LM317 Getting more Heat

I'm using LM317 for micro controller power supply . Output is 3.3 V . I draw around 120 mA current from LM317. Input is 12 V

and also I provide heat sink .

And In one stage 3.3 V automatically falling down by Like 3.29,3.28,3.27...

Sounds like it's going into thermal shutdown although a few mV of drift as it warms up is not entirely unexpected. It is dissipating (12 - 3.3) * 0.12 = 1.044W so it needs a heat sink and should get quite warm. Your only option is to increase the heatsink size, force cool it or use a different kind of regulator.


Brian.

Now I'm using two regulators for 3.3 V. LM2940 for 12 V to 5 V. and then LM3940 for 5 V to 3.3 V.

I want to reduce it as single regulator. How make it.

Kindly suggest which type heat sink sufficient for this.

You can add a resistor in series at the input.
To calculate its value, the possible minimum instantaneous voltage of the 12V source should be known first, unless 12V is regulated.

Note:
In this case, a capacitor will be needed between the regulator input and ground.

I also try do with switching regulator MC34063. In that case voltage travel through the track more distance the voltage automatically reduce Like from output capacitor point
3.3 V . If it is traveled 5 CM by track In that point 3.21 V only available.

I'm very Confused My spec is Input from SMPS= 12 V. Battery = 7.4V and Output 3.3 V @ 200 mA (max) for micro controller. How to do this.

So you have two voltage supplies; 12V and 7.4 V.
Please estimate the possible lowest and highest voltages of each source. Could you?

Highest voltage is 12.5 V From SMPS. The lowest possible voltage is From battery 7 V.

Battery Normal voltage 7.4 V. (Full charge 8.1 V , Low battery 7.0 V )

Good, now we have given data to work on.
Your regulated output (using LM317) is 3.3 V @200 mA.

The dropout drop of LM317 @200mA could be assumed not more than 2V (from datasheet).
Therefore, Vin should satisfy the inequality:
Vin ≧ Vin_lowest
where:
Vin_lowest = 3.3 + 2
or
Vin ≧ 5.3 V , [condition_1]

(1) We add two diode (as 1N400x); one for each supply, in series with its positive terminal.
The two diode cathodes are joined together with the LM317 input.

(2) In this case, the lowest input voltage (at the output of the two diodes) is:
Vin_min = 7 - 0.8 = 6.2 V
So Vin ≧ Vin_min, and Vin_min ≧ Vin_lowest. And [condition_1] is always satisfied (so far).

(3) Also in this case, at Vin_min the voltage excess is:
Vr1 = Vin_min - Vin_lowest
Vr1 = 6.2 - 5.3 = 0.9 V

(4) If we add a resistor (R1) between the positive terminal of the battery and the diode anode, its value could be:
R1 = Vr1 / I_load
where:
I_load = 0.2 mA , [given]
Vr1 = 0.9

R1 = 0.9 / 0.2 = 4.5 Ohm
The value of R1 could be then 4.7 Ohm, unless I_load could be greater than 200 mA.

(5)

Input and output side filter capacitors are ok? To prevent bad feedback.
Or try lower input voltage.

We will certainly talk about the capacitors

(5) Now let us calculate the dissipated powers of R1 and LM317 when Vbat_max = 8.1 V
About LM317:
Vin = Vbat_max - Vdiode - I_load_max * R1
Vin = 8.1 - 0.8 - 0.2 * 4.7 = 6.36 V
The voltage drop on LM317 is:
V_drop = Vin - Vout
V_drop = 6.36 - 3.3 = 3.06
Therefore the power dissipation of LM324 is:
P_dis = V_drop * I_load_max
P_dis = 3.06 * 0.2 = 0.612 W

Note:
To be sure that P_dis = 0.612 W is maximum, we should find P_dis in function of I_load first:
P_dis = V_drop * I_load
V_drop = Vin - Vout
Vin = Vbat_max - Vdiode - I_load * R1
Therefore:
P_dis = ( Vbat_max - Vdiode - I_load * R1 - Vout ) * I_load
P_dis = ( Vbat_max - Vdiode - Vout ) * I_load - R1 * I_load^2
The derivative of P_dis is
dP_dis / dI_load = ( Vbat_max - Vdiode - Vout ) - 2 * R1 * I_load
The maximum of P_dis occurs when dP_dis / dI_load = 0
0 = ( Vbat_max - Vdiode - Vout ) - 2 * R1 * I_load
I_load = ( Vbat_max - Vdiode - Vout ) / 2 / R1
I_load = ( 8.1 - 0.8 - 3.3 ) / 2 / 4.7
I_load = 0.425 A
Since I_load_max is 0.2 A only then P_dis = 0.612 W is maximum (in general it may not be so).

(6) About R1:
P_r1_max = I_load_max * I_load_max * R1
P_r1_max = 0.2 * 0.2 * 4.7 = 0.188 W
Note:
If R1 is not added this 188 mW would have to be also dissipated by the regulator.

(7) Now let us assume that the 12V SMPS is active. In this case, [condition_1] is always satisfied.
Note: Even if the battery is also connected, its diode voltage is reversed hence I_bat = 0

(8) SORRY, to go on we need to know the possible minimum voltage of the 12V SMPS. Is it 12V or lower?

I also try do with reduce the voltage using diode. But again I'm facing same problem . I reduce the using two diode for SMPS . No use . There no problem in output capcitor .

Sorry, you didn't answer my last question at (8) on post #10.

For instance, the diodes are not used here to reduce the input voltage but to select automatically the highest available one.

The added resistor in series with each source helps reducing the dissipation of LM317.
We did this for the battery source only since you couldn't tell me (yet) your estimated value of the possible lowest voltage for the 12V SMPS.
Could we assume it is 12V, or lower than 12V because of its output ripple @200 mA for example?

The SMPS minimum voltage is 11.5 V. If add resister in before input of the LM317. It reduces battery back up.

Replace input power source (SMPS) with some battery at 12V and try LM317. Maybe you have bad designed SMPS with high spikes which often increase temp even when idling.

Second options are that you have bad cloned chinese part or you didnt use part properly.



Best regards,
Peter

Thank you , Peter. I also changed input and battery. Again I'm facing same problem.

I have one more doubt I also try do with switching regulator MC34063. In that case voltage travel through the track more distance the voltage automatically reduce Like from output capacitor point 3.3 V . If it is traveled 5 CM by track In that point 3.21 V only available.

For precize voltage use precize dedicated regulator for 3,3V. I think this is better solution then regulated LM317. For mentioned current of 120mA you can use linear regulator or as better solution switcher solution.

LDO products for 3,3V :
https://www.tme.eu/en/katalog/#id_c...367,35,2572,364,10,365&used_params=364:24603;


Switcher products for 3,3V :
https://www.tme.eu/en/katalog/#id_c...24,365,364&used_params=364:38842,24603,76992;

Specially look switchers regulators in SO8, they can be very good and require little space on PCB.


Also MC34063 can give good result, just check circuit design and part values, keep tracks short as possible.


Without heatsink LM317 will be very hot to hold finger on it, you should measure temp on LM317 case, maybe temp is in ok range. 60C is hot for finger but can be in part functional range.


Best regards,
Peter

Here LM317 getting heat 180 C .That going to the Thermal shut down. I don't know how avoid it.

amayilsamy said:

The SMPS minimum voltage is 11.5 V. If add resister in before input of the LM317. It reduces battery back up.

Click to expand...

Sorry for my late reply, I was very sick (now I have a little headache only).

Please note that adding a resistor in series with the LM317 input reduces the voltage applied on LM317 only hence its dissipation.
The battery current is the same in the case a resistor is added or not, because this current is determined by the load only at the LM317 output.
If the series resistor is removed, its power dissipation will be added to LM317.

In other words, the life of the backup battery won’t be affected at all because the current, it supplies, is the same in both cases.

Please let me know if this answered your question or not.

Kerim

Thank you Mr. Kerim . I try reduce my output of SMPS by 9 V. then I come back to you

You don't need reducing it... I was just waiting to know if you get my answer about your last question so that I can go on. It seems you got it.


(8) In case of SMPS, the lowest input voltage (at the output of the two diodes) is:
Vin_min = 11.5 - 0.8 = 10.7 V
So Vin_min ≧ Vin_lowest, and [condition_1] is satisfied.

(9) Also in this case, at Vin_min the excess voltage is:
Vr2 = Vin_min - Vin_lowest
Vr2 = 10.7 - 5.3 = 5.4 V

(10) If we add a resistor (R2) between the positive terminal of the SMPS and the diode anode, its value could be:
R2 = Vr2 / I_load
where:
I_load = 0.2 mA , [given]
Vr2 = 5.4 V

R2 = 5.4 / 0.2 = 27 Ohm

(11) Now let us calculate the dissipated powers of R2 and LM317 when Vin_max = 12.5 V
About LM317:
Vin = Vin_max - Vdiode - I_load_max * R2
Vin = 12.5 - 0.8 - 0.2 * 27 = 6.3 V
The voltage drop on LM317 is:
V_drop = Vin - Vout
V_drop = 6.3 - 3.3 = 3 V
Therefore the power dissipation of LM324 is:
P_dis = V_drop * I_load_max
P_dis = 3 * 0.2 = 0.6 W

Note: (repeated from post #10)
To be sure that P_dis = 0.6 W is maximum, we should find P_dis in function of I_load first:
P_dis = V_drop * I_load
V_drop = Vin - Vout
Vin = Vin_max - Vdiode - I_load * R2
Therefore:
P_dis = ( Vin_max - Vdiode - I_load * R2 - Vout ) * I_load
P_dis = ( Vin_max - Vdiode - Vout ) * I_load – R2 * I_load^2
The derivative of P_dis is
dP_dis / dI_load = ( Vin_max - Vdiode - Vout ) - 2 * R2 * I_load
The maximum of P_dis occurs when dP_dis / dI_load = 0
0 = ( Vin_max - Vdiode - Vout ) - 2 * R2 * I_load
I_load = ( Vin_max - Vdiode - Vout ) / 2 / R2
I_load = ( 12.5 - 0.8 - 3.3 ) / 2 / 27
I_load = 0.156 A
Here I_load < I_load_max, so we must recalculate P_dis for 0.156 A
P_dis = ( 12.5 – 0.8 – 0.156 * 27 – 3.3 ) * 0.156
P_dis = 0.653 W (maximum dissipation of LM317).

(12) About R2:
P_r2_max = I_load_max * I_load_max * R2
P_r2_max = 0.2 * 0.2 * 27 = 1.08 W
Note:
If R2 is not added this 1.08 W would have to be also dissipated by the regulator.

For R2, I suggest using 4 500mW resistors of 27 Ohm connected to be equivalent to 27 Ohm, I guess you know how, right? In this case the equivalent resistor is rated at 2 W (0.5W * 4).

Please don't forget the two diodes.

Kerim