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I have few doubts regarding TL431.

Please correct if i am wrong..

TL431 internal architecure.

Intenal diode Reverse characterstic.

  1. TL431 can be used to generate voltages from Vref to 36V. Output voltage is taken across diode in reverse biassed condition. reverse characterstics says Reverse breakdown voltage is 2.5V, then how TL431 can generate different output voltages. Diode breakdown region allows more current(Ika) to be taken but reverse breakdown voltage(Vka) will be almost const. I didn't understand physically how it will generate different voltages.

  2. Please explain the operation of the other elements, like When Vref> 2.5V Opamp output will be high, then transistor will be ON and conducts current, but it will not effect diode behavior.

When Ref<2.5 Op-amp output is Low -->Transistor OFF, but it will not affect diode behavior. When the diode will be going to breakdown region?.

  1. Constant Current source using TL431

Constant Current Source

How to explain this circuit.?

One explanation says, REF always maintained at VREF->2.5V, so, constant current source Vref/Rs ----> How REF always maintained at 2.5V

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  • \$\begingroup\$ i didn't get answer to my questions...1. Based on REF voltage w.r.t 2.5V, BJT will turn ON or OFF, but no where data about the diode is mentioned.How diode will work here. ....................from Block diagram, across diode only output voltages taken, diode has Reverse breakdown voltage of 2.5V, then how output voltage can vary from 2.5 to 36V?.... \$\endgroup\$ – user19579 Apr 24 '14 at 5:25
  • \$\begingroup\$ Your complete premise is wrong. The BJT is ALWAYS conducting (on and off references have no application here) and the voltage at the high end of Rs is ALWAYS Vref. The Diode shown in the functional diagram has no meaning when the TL431 is in a normal mode (cathode more positive than anode terminal. The diode IS NOT a Zener, though I'm sure it's avalanche point is beyond the 36V rating. \$\endgroup\$ – Jack Creasey Sep 1 '18 at 23:19
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How to explain this circuit.?

enter image description here

Start from scratch imagining the transistor isn't conducting. Voltage across Rs is zero and therefore the TL431 is open circuit - but that's a problem because if it were open circuit (off) there would be decent base current into the transistor and Io would be large.

OK, look at it the other way, TL431 is conducting because Rs has enough voltage across it due to transistor being on - hey but this is wrong too because, if the TL431 were conducting there would be no voltage on the transistor base and therefore it isn't conducting and Rs has no voltage across it.

The goldilocks answer - the transistor is just conducting enough to allow 2.5 v to develop across Rs which in turn is just starting the turn the TL431 on and if it turned on too much, the current would reduce thru Rs and this would counteract the TL431 turning on too much.

It's called negative feedback.

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  1. The breakdown is more properly "forward breakdown" if you think of it as a zener- the cathode is negative. That's not the normal operating mode, in any case. Normally you want the cathode positive, and by dividing the voltage on the cathode down you can get the IC to regulate at any voltage from 2.5V (no divider) to 36V (maximum voltage).

  2. The voltage from cathode to anode should always be at least Vref, so the circuit will always have some voltage to function.

  3. In the case of the current sink, the voltage at the cathode will be Vref+Vbe (Vbe of the BJT) when it is regulating. The internal transistor shunts the current from the resistor to the BJT base to maintain the reference input of the TL431 at Vref.

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In general terms:

  • Vref --> Io: Having a known voltage reference Vref can be used to generate ANY current, if Vref somehow drives a known resistor (Io = Vref/R).

  • Vref --> Vo: Having a known voltage reference Vref can be used to generate ANY other voltage, if Vref can be somehow used to drive a known resistor (so the current is known: I = Vref/R) and then that current (or a mirror of it) goes through another resistor Ro, creating the new voltage: Vo = IRo.

The TL431 is simply the voltage reference and the feedback building block that can be applied in the scenarios mentioned above.

TL431 internal architecure.

The behavior is simply that when REF > (Vref+ANODE), the NPN-BJT will try to draw more current by decreasing its effective resistance (because the op-amp will provide more Vbe), and when REF < (Vref+ANODE), the NPN-BJT will try to draw less current by increasing its internal resistance (because the op-amp will provide less Vbe).

If this effect is used as a negative feedback loop that affects the REF input, then the system will try to reach an equilibrium where REF = Vref+ANODE.

In the cited case of generating an output current ("constant current-sink"):

Constant Current Source

[Note that the drop at Rs is REF, and ANODE is GND (defines 0V)]

It works because if the drop at Rs (REF) is less than Vref (2.5V), then the internal BJT will increase its resistance, shunt less of the base current of the external BJT, so the external BJT gets more base current, which increases its emmitter current, which causes the drop on Rs to increase (so REF increases). Summary: {REF > Vref} makes REF decrease.

On the other hand if the drop at Rs (REF) is higher than Vref (2.5V), then the internal BJT will decrease its resistance, shunting more current, so the base of the external BJT gets less current, so its emitter current is less, so less voltage is dropped at Rs, decreasing RES. Summary: {REF < Vref} makes REF increase.

Therefore, at equilibrium REF = Vref (2.5V). Since the drop on Rs is known, and Rs is known, then the collector current is known, which is your output current. So can basically set the output current by properly selecting Rs (Io = 2.5V/Rs).

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  • \$\begingroup\$ i didn't get answer to my questions...1. Based on REF voltage w.r.t 2.5V, BJT will turn ON or OFF, but no where data about the diode is mentioned.How diode will work here. ....................from Block diagram, across diode only output voltages taken, diode has Reverse breakdown voltage of 2.5V, then how output voltage can vary from 2.5 to 36V?. \$\endgroup\$ – user19579 Apr 24 '14 at 8:28
  • \$\begingroup\$ @user19579 If you are referring to the diode shown in the "Functional Block Diagram", its breakdown voltage is NOT 2.5V. It must be greater than 36V for sure. If you think of this device as a "Zener replacement", you can talk about reverse breakdown voltage, and is settable from 2.5V all the way up to 36V. If you're looking at the posted Figure 6 of IKA vs VKA, it is for the special test case of shorting REF with CATHODE, which sets the zener equivalent breakdown to the minimum: 2.5V. Let me know if this clears it up. \$\endgroup\$ – apalopohapa Apr 24 '14 at 9:12
  • \$\begingroup\$ @user19579 By the way, that diode just gives the "zener replacement" device a standard diode behavior when forward biased, as is also illustrated in your posted Figure 6 (for negative values of VKA, you can see the inverted diode curve, which would look side up if you plotted IAK vs VAK instead). \$\endgroup\$ – apalopohapa Apr 24 '14 at 9:20
  • \$\begingroup\$ Now i understand.i didn't observe VKA=Vref in Fig.6. you Really helped me. One point, it looks like,the diode will not contribute to the operation of TL431. i am curious to know, if diode is removed, TL431 has any effect on it's operation.? \$\endgroup\$ – user19579 Apr 25 '14 at 10:48
  • \$\begingroup\$ @user19579 I think it just comes into play when forward biased. \$\endgroup\$ – apalopohapa Apr 25 '14 at 22:37
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Actually there is NO breakdown in this device, unless you try to use it backwards and the protection diodes short the output or any other way that defeats its purpose which is to emulate a programmable Zener using what is called a Bandgap diode circuit. Using a clever symmetrical design developed in the mid 60's, this reference voltage has been used in 3 terminal regulators with a temperature compensated buffered 2.5 V with nearly an ideal response.

It uses some nifty mathematical ratios in geometry of the transistor areas , so that the NTC thermal voltage offset of one transistor matches a PTC thermal effect using a differential current ratio and geometry. I won't go into much detail here but using multiples of gain and transistor cells of a certain area and shape in parallel, size, this special ratio with "smoke" and current mirrors generates almost a perfect match in temperature coefficient and constant voltage at 2.495 V.

thanks to Boltzmann's constant and many Engineers a newer design with <1V reference voltages is also popular but then in 2010 and 2012, another Engineer improved the temperature coefficient to under 3.9uV/'C by adding another Op Amp and more clever ratios.

enter image description here

The above internal schematic shows the voltages at each node of the TL431.

TI has a spreadsheet for using this part in a 2nd order controlled feedback loop to select a voltage out for a high voltage flyback SMPS with all the Filter loop characteristics optimized for high stability.

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Few details from TI support group, which will help understanding..

important point i asked is What is the relation of diode with the operation of TL431.

http://e2e.ti.com/support/power_management/linear_regulators/f/321/t/338291.aspx/

In TL431. diode shown is not intentionally kept.

"The diode is unavoidable with the process technology of TL431. The diode also makes the device similar to a Zener diode that passes current in a forward biased mode. "

"actually in TL431, there is no Vz or breakdown, it is just BJT turning ON"

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