![]() ![]() ![]() So, we need to provide a means to dissipate the temperature from the regulator by using a suitable heat-sink.įrom the regulator datasheet (LM7805) we get the value of the junction to case thermal resistance (metal lead) for TO220 case and the maximum junction temperature as follows: Rjc = 1.7☌/W Jtmax=150☌ The Physical Thermal Model ![]() This also implies that the junction temperature of the regulator will dangerously rise up and the regulator will likely be damaged. The curve is shown below: Power Dissipation Characteristics for LM7805.Īt the maximum ambient temperature of 50☌ we can see that average power that regulator can dissipate without a heat-sink is well below the expected projected value of 6.13W. This gives us 6.32Vdc and 0.97Adc in the load which, for an output of 5Vdc and 1A nominal, implies that U1 will dissipate a power of P(W)=6,32V*0.97A=6.13Wįrom the datasheet we can get the maximum power dissipation at the needed ambient temperature. Proteus can help with this by using a DC Voltmeter and a DC Ammeter connected as shown in the picture above. ![]() To do this we will measure the voltage across the regulator itself between the input and output pins and the output current in the Load. Our first step is to figure out the power dissipated by the regulator U1. Linear Regulator Proteus Example DownloadĪ demo copy of Proteus is required to open the. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |