Problem with TL431 regulation, plse help me. Post #1
I had recently tried to add shunt regulation to my existing LM317 and LM337 regulated supply for my TDA1541 DAC. However, i encountered the problems below:
1. using the formula Vout= (1+R1/R2)*Vref , i worked out the R1/R2 for -5V & +5V to be 1K and -15V to be 1K/5K. And i adjust the Vo at the LM317 and LM337 to be +6V, -6V and -18V and connect directly to the shunt regulator directly. However, the Vout at the shunt regulators do not give the Vout i need. They simply give out whatever input the LM317 and LM337 feed them. Therefore to obtain the correct voltages after the shunt regulator, i need to adjust the LM317 and LM337.
2. the LM317 and LM337 runs hot! the same goes for the TL431!
Can anyone please enlighten?
I had recently tried to add shunt regulation to my existing LM317 and LM337 regulated supply for my TDA1541 DAC. However, i encountered the problems below:
1. using the formula Vout= (1+R1/R2)*Vref , i worked out the R1/R2 for -5V & +5V to be 1K and -15V to be 1K/5K. And i adjust the Vo at the LM317 and LM337 to be +6V, -6V and -18V and connect directly to the shunt regulator directly. However, the Vout at the shunt regulators do not give the Vout i need. They simply give out whatever input the LM317 and LM337 feed them. Therefore to obtain the correct voltages after the shunt regulator, i need to adjust the LM317 and LM337.
2. the LM317 and LM337 runs hot! the same goes for the TL431!
Can anyone please enlighten?
Read the datasheet
http://www.onsemi.com/pub/Collateral/TL431-D.PDF
Figure 18: a shunt needs a series resistive element between raw voltage supply and regulated output. The sum of shunt current and load current flow through that resistor and so make up the difference voltage between raw supply and regulated output.
In your case you connect a near zero-Ohm 6V supply right to the shunt. The TL431 tries to lower the output voltage by ever passing more current. As this current is happily passed on by the LM317, without voltage sagging, the two get into a deadly, erm, deadlock.
How to calculate the series resistor?
Suppose the load consumes 15mA. We add to this a shunt current of 10mA, for a total of 25mA. Vraw is 6V, Vreg is 5V:
R = (6V-5V)/55mA = 40.
But a better solution, BTW, might be to use the 317/337s as current sources, in which case they indeed act as voltage dropping elements for the 431s.
The Naked Truth article on the use of TL431s for audio.
http://www.onsemi.com/pub/Collateral/TL431-D.PDF
Figure 18: a shunt needs a series resistive element between raw voltage supply and regulated output. The sum of shunt current and load current flow through that resistor and so make up the difference voltage between raw supply and regulated output.
In your case you connect a near zero-Ohm 6V supply right to the shunt. The TL431 tries to lower the output voltage by ever passing more current. As this current is happily passed on by the LM317, without voltage sagging, the two get into a deadly, erm, deadlock.
How to calculate the series resistor?
Suppose the load consumes 15mA. We add to this a shunt current of 10mA, for a total of 25mA. Vraw is 6V, Vreg is 5V:
R = (6V-5V)/55mA = 40.
But a better solution, BTW, might be to use the 317/337s as current sources, in which case they indeed act as voltage dropping elements for the 431s.
The Naked Truth article on the use of TL431s for audio.
Hi!
I will post You today some Excell spreadsheet from (i belive) Texas Instruments with some aditional data . It should be easy to calculate the required voltage and series resistot. I am on work and I have that spreadsheet at my home pc - don't worry I won't forget.
Don't worry - in 10 hours I will post it to You.
regards
daniel
I will post You today some Excell spreadsheet from (i belive) Texas Instruments with some aditional data . It should be easy to calculate the required voltage and series resistot. I am on work and I have that spreadsheet at my home pc - don't worry I won't forget.
Don't worry - in 10 hours I will post it to You.
regards
daniel
What minimum current does TI's file use for the TL431?
I think it is 1mA, but I read on this forum that 10mA as a minimum current is recommended
I think it is 1mA, but I read on this forum that 10mA as a minimum current is recommended
peranders said:
That's the file I ment. You can also use a little lower value (for Rser) than calculated to increase thecurent through TL431 (that is for obtaining 10mA or so through TL431- check the temperature of the TL after You do so).
regards
daniel
checked out the Texas spreadsheet. very useful and convenient tool indeed.
what is the Idiv/Iref? i realise that changing this value will affect the value of R1/R2. What is the link here? how to determine this value to put into the table for calculation?
what is the Idiv/Iref? i realise that changing this value will affect the value of R1/R2. What is the link here? how to determine this value to put into the table for calculation?
commstech said:
That is the ratio between the currents of the resistor divider and the current throught the TL431 if I remember it right.
🙂
sparkle said:
hi sparkle.
so in my application, what is the appropriate value of Idiv/Iref i should enter?
You should read the datasheet for this because different brands are not exactly the same.Bricolo said:
The current is around 0.5 mA, but read the datasheet to be sure. I would recommend you to have at least 1-2 mA through the 431. Notic also the value of the decoupling cap. See the datasheet for this. If you have the wrong value you can get oscillations.
Werner said:Read the datasheet
http://www.onsemi.com/pub/Collateral/TL431-D.PDF
Figure 18: a shunt needs a series resistive element between raw voltage supply and regulated output. The sum of shunt current and load current flow through that resistor and so make up the difference voltage between raw supply and regulated output.
In your case you connect a near zero-Ohm 6V supply right to the shunt. The TL431 tries to lower the output voltage by ever passing more current. As this current is happily passed on by the LM317, without voltage sagging, the two get into a deadly, erm, deadlock.
How to calculate the series resistor?
Suppose the load consumes 15mA. We add to this a shunt current of 10mA, for a total of 25mA. Vraw is 6V, Vreg is 5V:
R = (6V-5V)/55mA = 40.
But a better solution, BTW, might be to use the 317/337s as current sources, in which case they indeed act as voltage dropping elements for the 431s.
The Naked Truth article on the use of TL431s for audio.
Werner,
Does it make sense if i simply adjust the LM317 and LM337 to the output voltage that i need ie. +5V, -5V and -15V and forego the use of the series resistor and feed directly to the TL431? Will that solve the problem at all?
commstech- look ...
if You have 10V output from the LM317 and want to have 5V after TL431 and Your load is , say 100mA just do that:
100mA+5mA (for TL431)=105mA
Series resistor is then 10V-5V= 5V; Rser= 5V (across the Rser)/105mA = 47R aproximate
Voltage divider is 5k for upper and 5k for lower resistor. If You feel that Your TL431 is cold You can lower Rser a little bit lower until You fell that TL431 is LITTLE bit warm. Thant should be it I think.
regards
daniel
if You have 10V output from the LM317 and want to have 5V after TL431 and Your load is , say 100mA just do that:
100mA+5mA (for TL431)=105mA
Series resistor is then 10V-5V= 5V; Rser= 5V (across the Rser)/105mA = 47R aproximate
Voltage divider is 5k for upper and 5k for lower resistor. If You feel that Your TL431 is cold You can lower Rser a little bit lower until You fell that TL431 is LITTLE bit warm. Thant should be it I think.
regards
daniel
If You tell me what You need I can calculate it for You to try- then You try it and correct if needed.
regards
daniel
Remember: TL should be little warm- not much- to ensure that enough current is flowing through TL- if not calculated.
Also the resistor divider determine the voltage not the series resistor. With Rser You determine current flowing through You load and TL (and resistor divider).
regards
daniel
regards
daniel
Remember: TL should be little warm- not much- to ensure that enough current is flowing through TL- if not calculated.
Also the resistor divider determine the voltage not the series resistor. With Rser You determine current flowing through You load and TL (and resistor divider).
regards
daniel
Hi sparkle,sparkle said:
after all the explanation here, and Assuming i have a 6V to feed to the shunt regulator; and i want 5V as the output. and the current consumption should be about 45mA as according to the TDA1541 datasheet. Basing on this, the Rser should be:
(6V-5V)/(45mA + 10mA) = 1V/55mA = 18ohms
I am using 1kohm for both R1 and R2 to achieve the Vout. is it OK?
Do i get it right at all? Thanks for your patience and help.
🙂
Hi!
It is almost there- everything right except that You have 2.5mA running through the voltage divider. So if You want better calculation then add that current to Your 55mA - it would not affect the value of the series resistor much. Also if the TL431 is running to hot (with 10mA) than put a little bit bigger series resistor and that is it.
You got the point of it.
Please don't thank me for the patience- You don't know how much mistakes and don't know's I make 🙁 . It is my duty to help You since a lot of guys also helped me. I AM GLAD THAT I CAN HELP YOU IN SOME WAY....😀
best regards
daniel
It is almost there- everything right except that You have 2.5mA running through the voltage divider. So if You want better calculation then add that current to Your 55mA - it would not affect the value of the series resistor much. Also if the TL431 is running to hot (with 10mA) than put a little bit bigger series resistor and that is it.
You got the point of it.
Please don't thank me for the patience- You don't know how much mistakes and don't know's I make 🙁 . It is my duty to help You since a lot of guys also helped me. I AM GLAD THAT I CAN HELP YOU IN SOME WAY....😀
best regards
daniel
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