USB multi-output power supply --- circuit diagram 1 (Boost circuit)
Create a low-power power module with components that can be directly plugged into a computer's USB port. Not only pull-down but also boost can be achieved to get overall voltage source we want, and even positive and negative voltage sources. According to this idea, start with circuit diagram. Everyone is welcome to join discussion. If it can give you pleasure and allow you to gain more knowledge, I will share various DIY gadgets that I have made in future.
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Do It Yourself
Do it yourself exercise. Build a multi-output USB power supply from scratch.
Step-by-step instructions for designing, welding and manufacturing from scratch.
Typically, USB output of computer we use is about 5V. Different computers have different output currents (desktops, standard voltage laptops, and low power laptops). Before I start, let me explain that output current of a typical computer USB power supply is 0.5A, which depends on configuration of computer. In our project, maximum load capacity of one power supply should be 0.5A.
Step 1 Schematic Design
The first step is to develop a circuit diagram.
Make some more commonly used DC voltages. First increase USB_5V voltage to 12V with the Boost circuit and then lower it to 5V and 3.3V respectively. We also made 2 negative voltage power supplies on this board, using USB_5V to step down to -12V and then use -12V to step up to -5V.
5V_USB --> 12V, 0.5A
As shown in figure above, we chose domestic MT3540 chip as BOOST chip, and manufacturer also gave recommended circuit for this chip, frequency of this chip is f = 1.2 MHz. First of all, it needs to be boosted with BOOST circuit to convert 5V to 12V.
Voltage divider resistance calculation
Parameters for several components require further confirmation. Since output voltage Vout is 12V, resistance value of voltage feedback contact FB must match output voltage. Looking at datasheet, we know that Vfb = 1.2V.
According to resistor voltage divider formula and loop impedance, we can calculate the resistance of resistors R1 and R2.
Volt-second balance increase
Based on topology of Boost circuit, let's analyze how a choke works, as shown in the figure below:
The figure above shows topology of Boost circuit. The reason Boost can increase voltage is because of charging and discharging of inductor. As shown in figure above, when switch is turned on, Vin charges inductor at same time; when switch is turned off, inductor discharges at this time and it moves to Vin, which is equivalent to lifting inductor to power supply.
We know that reason why output voltage Vo can reach balance is because charging current of inductor = discharging current. Then we know relationship between inductor current and inductance:
When inductor is charging, Δt is equal to ton, when inductor is discharging, Δt is equal to toff.
Similarly, let's analyze voltage across inductor during ton and toff, respectively.
As you can see from picture above,
We have just analyzed duty cycle of Boost circuit, so what is average current IL of inductor? How to take inductance?
According to law of conservation of energy, we can know:
Because average inductor current is average of area of the current triangle,
The inductance calculated above is 0.5A when operating in BCM mode, if we want to deepen Boost circuit's operating mode, we need to multiply it by a factor of 0.2, which means that the load enters 100mA (0.5 A *0.2), BCM mode is entered.
Here we can choose a nominal inductance of 4.7 µH 1 A.
Scheme of the Boost circuit after completion
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