The problem you described is roughly attributed to ASM1117 not installed enough heat sink, ASM1117 working time is too long, and there is no timely heat dissipation, resulting in a short circuit inside the chip, thus dragging down the input voltage (programmer end).
Two reasons. Base current is too small to saturate conduction, workThe consumption exceeds the rating. The load current is too large and exceeds its capacity. Better check it with a multimeter test. This is the reason why the S8050 with 8V voltage regulation to 5V burns.
The emitter resistance may be different from the original, the smaller the resistance, the higher the output voltage, 9014 magnification may also be different from the original, the smaller the method multiple, the higher the output voltage.
In this way, first understand the cause of 7805 heat, because 7805 is a linear power supply chip, its input voltage Ui and output voltage 5V between the pressure difference U this part of the energy is solved by heat dissipationV power supply '>7805 Voltage regulator chip How to supply 5V power to the system
When the voltage is input, the 7805 will detect the input voltage and stabilize the output voltage at about 5V through an internal control circuit. In this way, the MCU can obtain a stable power supply voltage to ensure normal operation.
The output voltage of the regulated power supply can be adjusted by adjusting the resistance in the circuit. Under normal circumstances, the 7805 voltage regulator can maintain 5V output voltage in the input voltage range _10%~ 35%.
Power supply without capacitor, reset circuitYou need to add capacitance. Fallen leaves chasing the autumn wind does not move | Posted on 2012-04-02 Report | Comments 1 0 12V is a battery? If it's a battery don't use any filter capacitor. However, due to the large pressure drop on the 7805, the heat will be large, and it is recommended to use heat sinks.
If the output voltage is lower than the reference voltage, the feedback circuit increases the current flowing through the regulator circuit, thus increasing the output voltage. This adjustment allows the 7805 to adjust the input voltage to an output voltage close to 5V, and to maintain this voltage unchanged even if the input voltage or load current changes. V power supply has been converted to 5V by 7805, they are just common ground. 12V is equivalent to the input voltage, 5V is its output voltage. Therefore, after the MCU, of course, it is 5V to supply power to the MCU. The output voltage Uo can be improved to a certain extent, and the output voltage Uo is the sum of the output voltage of the 78XX regulator and the voltage regulator diode VC1. VD2 is the output protection diode, once the output voltage is lower than the VD1 voltage regulator value, VD2 is switched on, the output current is bypassed, protecting the 7805 voltage regulator output stage from damage. 1, LM780AMS111LD1117. The LM7805 is a classic 5V voltage regulator chip, which has the characteristics of low working current and high stability. The AMS1117 is a chip suitable for low power requirements and can operate at low input voltages; LD1117 is a voltage regulator designed specifically for automotive electronics applications. 2, you can use highPrecision three-terminal voltage regulator integrated circuit LM338 to solve. LM338- Circuit parameters Adjustable voltage range: 25V - 25V Output current: rated 5A, instantaneous maximum current 8A Voltage adjustment rate: 0.003 It is recommended to use the form of 9 volts transformer rectifier filter LM338 to achieve. 3, if this mobile power supply is used for gsm mobile station, 5a is very worried, or barely enough. According to your "possible 2a", it seems that it is really done for gsm mobile stations, good, even if it is powered by AC, 7805 will not work. We used to do it at allI didn't think about 7805. 4, the general use of three-end regulator, 7805, input voltage 5V-35V, output voltage 5v, small current with 78l05. LM1117-5 low voltage differential chip is also available. Also available W3520, depending on application and current. You can use a 5v regulator diode. 5, this is a positive output 5V DC voltage regulator power supply circuit. IC adopts integrated voltage regulator 7805, CC2 is the input and output filter capacitor, RL is the load resistance. When the output current is large, the 7805 should be equipped with heat dissipationBoard. The following diagram shows the application circuit for increasing the output voltage.