Analysis of the Principle of Power Switching Circuit

xxhhzz

Analysis of the Principle of Power Switching Circuit [Copy link]

 

Analysis of the principle of power switching circuit:

1 input power is 12V power adapter

2-way input power supply is 12V lithium battery

Requirements: The power adapter is the main power source, and the lithium battery is the backup power source. Please analyze the circuit principle and correct it. Thank you

maychang

It is speculated that the purpose of this circuit is: (1) When the power adapter loses power, the gate of the P-channel MOS tube is pulled down through Q7 and Q8, and the MOS tube is turned on to supply power to VDD_12V_IN. At the same time, when both lines have power, it can be controlled by STM32_POWER.

But in this circuit, the power adapter (1) does not drop from 12V to zero immediately as a switch does. It is likely that the voltage drops gradually, and this process may be relatively slow (if a large capacitor is connected in parallel to the output of the power adapter). According to this diagram, when the voltage of (1) drops to only 8V or even 6V, Q7 will not turn off, and of course Q8 will not turn on, and the MOS tube will not turn on either. Does the load allow such a low supply voltage?

xxhhzz

maychang published on 2020-4-10 17:27 It is speculated that the purpose of this circuit is: (1) When the power adapter loses power, the gate of the P-channel MOS tube is pulled low through Q7 and Q8, and the MOS tube is turned on to supply power to VDD_12V_IN, ...

The back-end load voltage cannot be lower than 9V, otherwise there will be insufficient power supply. How should it be modified and improved?

maychang

xxhhzz posted on 2020-4-10 17:30 The rear-end load voltage cannot be lower than 9V, otherwise there will be insufficient power supply. How should it be modified and improved?

If the STM32_POWER control function can be removed, then it is enough to keep D15 and D16. Let the adapter output voltage be slightly higher than the battery voltage. When the adapter has power, it will be powered by the adapter. When the adapter output voltage is insufficient, the battery will automatically take over. This is a seamless connection.

maychang

xxhhzz posted on 2020-4-10 17:30 The rear-end load voltage cannot be lower than 9V, otherwise there will be insufficient power supply. How should it be modified and improved?

If the STM32_POWER control function must be retained, a voltage comparator can be used. The voltage at (1) is divided by resistors and compared with a reference voltage. If the voltage at (1) is lower than a certain value, the voltage comparator outputs a low level, turning on the MOS tube.

In the current circuit, the voltage at (1) is actually divided by resistors and compared with the gate on voltage of Q7. However, the gate on voltage of Q7 has a large dispersion and cannot be controlled.

If Q7 is changed into a bipolar NPN transistor, a voltage regulator is connected in series with a resistor at (1), it is easier to control the voltage drop at (1) to turn on the P-channel MOS tube.

maychang

xxhhzz posted on 2020-4-10 17:30 The rear-end load voltage cannot be lower than 9V, otherwise there will be insufficient power supply. How should it be modified and improved?

The original circuit has an STM32_POWER control terminal, which is probably an MCU output pin. The voltage at (1) is 12V. If it is connected to this pin through a 3.3k resistor, the pin may not be able to withstand such a high voltage.

maychang

xxhhzz posted on 2020-4-10 17:30 The rear-end load voltage cannot be lower than 9V, otherwise there will be insufficient power supply. How should it be modified and improved?

If Q7 uses an NPN bipolar transistor, then Q8 is better to also use an NPN low-power bipolar transistor to reduce the number of component types.

xxhhzz

maychang posted on 2020-4-10 18:21 If Q7 uses NPN bipolar transistors, then Q8 is better to use NPN low-power bipolar transistors to reduce the number of components.

Now I want to design a dual-input power input (one power supply is 9V, the other power supply is 12V). By default, 9V is used for power supply, and 12V is used for power supply when 9V is not available. However, the 9V power supply can control the 12V power supply through the MOS tube, so that when both power supplies exist at the same time, 9V is also used for power supply. How should this control circuit be designed?

xxhhzz

xxhhzz Posted on 2020-4-11 08:37 Now I want to design a dual-input power input (one power supply is 9V, the other power supply is 12V). By default, it is powered by 9V. When there is no 9V, it is powered by 12V. But I want to use...

Now I want to use a P-channel MOS tube, as shown below

maychang

xxhhzz Posted on 2020-4-11 08:37 Now I want to design a dual-input power input (one power supply is 9V, the other power supply is 12V). By default, it is powered by 9V. When there is no 9V, it is powered by 12V. But I want to use...

Then use the first circuit, with 9V power supply at (1) and 12V power supply at (2). However, the two low-power MOS tubes Q7 and Q8 are replaced with low-power bipolar NPN transistors.

The base of Q7 is connected in series with a voltage regulator and a resistor to 9V (it is better to connect a resistor to the ground in parallel to the base of Q7 to discharge some current). The rated voltage of the voltage regulator should be roughly equal to the operating voltage of the 9V power supply. The collectors of Q7 and Q8 use 12V as the power supply.

xxhhzz

maychang posted on 2020-4-11 10:55 Then use the circuit in the first post, with 9V power supply at (1) and 12V power supply at (2). But the two low-power MOS tubes Q7 and Q8 are replaced with low-power bipolar NPN transistors...

This circuit is a bit troublesome. Will the MOS tube circuit you posted work?

maychang

xxhhzz posted on 2020-4-11 11:12 This circuit is a bit troublesome. Will the MOS tube circuit you posted work?

I have already said on the 5th floor: "The current circuit actually compares the voltage at (1) with the gate on voltage of Q7 after being divided by resistors. However, the dispersion of the gate on voltage of Q7 is quite large and cannot be controlled."

In fact, using two NPN low-power tubes only adds one voltage regulator tube compared to the first circuit.

xxhhzz

maychang posted on 2020-4-11 11:23 I have already said on the 5th floor: "The current circuit is actually the voltage at (1) divided by a resistor and compared with the gate conduction voltage of Q7. But the gate of Q7...

I know, but this circuit is a bit complicated and the cost is high. Now I want to use a MOS to realize this function. I don't know if it is possible. The 12V below is a control signal

maychang

xxhhzz posted on 2020-4-11 11:35 I know, but this circuit is a bit complicated and the cost is high. Now I want to use a MOS to realize this function. I don’t know if it’s possible. The 12V below is a control...

What you want is: dual input power input (one power supply is 9V, the other power supply is 12V), by default, 9V power supply, 12V power supply when there is no 9V, but the 9V power supply can control the 12V power supply through the MOS tube, so that when both power supplies exist at the same time, it is also 9V power supply

I don't see a 12V supply in this picture.

xxhhzz

maychang posted on 2020-4-11 11:41 What you want is: dual input power input (one power supply 9V, one power supply 12V), 9V power supply by default, 12V power supply when no 9V, but if you want to use 9V...

maychang

xxhhzz 发表于 2020-4-11 13:38

This circuit cannot realize "dual input power supply input (one power supply is 9V, the other power supply is 12V). By default, it is powered by 9V. When there is no 9V, it is powered by 12V. However, the 9V power supply can control the 12V power supply through the MOS tube, so that when both power supplies exist at the same time, it is also powered by 9V."

maychang

xxhhzz 发表于 2020-4-11 13:38

When there is 12V voltage at +12V_POWRE in this circuit, the 12V path supplies power to the load. The 9V path is a "backflow" of current, that is, the current flows from the drain of the P-channel MOS tube, through the diode inside the MOS tube (not shown in the figure) and flows into the 9V power supply, even if the MOS tube is turned off.

xxhhzz

maychang published on 2020-4-11 14:15 When there is 12V voltage at +12V_POWRE in this circuit, the 12V path supplies power to the load. The 9V path is the current "backflow", that is, the current...

The desired function is that when both power supplies are present, the default power supply is 9V, and when there is no 9V, the default power supply is 12V

Can this circuit be modified to achieve this function?

maychang

xxhhzz Posted on 2020-4-11 14:21 The desired function is that when both power supplies are present, the default power supply is 9V, and when there is no 9V, the power supply is 12V. Can this function be modified based on this circuit...

"Can this circuit be modified to achieve this function?"

cannot.

People go to high places, and water flows to low places.

xxhhzz

maychang posted on 2020-4-11 14:29 "Can this circuit be modified to achieve this function?" No. People go to high places, and water flows to low places.

I just checked the circuit and it doesn't work

"The desired function is that when both power sources are present, the default power supply is 9V, and when there is no 9V, the default power supply is 12V."

If you want to realize this function, how should you make the circuit? Can you give me some ideas?