Everyone, the 12V in the picture is to power three 12V relays, 78M05 (TO-220 package). What is the general input voltage of 78M05? In addition, where is the 100R in the figure placed to limit the current instead of reducing the pressure?
Taina postato su October 14, 2020
Resistors are shunted in parallel, and connected in series will only step down.
Henna postato su October 14, 2020
What I want is to reduce the heat of 7805, which increases the input voltage fluctuation range. The total heat is the same and the efficiency is low.
Salomon postato su October 14, 2020
Although string resistors and string diodes reduce the heating of the 78M05, they greatly reduce the 78M05's voltage stabilization performance. 78M05 heating problem can be solved by adding heat sink. Or use two three-terminal regulator IC cascade to solve.
Esa postato su October 14, 2020
1. The first sentence of your question, 12V, relay, what is the relationship with 7805? I didn't understand
Second, you must have a voltage drop when adding resistance to the power supply loop, which is an inevitable problem
3. If the power supply voltage is too high, if it is to change the circuit, if it is to change as little as possible, (when the above three premises are met) the string diode may be stronger than the string resistance
4. If none of the above three points is the answer you want, then you can only use this circuit like this. Without knowing the range of your load current at the 5V terminal, it is impossible to determine how much resistance you need to string. The key to the problem is that in the case of the maximum load current, the 12V power supply removes the 5V of the load, removes the voltage drop of the series resistance, and what is the voltage drop of the IO terminal of 7805. This must be within the effective range of 7805. Of course, Is the minimum allowed value of 7805
