input voltage of LM317A

Hi Takahiro,

The input voltage at which your output starts to rise will depend on the load current as you can see from Figures 5 and 6 in the datasheet. At 1 A load current the output voltage starts rising when Vin = 1.5 V regardless of the output voltage. 

This fact can be used to approximate the input voltage for your application's load current with the following procedure:

1. Determine the dropout voltage, Vdo, based on your application's load current using Figure 4 in the datasheet.
2. When the load current is 1 A, the output starts rising when Vin = 1.5 V. As such, the slope of the linear portion of the Vout vs Vin curve can be calculated as m = dVout/dVin = (Vout - 0 V)/(Vin - 1.5 V) = (Vout - 0 V)/[(Vout + Vdo@1A) - 1.5 V] 
3. Since the Vout/Vin lines are parallel regardless of load current, the slope equation can be used to calculate the input voltage at a different load current:

m = (Vout - 0 V)/(Vout + Vdo@1A - 1.5 V) = (Vout - 0 V)/(Vout + Vdo@Iout - Vstart)

Where Vstart is the input voltage when the output starts to rise for the specific load current, Iout. 

Finally, you can estimate the input voltage at which the output starts to rise by simplifying the equation from step 3: 

Vstart = Vdo@Iout - Vdo@1A + 1.5 V = Vdo@Iout - 0.5 V

Thanks,

Gerard

Thank you for your answer!

We understand Vstart.

We want to know about input voltage.
We think Vstart is output voltage that starts to rise.

Please tell us the minimum input voltage when output voltage changes from 0V to Vstart.
We Continuously change input voltage from0V to 24V.

Best regards,


Takahiro Nishizawa

Hi Takahiro,

Vstart is the input voltage when the output starts to rise. We can use Figure 5 as an example for Rout = 1.2 ohms. This gives a load current of approximately 1 A. Using Figure 4, the dropout voltage is typically 2 V@1 A load current. Now using the equation from my above post:

Vstart = Vdo@Iout - 0.5 V = 2 V - 0.5 V = 1.5 V

As shown by the graph, this is the minimum input voltage when the output voltage starts rising:

We can use the other curve on the same figure as another example, where Rout = 243 + 0 ohms. In this case the dropout voltage is calculated as Vin - Vout when the output starts falling out of regulation.

Vdo@5 mA = 2.6 V - 1.25 V = 1.35 V

Vstart = Vdo@Iout - 0.5 V = 1.35 V - 0.5 V = 0.85 V

0.85 V is approximately the input voltage at which the output starts rising for 5 mA load condition:

Keep in mind that these are approximations based on the typical Vout vs. Vin characteristics for the LM317.  

Thanks,

Gerard

Thank you for your answer!

We have new questions about output capacitor of LM317A.

Vin=24V Vout=3.3V

We try to use tantalum capacitor(10µF) and ceramic capacitor(1µF) in output.
Is this configuration correct?

And We are thinking about another way.
Ceramic capacitor(10µF) + Ceramic capacitor(0.1µF)
Are there any problems with this usage?

Best regards,


Takahiro Nishizawa

Thank your for your support!


How about your situation?
We are waiting for your answer!
Please tell us your answer about this question.


Vin=24V Vout=3.3V

We try to use tantalum capacitor(10µF) and ceramic capacitor(1µF) in output.
Is this configuration correct?

And We are thinking about another way.
Ceramic capacitor(10µF) + Ceramic capacitor(0.1µF)
Are there any problems with this usage?


Best regards,


Takahiro Nishizawa

Hi Takahiro,

I apologize for the delay. There are no problems with either configuration you listed. The LM317A is stable without any output capacitor, but TI recommends using at least 1 uF solid tantalum. Using 10 uF tantalum with an additional ceramic capacitor will further improve stability and reduce transient ringing. Keep in mind that an input bypass capacitor is also recommended for the LM317A, especially when using an output capacitor. At least 0.1 uF ceramic or 1 uF tantalum on the input will ensure stability.

Thanks,
Gerard

Thank you for your answer!

I understood that Either tantalum and ceramic capacitor do not matter.

We want to use ceramic capacitor 10µF.
Vin is 24V , Vout is 3.3V and Iout=100mA


We are concerned about quiescent current at 3.3 V.
Please tell us the maximum output current when we use 10µF ceramic capacitor.
Can this IC correspond to output current 100mA when we use 10µF ceramic capacitor?

Best regards,


Takahiro Nishizawa

I asked new question.
Please tell us your answer.

I understood that Either tantalum and ceramic capacitor do not matter.

We want to use ceramic capacitor 10µF.
Vin is 24V , Vout is 3.3V and Iout=100mA


We are concerned about quiescent current at 3.3 V.
Please tell us the maximum output current when we use 10µF ceramic capacitor.
Can this IC correspond to output current 100mA when we use 10µF ceramic capacitor?