I'm really not comfortable with things heating up, including this LM7805 integrated circuit wired below.
Why does this mighty LM7805 heat up?
Thank you for the help!
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4\$\begingroup\$ also bad, but unrelated here: the LM7805 needs decoupling caps. You must not use it without. \$\endgroup\$– Marcus MüllerCommented Feb 25, 2022 at 10:15
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1\$\begingroup\$ not to mention that cable twisting is not an appropriate way of connecting wire to an IC, or a motor. \$\endgroup\$– Marcus MüllerCommented Feb 25, 2022 at 10:16
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\$\begingroup\$ @MarcusMüller I thought using LM7805 is a straightforward process. Boy I was wrong.... \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:20
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2\$\begingroup\$ no, you'd simply need to use something else than a linear regulator when you're dropping a non-negligible amount of the supply voltage; 1V difference between in- and output might not be enough for the LM7805 to work properly. Also, using a 5V regulator with a 3V motor was ... questionable, to begin with. \$\endgroup\$– Marcus MüllerCommented Feb 25, 2022 at 10:48
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1\$\begingroup\$ 9V battery supplying 5V regulator supplying 3V motor. This seems logical to you! And you are worrying about the regulator overheating. I'm surprised the 3V motor is not smoking! Consider a 3.3V switcher driven by your 9V battery. SPCLVR - Society for Prevention of Cruelity to Linear Voltage Regulators. \$\endgroup\$– StainlessSteelRatCommented Feb 25, 2022 at 16:51
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3 Answers
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You are feeding the 7805 from a 9 volt battery, so there will be a 4 volt voltage drop over it. If your motor draws 500 mA, it results in power dissipation of:
4 V * 0.5 A = 2 W
In the 7805. Depending on the specific mfg PN the device might have a junction to ambient thermal resistance of 50 C/W. That would result the 7805 heating up
2 W * 50 C/W = 100 C
above the ambient temperature.
Adding a heat sink might drop the thermal resistance for example to 25 C/W, which would result in heating
2 W * 25 C/W = 50 C
You need to measure or check the datasheet to know how much current the motor is drawing and get a heat sink large enough to dissipate it.
Another option is to change to a switching mode regulator. Some of them are pin compatible to 7805 (for example CUI V7805-1500R). They will dissipate substantially less power to heat so your battery will last longer too.
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\$\begingroup\$ ouch, 2 watts sounds hot, especially how small the LM7805 is.... \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:33
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\$\begingroup\$ Thanks for the assistance, comrade! I think I must weaken the power of 9 V battery and turn it into 6 V so the LM7805 will be usable. Thanks! \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:43
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\$\begingroup\$ You can also consider driving it directly from 3 x 1.5 V batteries or 4 x rechargable 1.2 V batteries. You might end up having problems with the switch lifetime, depending on how you do the power switching. \$\endgroup\$– RalphCommented Feb 25, 2022 at 10:49
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The DC motor is stalled due to the prop being prevented from rotating hence, the load on the regulator is going to be around 1 ohm (the resistance of a small DC motor that is stalled). That means if the 7805 could produce 5 volts it would also be feeding 5 amps to the load and, we know that isn't happening.
So, the output voltage will be much less than a volt and, the input voltage will be about 9 volts and, the current might be around 100 mA. That means about 1.8 watts of heat produced by the 7805.
Simple answer: circuit abuse.
answered Feb 25, 2022 at 10:14
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\$\begingroup\$ I wonder how much power this DC motor eats up. I knew it needs 3 V to make this DC turn but I wonder how much current it needed to complete turn. \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:35
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\$\begingroup\$ Data sheets and ammeters can give you those answers @Shobeh \$\endgroup\$– Andy akaCommented Feb 25, 2022 at 10:40
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\$\begingroup\$ Thanks for the assistance, comrade! I think I must weaken the power of 9 V battery and turn it into 6 V so the LM7805 will be usable. Thanks! \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:43
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1\$\begingroup\$ A 9 volt battery of the type in your picture is barely acceptable for DC motors and, a 6 volt supply to a 7805 is well-below the minimum needed to provide regulation. I urge you to use a more powerful battery arrangement like three AAs in series (4.5 volts) and have them directly connected to your motor. \$\endgroup\$– Andy akaCommented Feb 25, 2022 at 11:13
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\$\begingroup\$ If you look closely at the photograph, the motor is not stalled. The blades of the propeller are not in contact with the surface, and the battery isn't connected on one side. \$\endgroup\$ Commented Feb 25, 2022 at 12:52
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It might have to dissipate too much power or it may be unstable and waste energy into high frequency oscillations as it does not have the required bypass capacitors for stability.
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\$\begingroup\$ But why my boy LM7805 has to dissipate that much heat if the LM7805 just eat the power of the 9V battery? \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:13
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3\$\begingroup\$ Any power it has to eat will be converted to heat which will make the temperature rise. \$\endgroup\$– JustmeCommented Feb 25, 2022 at 10:16
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\$\begingroup\$ Thanks for the assistance, comrade! I think I must weaken the power of 9 V battery and turn it into 6 V so the LM7805 will be usable. Thanks! \$\endgroup\$– ShobehCommented Feb 25, 2022 at 10:43
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\$\begingroup\$ The recommended minimum input voltage is 7 to 8 volts, depending maybe on exact 7805 model. 6V is too little. \$\endgroup\$– JustmeCommented Feb 25, 2022 at 11:54