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xelos
Joined: 01 May 2004
Posts: 11
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 09 Oct 2004 21:45 Power dissipation |
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Hello, I have a question on voltage stabilization and power dissipation.
I'm working on a little device, which will be powered by alkaline battery 9V. For powering my PIC, I use LM7805. And my question is about heating of LM7805. I did not yet do my PCB and I want to know must I use some radiator to dissipate the power? The current consumption of my board is about 100mA (2LEDs, PIC, CompactFlash - in writing CF can conusumpt till 90mA). Or may be I can use a resistor, before stabilisator, which will dissipate the power?
Sorry for my English.
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Borber
Joined: 01 Jan 1970
Posts: 1485
Helped: 111
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| 09 Oct 2004 22:55 Re: Power dissipation |
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| Dissipation on your LM7805 at battery voltage 9V and current .1A will be .4W. So there is no need for radiator. Maximum allowed dissipation with no heatsink and TO220 case at ambient temperature 50 deg. Celsius is about 1W.
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djalli
Joined: 10 Nov 2001
Posts: 887
Helped: 15
Location: 1600 Pennsylvania Avenue, Washington DC 20500
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| 10 Oct 2004 0:08 Power dissipation |
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| 50 deg. Celsius? Where do you live ambient temperature is 50 degree celcius?
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Borber
Joined: 01 Jan 1970
Posts: 1485
Helped: 111
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| 10 Oct 2004 0:39 Re: Power dissipation |
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It make no sense to collect points like this. You must know that allowed power dissipation of a device derates with ambient temperature. 50 deg. C is an example temperature.
In a car on the sun temperatures can reach 50 deg.
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nicleo
Joined: 06 Sep 2004
Posts: 717
Helped: 60
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| 10 Oct 2004 6:11 Power dissipation |
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| If the package of your LM7805 is TO-220 (or it's LM7805CT, 'T' package can carry higher amperes), it should be safe to operate at 0.4W. If you feel not comfortable with the heat, then you can always attach a small heatsink to the regulator.
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xelos
Joined: 01 May 2004
Posts: 11
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| 10 Oct 2004 8:48 Re: Power dissipation |
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thanks everybody, I'll make my circuit without heatsink.
But I saw somewhere, that I can use a resistor before LM7805 to decrease voltage on stabilisator, so it will decrease the power? What do you think about it?
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nicleo
Joined: 06 Sep 2004
Posts: 717
Helped: 60
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| 10 Oct 2004 10:36 Power dissipation |
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LM7805 is a linear voltage regulator. The heat dissipation of this regulator is proportional to (Vout-Vin) x Iout. So, for your application (9V battery, 5V output, 100mA), as mentioned by Borber, the power dissipation is 0.4W.
You can reduce the heat generated by LM7805 if you put a resistor in (front) series with the regulator. Due to voltage drops across the resistor, the 'Vin' to the regulator will be lower than 9V. Say, the voltage drops across resistor is 1.5V, so the 'Vin' to regulator will be 7.5V, and the power dissipation will be (7.5-5.0)*0.100 = 0.25W.
For LM7805 (frm Fairchild Semiconductors), the dropout voltage (Vdo) required for proper operation is 2V.
Dropout Voltage (Vdo), at Iout = 1A, Tj=+25C, =2V
For safety margin, we choose a resistor value of 15R so that when your circuit draws 100mA, the voltage drops across resistor is 1.5V and the 'Vin' to the regulator is 7.5V. In this case, the power dissipated by resistor will be 15*0.1*0.1=0.15W. So, instead of using a heatsink, the resistor actually takes some heat from the regulator.
If you choose Low Dropout (LDO) type of regulator, e.g. LM2940 (Vdo = 0.5V), you can 'shift' more power to resistor. Again, for safety margin, we choose 'Vin' to be 6V. So, R series should be 30R. The power dissipation from LM2904 is 1.0*0.1 =0.1W only! However, the resistor should be able to withstand for 30*0.1*0.1=0.3W (0.5W resistor is quite common). Another advantage to use LDO is you might opt for 6V battery instead of 9V. If you choose 6V, then you no need the series resistor anymore.
So, series resistor will not reduce the overall power consumption of the circuit. It just shifts some heat from regulator to itself. I do not know how hot your regulator is now. But it your circuit is not exposured, so just let it be that without series resistor. It won't burn as it can withstand 1W as claimed by Borber.
Last edited by nicleo on 10 Oct 2004 11:08; edited 1 time in total |
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Borber
Joined: 01 Jan 1970
Posts: 1485
Helped: 111
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| 10 Oct 2004 11:04 Re: Power dissipation |
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| By the way I suggest to use simple stepdown switcher instead linear regulator. It will prolong your battery life.
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nicleo
Joined: 06 Sep 2004
Posts: 717
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| 10 Oct 2004 11:10 Re: Power dissipation |
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| Borber wrote: |
| By the way I suggest to use simple stepdown switcher instead linear regulator. It will prolong your battery life. |
Interesting ...
Would you please advise how a switcher could prolong the battery life?
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Borber
Joined: 01 Jan 1970
Posts: 1485
Helped: 111
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| 10 Oct 2004 12:03 Re: Power dissipation |
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| Actually I meant DC-DC stepdown conwerter. Circuits like this have high efficiency as 80% or more. Linear regulator in this case waste as much as .4W. Power consumption with linear regulator is 9V*.1A=.9W. Using MC34063 for example, power consumption will be about .6W.
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nicleo
Joined: 06 Sep 2004
Posts: 717
Helped: 60
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| 10 Oct 2004 13:45 Power dissipation |
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Initially, I guess the 'battery life' in your message is the number of charge-recharge cycle. How come I could relate the 'battery' to re-chargable battery, a bit off-topic, sorry. 
In the case of linear regulator, 7805, the efficiency only 55%. For a switching regulator with 80% efficiency, the current drawn from 9V battery will be 70mA only. So, the battery-powered portable device with the switching regulator can last longer. Latest switching regulator can push the efficiency higher (up to 90% or more).
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cetc1525
Joined: 08 Oct 2004
Posts: 177
Helped: 3
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| 10 Oct 2004 14:48 Re: Power dissipation |
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| if you have no more area to locate the device,i suggest you to choose linear regulator.otherwise ,you can use switch regulator
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