What is the purpose of LT1431 in this circuit?

Thread Starter

jm233h12

Joined Jan 9, 2020
10
I have took a simulation from Analog for a forward converter, which generates 54V@1.5A.

In isolated power supplies I often see the opto-coupler technique, for feedback and control system. However, in the schematic, an additional component called LT1431 https://www.analog.com/media/en/technical-documentation/data-sheets/1431fe.pdf. The data sheet says it is a programmable reference, shunt regulator. But why is it in the circuit? What does it improve upon compared to using just opto-isolator alone? Does it improve load regulation upon transient events, or something along those lines?

Photo attached of circuit:
 

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Alec_t

Joined Sep 17, 2013
11,665
But why is it in the circuit?
It allows the output voltage to be set to a chosen voltage. A fraction of the output voltage is derived via voltage divider R6-R7. If that fraction exceeds the internal reference 2.5V, the Coll pin goes low to switch on the opto-LED, which in turn provides feedback to the LT8310 to cause the output voltage to reduce.
 

Thread Starter

jm233h12

Joined Jan 9, 2020
10
It allows the output voltage to be set to a chosen voltage. A fraction of the output voltage is derived via voltage divider R6-R7. If that fraction exceeds the reference 2.5V, the Coll pin goes low to switch on the opto-LED, which in turn provides feedback to the LT8310 to cause the output voltage to reduce.
So it is needed in all opto-isolated supplies?
 

crutschow

Joined Mar 14, 2008
25,460
So it is needed in all opto-isolated supplies?
Yes, as a reference voltage is needed at the output so the feedback signal can regulate to an accurate output voltage.

It's also commonly done with the inexpensive 2.5V, TL431 reference controlling the optocoupler (as below), albeit with likely poorer temperature stability.
The 5V is the regulated output voltage, which can be any voltage as determined by the relative values of R1 and R2 to give 2.5V at the TL431 control input (the regulation voltage).

1578590961415.png
 
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Alec_t

Joined Sep 17, 2013
11,665
So it is needed in all opto-isolated supplies?
If not the LT1431, then some equivalent is needed, such as that shown by Crutschow. The output voltage must be compared with a reference voltage to get an error signal which the main IC uses to regulate the output voltage.
 

MisterBill2

Joined Jan 23, 2018
7,048
The opto isolator and the shunt reference are there for separate purposes, but some great engineering allows them to serve one common goal. The isolator because the power out must not be commoned with the input power source. The voltage reference because a stable regulator circuit needs a stable reference to compare with.
 

MisterBill2

Joined Jan 23, 2018
7,048
I realize now that I ignored those switchers that use a transformer, not an inductor. Most of them still need an opto isolator in the feedback loop, although a few use some other means. But every regulated supply needs a voltage reference to compare to. There are a lot of different references, though.
 

Thread Starter

jm233h12

Joined Jan 9, 2020
10
I realize now that I ignored those switchers that use a transformer, not an inductor. Most of them still need an opto isolator in the feedback loop, although a few use some other means. But every regulated supply needs a voltage reference to compare to. There are a lot of different references, though.
I understand. But why does this technique use the shunt regulator, rather than just a normal potential resistive divider that I usually see. Is there a benefit?

My assumption was that the transient response is better. The shunt regulator would deal with any transients better than would a simple resistive divider?
 

ci139

Joined Jul 11, 2016
1,696
i'm not shure if this extreamly similar ON semi AN is the one i read , byt they showed there how the active zener "improves!"... (←that along with the other components chosen right (which may be not 1 pass solution)) ...the feedback phase / frequency response (it's for a feedback loop improvement)
 

crutschow

Joined Mar 14, 2008
25,460
I understand.
Apparently not, based on your next question.
why does this technique use the shunt regulator, rather than just a normal potential resistive divider that I usually see. Is there a benefit?
Those are two functions that do entirely different things.
You need a reference voltage so that the output can be regulated to the desired voltage.
All a resistive divider does is reduce the voltage, it does not serve as a reference.

Say you want a 5V output voltage.
With a 2.5V reference, such as the TL432, a resistive divider is used to reduce the output 5V to 2.5V at the reference input.
The reference than increases or decreases the feedback voltage so that the 2.5V from the output divider equals the 2.5V internal reference voltage.
 

Thread Starter

jm233h12

Joined Jan 9, 2020
10
Apparently not, based on your next question.
Those are two functions that do entirely different things.
You need a reference voltage so that the output can be regulated to the desired voltage.
All a resistive divider does is reduce the voltage, it does not serve as a reference.

Say you want a 5V output voltage.
With a 2.5V reference, such as the TL432, a resistive divider is used to reduce the output 5V to 2.5V at the reference input.
The reference than increases or decreases the feedback voltage so that the 2.5V from the output divider equals the 2.5V internal reference voltage.
Well, thank you for clarifying anyway. I apologise if I got it wrong. I think maybe I am getting confused because when I see feedback circuit with "Vref", this is usually just a voltage source in the schematic rather than any kind of shunt regulator or diode.

What I still don't understand is what is the benefit of the actual shunt regulator. It is to my understanding that a shunt regulator will attempt to drive its output voltage to the same value irrespective of the input. So if the output voltage changes, say from 5V to 6V, and we need to drive it back down to 5V but the shunt voltage regulator gives an output voltage of 2.5V - how exactly can the primary side know the output voltage has changed?
 

MisterBill2

Joined Jan 23, 2018
7,048
The explanation given is correct. The whole purpose of a regulator is to provide a stable output voltage, and thus iit needs a stable reference voltage. In the case of the circuit shown, and deviation of the voltage alters the output of the optoisolator, which causes an adjustment to be made to correct the output voltage.
 

crutschow

Joined Mar 14, 2008
25,460
What I still don't understand is what is the benefit of the actual shunt regulator. It is to my understanding that a shunt regulator will attempt to drive its output voltage to the same value irrespective of the input. So if the output voltage changes, say from 5V to 6V, and we need to drive it back down to 5V but the shunt voltage regulator gives an output voltage of 2.5V - how exactly can the primary side know the output voltage has changed?
You need to understand how the regulator is part of the complete feedback loop.

In this case the shunt "regulator" is controlling the feedback current through the opt isolator, not generating a 2.5V output.
Thus if the output causes the regulator control input to go above 2.5V, the TL431 shunt current increases, thus increasing the current through the opto isolator control LED.

This is then coupled to the opto output transistor, increasing its current, and thus the voltage to the switcher control input circuit.
This reduces its duty-cycle and thus lowers the output voltage.

Similarly, if the regulator input goes below 2.5V, the shunt current is reduced, thus increasing the switcher duty-cycle and output voltage.
In this way the shunt/opto current settles at a value that just maintains the PWM duty-cycle at the proper point to keep the output at the desired voltage.

Make sense?
 
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ci139

Joined Jul 11, 2016
1,696
yes, "The LT®1431 is an adjustable shunt voltage regulator with 100mA sink capability, 0.4% initial reference voltage tolerance and 0.3% typical temperature stability. On-chip divider resistors allow the LT1431 to be configured as a 5V shunt regulator, with 1% initial voltage tolerance and requiring no additional external components. By adding two external resistors, the output voltage may be set to any value between 2.5V and 36V. The nominal internal current limit of 100mA may be decreased by including one external resistor. A simplified 3-pin version, the LT1431CZ/LT1431IZ, is available for applications as an adjustable reference and is pin compatible with the TL431. "
versus /
"The TL431, A, B integrated circuits are three–terminal programmable shunt regulator diodes. These monolithic IC voltage references operate as a low temperature coefficient zener which is programmable from Vref to 36 V with two external resistors. These devices exhibit a wide operating current range of 1.0 mA to 100 mA with a typical dynamic impedance of 0.22 Ω. The characteristics of these references make them excellent replacements for zener diodes in many applications such as digital voltmeters, power supplies, and op amp circuitry. The 2.5 V reference makes it convenient to obtain a stable reference from 5.0 V logic supplies, and since the TL431, A, B operates as a shunt regulator, it can be used as either a positive or negative voltage reference. • Programmable Output Voltage to 36 V • Voltage Reference Tolerance: ±0.4%, Typ @ 25°C (TL431B) • Low Dynamic Output Impedance, 0.22 Ω Typical • Sink Current Capability of 1.0 mA to 100 mA • Equivalent Full–Range Temperature Coefficient of 50 ppm/°C (0.6% -40÷85°C) Typical • Temperature Compensated for Operation over Full Rated Operating Temperature Range • Low Output Noise Voltage"

*
-- seems the LT variant has a bit lower upper frequency range by dynamic impedance . . . even when considering (MHz) as a typo 431v-s.png
 
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