A linear three-terminal voltage regulator current expansion circuit

Relevant key device information applied in this article: LM7805 is a linear three-terminal voltage regulator current expansion circuit. This circuit is a very common linear three-terminal voltage regulator current expansion circuit. 1. Disadvantages of the power supply 1.1 This power supply is a linear stabilizer Voltage circuits all have their own unique internal power losses, and all voltage drops are converted into heat losses, resulting in low efficiency. Therefore, special attention should be paid to heat dissipation issues. 1.2 Since the core component 7805 does not work at a high speed, the input voltage Or the response to sudden changes in load current is slow. 1.3 This circuit does not have a power protection circuit. The 7805 itself has overcurrent and temperature protection, but the current expansion transistor TIP32C is not protected, so there is a big shortcoming. If the 7805 is in the protection state , the output of the circuit will be Vin-Vce. If the output of the circuit exceeds the expected value, special attention should be paid to this. 2. Advantages of the power supply. 2.1 The circuit is simple and stable. It is easy to debug (almost no debugging required). 2.2 The price is cheap and suitable for cost control. Products with demanding requirements. 2.3 There are almost no components that generate high-frequency or low-frequency radiation signals in the circuit, the operating frequency is low, and EMI is easy to control. 3. Circuit working principle. Io = Ioxx + Ic. Ioxx = IREG – IQ (IQ is Quiescent operating current of 7805, usually 4-8mA) IREG = IR + Ib = IR + Ic/β (β is the current amplification factor of TIP32C) IR = VBE/R1 (VBE is the base conduction voltage of TIP32) So Ioxx = IREG – IQ = IR + Ib – IQ = VBE/R1 + IC/β- IQ Since IQ is very small, it can be omitted, then: Ioxx = VBE/R1 + IC/β Check the TIP32C manual, VBE = 1.2V, where β can be 10 Ioxx = 1.2/R + Ic/β = 1.2/22 + Ic/10 = 0.0545 + Ic/10 (here we take 22 OHM in the main map) Ic = 10 * (Ioxx – 0.0545) Assume Ioxx = 100mA , Ic = 10 * ( 100 - 0.0545 * 1000 ) = 455(mA) then Io = Ioxx + Ic = 100 + 455 = 555 mA. Assume Ioxx = 200A, Ic = 10 * ( 200 - 0.0545 * 1000 ) = 1955mA Io = Ioxx + Ic = 200 + 1955 = 2155mA From the above two examples, it can be seen that the output current has been greatly improved. The above calculation has been described in many posts, and it can be deduced carefully. 3.2 The size of the resistor R The size has a great relationship with adjusting the current passing through 7805. It can be seen by taking different values ​​and bringing them into the above formula. The larger R is, the smaller the current flowing through 7805 will be when the same current is output, and vice versa. Usually in such a circuit, a heat sink is added to the current expansion transistor TIP32, but there is no need for the 7805. However, the value of R cannot be too large. The condition is: R < VBE / (IREG – IB). 3.3 The input terminal of the 7805 in the circuit The value of the capacitor is a mistake. Friends have analyzed it before. It mainly causes surges. The output is much greater than 5V at the moment of power-on, causing damage to subsequent circuits. In actual use, in order to suppress the self-excitation of the 7805 Oscillation, this capacitor is usually 0.33uF (most common specs recommend this parameter)