LDO regulator problem

[purushothama]

Hi,
i am having problem with my LDO regulator, whenever i connect input of LDO with Switching regulator, LDO is not able to drive the load but if i connect LDO with external power supply it is able to drive the rated current, can anyone explain why this is happening.

 

[KerimF]

What is the average voltage of your switching regulator output?
What is the capacitor value at the LDO input?
Is there a resistor in series before the LDO input?
What is the estimated minumum input voltage of your LDO, for the load current?
What do you mean by "LDO is not able to drive the load"? Do you mean its output becomes zero for example?

 

[purushothama]

What is the average voltage of your switching regulator output?
12V DC
What is the capacitor value at the LDO input?
10uF
Is there a resistor in series before the LDO input?
No
What is the estimated minumum input voltage of your LDO, for the load current?
12V
What do you mean by "LDO is not able to drive the load"? Do you mean its output becomes zero for example?

no it is heating and voltage is getting reduced. we are expecting 5V it is reducing to 2.5V. but if input of LDO regulator is connected with extenal linear supply it is working perfect. also i checked the output of linear regulator it able to source sufficient current. following is the conversion procedure.

-->48V--Switching regulator->12V(2A)--LDO--->5V(400mA)

 

[KerimF]

Since your LDO is 5V then 6V at its input should be good to let it deliver 400mA. (Better estimation could be made if the LDO name is known)
The first experiment we can think of is to insert a resistor Rs between the 12V (of SMPS) and the input of the LDO.
Rs = (12 - 6)/0.4 = 15Ω
It will dissipate:
P_rs = (12 - 6) * 0.4 = 2.4 W
if it is hard to find 5W 15Ω you can replace it with 10 resistors of 150Ω 500mW connected in parallel.
This resistor (or resistors) will help reducing the heat dissipated by the LDO.

By the way you didn't tell me the output voltage of your linear external regulator.

Kerim

 

[purushothama]

Since your LDO is 5V then 6V at its input should be good to let it deliver 400mA. (Better estimation could be made if the LDO name is known)
The first experiment we can think of is to insert a resistor Rs between the 12V (of SMPS) and the input of the LDO.
Rs = (12 - 6)/0.4 = 15Ω
It will dissipate:
P_rs = (12 - 6) * 0.4 = 2.4 W
if it is hard to find 5W 15Ω you can replace it with 10 resistors of 150Ω 500mW connected in parallel.
This resistor (or resistors) will help reducing the heat dissipated by the LDO.

By the way you didn't tell me the output voltage of your linear external regulator.

Kerim

Hi,
thanks for your reply. i am using LM1117-5.0V Regulator. input to LDO is 12V and O/P is 5V. can you explain why if i give 12V input to LDO from externally it is working fine.

 

[KerimF]

Hi,
thanks for your reply. i am using LM1117-5.0V Regulator. input to LDO is 12V and O/P is 5V. can you explain why if i give 12V input to LDO from externally it is working fine.

I will download the datasheet to have a better idea how to answer you ;-)

It seems that the high frequency ripple of the switching regulator introduces a state of instability in the internal circuitry of the LDO. The 10uF at the input may not be effective at this frequency. Perhaps if increased to 100uF, it will be able to attenuate this ripple to a good extent.

======================
Edited:
But loading the output of a switching supply by a relatively large capacitor may disturb its function.
And if the ripple frequency is relatively high, it is better to add at the LDO input 100nF and perhaps 10nF to bypass it.
======================

Of course, even a small resistor between the 12V and the LDO input can better attenuate the ripple.

In any case, since your input is now 12V and your load is 400mA, your LDO has to dissipate 7*0.4 = 2.8 W. This needs a good heatsink. I personally give my LDO a hand to be cooler by inserting at its input a resistor to take some heat from it ;-)

Added:
The length of wires between the 12V and LDO if long enough may cause a state of resonance at the LDO input. Obviously when one has an oscilloscope most problems can be seen and solved more quickly.

Kerim