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# 24 DCV Power supply, 7824 over heating problem

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#### eng_ahmed22

##### Full Member level 3
hi all

I have build the shown circuit, The problem is the 7824 voltage regulator is suffering from high temp, and I am afraid this high temperature damage the regulator,
how can I overcome this??

according to the datasheet the Max. input voltage is 40 Vdc. the device is supposed to work 24/24, the max. current is about 800mA

Thanks and best regards

Try using a heatsink, you need to calculate the thermal resistance, see some example

Ta - Ambient temperature (example 30°C)
Th - Heatsink temperature (example 50°C)
Vh - Voltage to heater 16V
Ih - Current to heater 0.8A
Tr = Temperature rise =Th - Ta = 50 - 30 = 20°C
Ph - Power applied to heatsink = Vh * Ih = 16 * 0.8 = 12.8W
Rth - Thermal resistance (in °C/W) = Tr / Ph = 20 / 12.8 = 1.57 °C/W

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Didn't you miss a capacitor on the schematic?

You said your maximum current 800mA.
What is the possible value of the maximum mains voltage?
Also what is the possible value of the minimum mains voltage?

Kerim

Didn't you miss a capacitor on the schematic?

You said your maximum current 800mA.
What is the possible value of the maximum mains voltage?
Also what is the possible value of the minimum mains voltage?

Kerim

Yes, I'm using 1000uF/50 v after the bridge
The maximum mains voltage is 240V
The minimum mains voltage is 210V

I have suffered from the same problem, output of my secondary coil is about 26VAC, it is rectified and supplied to the regulator(7824-TO220). Although 400-500ma current is drawn, the regulator gets such hot that I cannot touch it. I have tried a little heat sink, it did not resolved the problem. I m using 12V little cooling fan now, it resolved the problem surprisingly. Nevertheless, this solution is not a good solution, I m looking for other linear 24V regulators whose current sourcing capability is better. Suggestions will be welcome. Thanks.

It’s not a problem of the current efficiency, just too much heat dissipated on the power circuit. The regulator must always be fixed to a suitable heat sink.

use a good quality LM7824 made by ST electronics
also connect a 0.1 uF capacitor both at input and output pins of 7824 to ensure stability
you may increase the size of the heat sink

Is there any way to reduce the input voltage before the regulator?

You can use a pre-regulator in the input before the 7824.
For example an LM317 but the resulting heat will be the same as before but shared in two devices, you can only lower it using a switching regulator either as stand alone either as a pre-regulator.

For example check the MC34063

MC34063A design tool
MC34063 DC-DC calculator

Alex

Yes, I'm using 1000uF/50 v after the bridge
The maximum mains voltage is 240V
The minimum mains voltage is 210V

Here are the results I got:

C=1680uF to ensure 26V at the regulator input
Pdis of LM7824 = 3.3 W

Pdis of LM7824 = 4.5 W

Pdis of LM7824 = 6.9 W

Maximum Operating Temperature = 125 deg
Junction-Case Thermal Resistance = 5 deg/W
Estimated max air temperature = 65 deg
Required Junction-Air Thermal Resistance < (125-65) / 6.9 = 8.7 deg/W
Heatsink Thermal Resistance < (8.7 – 5) = 3.7 deg/W

For economical reason, I use as a heatsink a piece cut from a square or rectangular aluminum bar (2x2cm or 2x4cm) of a suitable length. I fix the aluminum bar at the opposite side using a dummy TO220 package or just a heavy copper wire (as a high jumper).

Kerim

eng_ahmed22

### eng_ahmed22

Points: 2
The reason why it's getting hot is because it is a linear regulator. With 34v input and 24v output (load current = 0.8A), the remaining power (10v * 0.8A = 8W) is dissipated by the regulator as heat. That's a lot of heat dissipated. You need to use a good heatsink as has been said above. Alternately you could use a switching regulator which will be much more efficient and will not get so hot. Take a look at LM2575 and L4971.

http://www.datasheetcatalog.org/datasheet2/7/0ychsc59fe061qqohd1suqlit8ky.pdf
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/5598.pdf

Hope this helps.
Tahmid.

Reduce the input voltage to regulator. 27V should be sufficient. This will reduce the loss, otherwise you want to go far Off line switcher. They are as simple as linear regulator in usages but do not suffer from higher power loss issue like linear regulator.

Is this solution solve the problem?.
The current will divided between the two regulators.
i.e. the heat will be divided too.
and that will reduce the size of the heat sink

In theory, it works but how good the current balance between the two regulators couldn't be known for sure. In other words, one regulator might source more current than the other if there is a small difference between their regulated output voltage. And this balance may also differ with temperature. So one cannot be sure that the two regulator will dissipate equal powers. For instance I think you cannot split your load into two halves.

If it is just for one circuit you can test several regulators and choose the best pair. In any case it is good to mount them on the same heatsink.

As mister_rf has already pointed it out, adding equal resistors in series (one at each regulator output) helps in the current balance. Obvioulsy a higher value gives better sharing but the voltage drop also increases hence the output voltage will depend more on the load current. So a tradeoff should be made.

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No, the regulators will not work properly in parallel because the will not share the load equally. An even simpler way to parallel multiple voltage regulators together is to add very low below 0.1- 0.2 Ohm ballast/equalising resistors in series with the regulator outputs. Doing so should make the regulators share current equally and therefore operate together without problems. However that this approach to load balancing makes regulation a bit less accurate

I think the solution now is to use a suitable heat sink.
and next time I have to use a switched mode power supply instead of ordinary linear regulators.