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Forum Index : Electronics : A simpler way to think about the TL431?
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LadyN Guru  Joined: 26/01/2019 Location: United StatesPosts: 408 |
I am really starting to like the TL431 after reading about it and would prefer a simpler way to think about it when trying to use it in my design. Would I be making any mistakes if I said: 1. It's a two input opamp that drives an output transistor 2. It regulates the base current to this transistor in proportion to min(Rb/(2.5 - Vinput_at_ref), Ib_sat) where Rb is the base current limit resistor between the output of the opamp and the output transistor Ib_sat is the saturation current necessary at the base of the output transistor so it goes beyong linear mode Does this way work for a person who does not understand opamps (yet)? |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Yes its an opamp with one input tied to a very temperature stable 2.5v voltage reference, and the output driving an open collector transistor. These are normally used as a linear shunt voltage regulator, but can also be used as a high gain voltage comparator, that switches on and off very abruptly at a 2.5v threshold. Very useful device, because it can sink 100mA on the output, and could easily drive a LED or small relay directly. Cheers, Tony. |
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| LadyN Guru Joined: 26/01/2019 Location: United StatesPosts: 408 |
The Vref input can be used to sink current too I believe. |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
All the information I have suggests the input is high impedance and connected internally to the input of an op amp. http://www.ti.com/lit/an/slva987/slva987.pdf Cheers, Tony. |
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wiseguy Guru Joined: 21/06/2018 Location: AustraliaPosts: 1156 |
Definitely a high impedance input - maybe when tied directly to the output (collector)for a 2,5V shunt reference it gives the impression it is sinking , but its only sensing. When driving an opto from the output (in an analog isolated feedback control circuit) its always a good idea to have a resistor in parallel with the opto as the small current (400uA) drawn by the TL431 to operate can cause unexpected performance, a value of ~ 4K7 or less does the trick. Using the ATL431 that only draws 20uA is also a solution. Dont forget the minimum cathode current requirement in the spec of 1mA. The TL431 also makes a great stable two terminal zener of whatever voltage you need by adding 2 resistors and only needs the minimum standing current of 1mA. This device is one I will always have in my parts store - I use them often. It is easier to model them in your head that when Vref is ~ 2.5V the output is sinking current, but a fraction below 2.5 the output is a high impedance. So your choice of resistors should present 2.5V at the reference when the output is the voltage you need. If at first you dont succeed, I suggest you avoid sky diving.... Cheers Mike |
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| hotwater Senior Member  Joined: 29/08/2017 Location: United StatesPosts: 120 |
Definitely not a high impedance input all the time. Think of the input tied to the collector thru a diode. When the REF input goes above about 2.5V, that substrate starts conducting. The collector will be driven to under 2V. The whole dam thing becomes a poor zener. The spec sheet in a way warns of this saying the REF input can not take more than 10ma. For this reason you can not daisy chain several TL431 in the same voltage divider and have accurate results if there is more than 50my between them. As an example, you can not directly connect a TL431 to a lithium battery and have it signal when the voltage drops below 2.5V. |
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| wiseguy Guru Joined: 21/06/2018 Location: AustraliaPosts: 1156 |
Definitely a high impedance all the time when you are using them as they were intended to be used. The data sheet specifies typ 2uA of reference current. The absolute maximum ratings you referred to where ref current of 10mA is specified was aimed at informing people how far they can torture the part just before they die. I would use a TLV431 for a Lithium battery sense below 2.5 V its reference is 1.24V and they consume just 55uA to work. By the way I liked your somewhat non standard use for the IR2153 and switching HVDC without arcing - quite creative. If at first you dont succeed, I suggest you avoid sky diving.... Cheers Mike |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
I think that is the key right there. The most common usage has some kind of voltage feedback that holds the input fairly close to its switching point. As with many analog chips, there are often distinct voltage limits to how far you can drive normally high impedance inputs before something starts to conduct and cause problems. Cheers, Tony. |
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| hotwater Senior Member Joined: 29/08/2017 Location: United StatesPosts: 120 |
The problem is similar to the fact that a microprocessor can be powered by any input pin. People are sometimes amazed to see a microprocessor keep running after they removed the power because one pin was still feeding power in from an external source. The substrate acts as a diode. |
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| LadyN Guru Joined: 26/01/2019 Location: United StatesPosts: 408 |
My current understanding is that it biases the bypass transistor in the active region to dissipate the excess power that we don't want to send to the load that we are trying to regulate. Above the threshold, the biasing is high enough that the bypass transistor enters saturation and diverts most of the power from the load that we are trying to regulate. At that point, the power supply can droop, essentially forcing the PSU into the expected regulation, or, if the PSU does not droop, the transistor just gets as hot as it can in saturation mode. This tells me that the TL431 has an upper limit to the current it can sink, possibly limited by the Ic_max of the bypass transistor? Hence, an external bypass transistor, with higher Ic_max can be added, that will allow the pair to handle more power. Did I understand it right? |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
There are two separate issues here. The first is the behavior of the input voltage sensing pin, pin one. In its normal mode of operation this pin usually becomes part of a feedback loop which holds pin one very close to 2.5v, where the input bias current is very low, and the pin a high impedance voltage sensing point. If you try to force pin one well above 2.5v, the input current rises quite steeply, and its no longer a high impedance. In fact you can blow the chip up if you try to pull the input pin up to a much higher voltage without using a series current limiting resistor. And yes you can add an external transistor, (usually PNP) to increase the output current. Cheers, Tony. |
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| LadyN Guru Joined: 26/01/2019 Location: United StatesPosts: 408 |
Thank you Tony! I want to take a step back to ensure I am on a solid foundation. Zener regulators are a type of shunt regulator which are linear regulators. Abstractly, a Zener works like a variable resistor that ensures that the voltage drop (Vdrop) across the load (Rl) never exceeds the zener (or avalanche) voltage (Vz). The way it does this is it draws very little current when Vin < Vz, so that Vdrop ~= Vin. AT just beyond Vin = Vz, the zener/avalanche effects kicks in causing the diode to shunt out as much as it can (given the limiting resistor that's in series) so Vz = Vdrop is still maintained. So it's dissipating (Vin - Vz) * Iz such that Il * Rl = Vdrop = Vz Is all of this absolutely accurate? |
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| Warpspeed Guru Joined: 09/08/2007 Location: AustraliaPosts: 4406 |
Yes. Cheers, Tony. |
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