Hello everyone,
now i want to use thvoltage regulator TPS72325 to generate fest output -2.5v as negative voltage source for the amplifier.
But i have simulated it as the data sheet, but i don't know what is the problem.
As below is the circuit from the data sheet.

I have selected the 5V for the enable pin. Input is -6V, and output is very strange. If the enable pin is open and not connected, the result is also wrong.
Could you tell me what's my problem. Thanks so much in advance!

 

Have you tried to let it simulate for longer? It might just take some time to charge the output cap. Also, can you post your asc and lib file?

 

Looking at the datasheet I don´t see why it should not work with 5V at the enable input (as long as it is less than 5.5V). Do you have all the three caps with the right values and right next to the chip?

 

Looking at the datasheet I don´t see why it should not work with 5V at the enable input (as long as it is less than 5.5V). Do you have all the three caps with the right values and right next to the chip?

I have checked and all the right caps are in the right position. Maybe because i used the voltage divider to generate -6V from -15V at the input pin?
Is it possible to use this regulator with the enable pin unconnected? unconnected means the voltage on this pin is 0V?

 

Unconnected will turn the regulator off. What voltage divider? Can you show your schematic? My guess is that the regulator is trying to work, but you don´t provide it with enough voltage under load. What is your load and how much current is it drawing?

 

I don´t think the enable pin is your problem. You don´t say what is the load, but the resistor divider won´t let you draw more than about 20mA from the output without the voltage collapsing.
You should use some better arrangement to get the input of the TPS below 10V instead of the resistors. A 6.8V zener instead of R1 and some higher value like 47k instead of R2 should work.

 

Datasheet says that the enable pin should be tied to in if you don´t want to use the shutdown functionality. I would not try using with the pin floating as any interference will couple onto it and turn the chip on and of randomly.

The 15V source, R1 and R2 are equivalent to a 6.07V source with 404 ohms in series. The regulator has droput voltage of 250mV. From this you can see that the in pin has to be at least 2.75V in order for the regulate to keep working.
The drop on the equivalent resistance therefore has to be less than 6.07V-2.75V=3.32V, so the max current your regulator can supply is 3,32V/404Ω=8.2mA. Your six amplifiers draw 30mA just to power them, you still didn't show your actual schematic, but I expect there to be some load on those and the demand even higher.
As you can see the divider is silly and not a practical solution. Why don´t you use a regulator that can work with 15V on the input and be done with it?

 

Datasheet says that the enable pin should be tied to in if you don´t want to use the shutdown functionality. I would not try using with the pin floating as any interference will couple onto it and turn the chip on and of randomly.

The 15V source, R1 and R2 are equivalent to a 6.07V source with 404 ohms in series. The regulator has droput voltage of 250mV. From this you can see that the in pin has to be at least 2.75V in order for the regulate to keep working.
The drop on the equivalent resistance therefore has to be less than 6.07V-2.75V=3.32V, so the max current your regulator can supply is 3,32V/404Ω=8.2mA. Your six amplifiers draw 30mA just to power them, you still didn't show your actual schematic, but I expect there to be some load on those and the demand even higher.
As you can see the divider is silly and not a practical solution. Why don´t you use a regulator that can work with 15V on the input and be done with it?

Thanks so much for your detailed explanation.
Because i don't have time to change the PCB design. So i have to think of a solution to solve this problem but don't change the PCB. Before it's my fault to forget to think about the current. I have chosen this regulator because the output is fest -2.5V. But now i think it's a wrong choice. My load is only the six amplifier, three as adder, three as inverting amplifier.
I have tried if i don't change anything, the regulator can power three amplifier. But if more, then failed.
Is it possible to change the value of divider resistor to fix this problem? For example, changing the position of 1K and 680R to get -8.93V as input for the regulator. And then the current can also double.
Or do you have any other advice to solve it without changing PCB design?

 

Yes you could adjust the ratio and the total value of the resistor to make it work, but be aware that you will be wasting a lot of power in that divider.

 

Yes you could adjust the ratio and the total value of the resistor to make it work, but be aware that you will be wasting a lot of power in that divider.

Yes, I have tried in the simulation in LTSpice. Now i have seen it works with 110R and 200R as voltage divider. The voltage should be -9.68V but in the simulation is -7.37V. The output is now -2.5V and i will buy the resistor and try it in the PCB experiment next Monday.
Thanks for your advice!