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### 7805 Pin Configuration and Voltage Regulator Circuit

Information Posted in Electronics
7805 is an easy to use voltage regulator IC which outputs 5 volts rated at 1A max. It takes an unregulated DC voltage input which can be fluctuating within its input limits and converts this fluctuating voltage input into a perfectly regulated 5 volts power output. For example, a 12 volt lead acid battery when fully charged gives out approximately 12.70 volts and when fully discharged, it gives out 10.50 volts. This difference can be even more under load or under charging state. If we use this battery as an input source for our 7805, output voltage will remain 5 regardless of that voltage difference of battery during charging and discharging phases.

7805 circuit diagram

I have attached an image of 7805 IC to describe its pin configuration. In addition to that I have also attached a small fully working voltage regulator circuit diagram with this tutorial. Please note that the two capacitors used in this circuit are not mandatory but they are good to maximize voltage regulation. The capacitor values I have used in this circuit are not written on stone, you can change them slightly.
7805 IC has a thermal shutdown feature to protect the IC in case of overheating so it should be safe to use 7805 without a heatsink plate for less than 200mA load. However should your load cross 200mA, you should consider using a heatsink plate. Heatsink should be large enough to bring 7805 heat to such a level that you can comfortably touch it. 7805 is a linear voltage regulator, so it is not very efficient and it has drop out voltage problem. It wastes a lot of energy in the form of heat. You can calculate the wasted energy with following formula. This formula will also help you to estimate the size of heatsink plate you will require to disperse amount of heat generated by 7805. (Input Voltage – 5) x Output Current
Suppose the input 15 volt and output current you require is .5 Amp by using above formula
(15 – 5) x 0.5
=10x0.5
=5W
5W energy is being wasted as heat and you will need a decent sized heatsink plate to disperse this heat to ease your 7805. On the other hand the energy you are actually using is only (5 x 0.5Amp) =2.5W. So you are going to waste twice energy then you are actually utilizing. On the other hand, if you lower your input voltage to 9V then at the same amount of load, only 2W { (9-5) x 0.5 } energy will be wasted as heat.
So the conclusion is, the higher the input voltage get, the less efficient your 7805 will be. You should try to stay slightly over 7.5V. However don’t get bellow 7.5V as your 7805 won’t give a regulated output if the input voltage get bellow 7.5V. If your input voltage are less than 7.5 like 6V then you should consider using a low dropout voltage regulator such as LM2940. Pin connections of LM2940 are also same like 7805.
You don’t need extra components to create a 5 volt regulated power supply with 7805. However it is a good idea to use one capacitor on input and one on output pins to make output voltage smooth but then again, they are not necessary to be used.
As per specs, 7805 input voltage should range between 7.5V to 35V but personally I haven’t tried more than 15V yet. Maximum output current of 7805 is 1A with a good sized heatsink plate. If you are planning to use more than one 7805 in parallel to get more current (above 1A) then you better try to put .47Ohm (15W) resistor on the output pin of each IC to cover the slight voltage difference as technically no two 7805 can provide exactly same output voltage. 15W resister is used because large resistors can withstand more heat however you can use any resistor as long as it is above 5W. Bellow is the schema in which two LM7805 are used in parallel.

7805 parallel use

Please note that LM7805 is not designed to be used in parallel that is because no two LM7805 can output same volts.One of them may output 4.99V while the other will output 5.01V. Due to higher voltage, the later will try to take more load than the first one. Also another issue is the higher voltage IC (5.01V) will try push the low voltage IC above it's maximum volts (4.99V). The lower voltage IC will sense the voltage higher than its upper limit and will shutdown putting all the load to higher voltage IC. It will then power-up again once the volt drop as the higher voltage IC wont be able to take all the load but still this can start an undesirable oscillation especially if the voltage difference is significant.
Above circuit is actually an attempt to balance slight voltage difference between both regulator ICs using two resistors. As soon as a load is applied to this circuit, the load will push the volts after resistor to slightly less than 5 which will stop any oscillation caused by voltage difference and will also balance the load a bit more evenly. This way you may be able to extract slightly less than 2A current with the price of a very slight voltage drop bellow 5V in final output. Although i have never came across any issue while using this circuit but still i do not recommend you to use it to power-up expensive devices.

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