Linear three-terminal voltage regulator current expansion circuit diagram

The circuit is a very common linear three-terminal voltage regulator current expansion circuit . When we actually used it, we encountered some faults due to lack of thorough consideration or low-level errors.

1. Let’s first talk about the shortcomings of this power supply.

1.1 This power supply is a linear voltage regulator circuit, which has its own unique internal power loss. All voltage drops are converted into heat losses, and the efficiency is low. Therefore, special attention should be paid to heat dissipation issues.

1.2 Since the working speed of the core component 7805 is not very high, the response to sudden changes in input voltage or 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. The circuit output exceeds the expected value, so special attention should be paid to this point.

2. Advantages of power supply

2.1 The circuit is simple, stable and easy to debug (almost no debugging required).

2.2 The price is cheap and suitable for products with strict cost 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 and other aspects are easy to control.

3. Circuit working principle

Io = Ioxx + Ic.

Ioxx = IREG – IQ (IQ is the quiescent operating current of the 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, its β can be 10

Ioxx = 1.2/R + Ic/β = 1.2/22 + Ic/10 = 0.0545 + Ic/10 (take the 22 OHM in the main map here)

Ic = 10 * (Ioxx – 0.0545 )

Assuming Ioxx = 100mA, Ic = 10 * ( 100 - 0.0545 * 1000 ) = 455(mA)

Then Io = Ioxx + Ic = 100 + 455 = 555 mA.

Assume again that Ioxx = 200A, Ic = 10 * ( 200 – 0.0545 * 1000 ) = 1955mA

Io = Ioxx + Ic = 200 + 1955 = 2155mA

It can be seen from the above two examples that the output current is greatly improved.

3.2 Size of resistor R

The size of R has a great relationship with adjusting the current passing through the 7805. It can be seen by taking different values ​​and adding them to the above formula. The larger the R, the smaller the current flowing through the 7805 when outputting the same current, and vice versa. However. Usually in such a circuit, a heat sink is required for 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 value of the capacitor at the 7805 input end in the circuit 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-oscillation of 7805, this capacitor is usually 0.33uF (most common specs recommend this parameter)

This circuit is used in a certain commercial equipment. Except that the capacitance parameter is not 100uF, the real circuit is the same as the parameters in the main post. Thousands of units of the product have been put into the market, which proves that it can be used. The reason for this discussion is that When a colleague used a new model of product, he changed the capacitor parameters, causing surge problems and burning many peripherals, so we analyzed it again.