current source or a voltage source such as

*ν*behind

_{g}*R*. Its contribution to

_{L}*ν*causes the resulting current into the node to be that of the closed-loop

_{o}*r*. Substituting the closed-loop

_{out}*ν*into the closed-loop

_{o}*i*,

_{o}from which

The closed-loop contribution of *ν _{g} * to

*ν*is the amplifier closed-loop gain with

_{o}*Τ*x

_{g}*ν*applied to the output node.

_{g}**G _{m and ro from Test Data} **

Returning to the overall problem of finding *G _{m} * and

*r*, we now have two equations that can be solved for them simultaneously. The first is

_{o}and the second is

One of the TL431 specs gives *r _{out} * (cl) = 0.2 Ω, though it varies considerably, from 0.13 Ω to 0.5 Ω, among manufacturers. The value of

*r*can be found by substituting the first into the second equation and solving;

_{o}Open-loop *r _{o} * 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

*V*, and then

_{R}*r*measured, though

_{out}*V*is not an available node. Consequently, in most circuit analysis,

_{R}*r*is considered negligible in its effects and is ignored.

_{o}In the output-resistance test circuit given by TI, *R _{L} * = 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

*R*= 1025 Ω.

_{L}Substituting *G _{0} * = 600,

*R*= 230 Ω,

_{LG}*r*(cl) = 0.2 Ω, and

_{out}*R*= 1025 Ω,

_{L}For *r _{out} * >> 1/

*G*,

_{m}*r*(cl) ≈ 1/G

_{out}_{m}. Most circuit designs can make this approximation though some value of

*r*is needed when including the

_{o}*R*path in the analysis. An accurate value of

_{f}*r*is difficult to determine, however. On the low end of the range,

_{o}*G*= 400 and

_{0}*r*= 123 Ω, still much larger than 1/

_{o}*G*. For

_{m}*G*= 600, if

*r*(cl) ≈ 0.383, then

_{out}*r*→ ∞ , with values within the specified range. The output BJT of the TL431 can be expected to have

_{o}*r*of 100 kΩ or more. What reduces

_{o}*r*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

_{o}*r*.

_{o}
Looks Awesome

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.

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

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!

Richtige und erstaunliche Artikel, teilten mir sehr! Vielen Dank

What in particular caught your attention?

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

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

Bitte schön

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

“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.)

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

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!