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| Reducing Current Consumption in the PCB |
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| A.Ersoz:
Hi, I tested my custom design PCB with four SPST analog switches and four opamps and total current consumption seems on power supply around 80mA. Now, I added 28 analog switches and 28 opamps on PCB. Now, the current level is 700mA. The linear voltage regulator is heating too much. So, why the problem can be occur? Output of linear regulator is 150mA. Thanks |
| TimNJ:
Are you asking why your linear regulator is getting too hot? It seems you've already kind of answered that! |
| Kleinstein:
Usually the CMOS SPDT switches use essentially no current (e.g. < 100 µA). For OPs using 80 mA is quite a lot - there must be quite some load to them. With 7 times the circuit one can expect 7 times the power supply: 7 times 80 mA would be 560 mA so already quite a bit. 700 mA is little more, but not that much. Chance are the circuit is bad from the beginning - either a poor plan, a wrong inserted part or a broken one. So back to the simpler 4 OP circuit and bring the current down to some 10-20 mA or less. |
| iMo:
4kg of apples cost $80. How much (4+28)kg of apples cost? 32/4*80=$640. You can hardly get 640mA out of a 150mA voltage regulator. The temperature of a LINEAR voltage regulator package could be calculated as follows: T = Tamb + (Vinput-Voutput)*Ioutput*thermal_resistance_of_the_package Example: Tamb = 25degC Vinput=15V Voutput=12V Ioutput=150mA thermal_resistance_of_the_package=60degC/Watt T=25C+(15V-12V)*0.15A*60C/W=25C+27C=52C The thermal resistance of the package you will find in the datasheet, it depends on the copper area beneath the smd package, or the type of a heatsink attached. |
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