Help analyze DCDC circuit

kal9623287

Help analyze DCDC circuit [Copy link]

 

The following is a DC24V DC input circuit. What are the functions of the circuits in the red boxes in this circuit?

Is it correct that U1 is for reverse connection protection?

maychang

【Help analyze DCDC circuit】

It doesn't look like a DC-DC conversion circuit. It looks like a protection circuit for reverse connection, overcurrent, etc. Especially when you see that the Vout terminal is connected to the base of Q2 through R1D2, etc.

mingplus

1. Q1: OCP

2. Q3: Latch, if OCP, Q1 ON =》Q2 ON =》Q3 ON => U1&U2 OFF. Once latched, OCP is released, U1&U2 will not conduct unless Vin is unplugged and plugged in again.

3. Q4: turns on U1 & U2.

4. U1: When U1&U2 are OFF, prevent Vout from flowing back into Vin.

However, when Q3 is turned on, Q3 will burn out, so a resistor should be added to the Q2 collector.

maychang

In this figure, D6 is labeled DFLZ33 and its breakdown voltage is estimated to be 33V.

If the breakdown voltage of D6 is 33V, then after power is applied to J1, the diode in U1 is turned on, the upper end of R10 is 24V, R10 and R11 divide the voltage, and 10.3V is divided on R10. This voltage is enough to make U2 and U1 saturated and turned on (the voltage drop of the MOS tube is smaller than the diode in the body), and the Vout end is powered. The load after Vout is powered and works normally.

maychang

If the voltage at J1 rises due to some fault and exceeds 33V, then as long as the current flowing through R12 exceeds 7uA, Q4 will be turned on. Since Q4 is connected in parallel with R10, the voltage across R10 will decrease, which will reduce the voltage between the gate and source of U2. After the voltage between the gate and source of U2 decreases to the point where it cannot maintain the saturated conduction of U2, U2 tends to turn off, so that the voltage at the Vout terminal no longer continues to rise. If we talk about the role, then the role of Q4 and its auxiliary components is to turn off U2 when the input voltage is too high to protect the load circuit from damage.

maychang

There is a 0.12 ohm resistor between the emitter and base of Q1. If the load current is too large, exceeding 0.7 volts/0.12 ohms = 5.8 amps, then Q1 will turn on. After Q1 turns on, its collector current is not zero, and this current will produce a certain voltage drop on R6. If the voltage drop on R6 exceeds the conduction voltage of the emitter junction of Q2, Q2 will also turn on. The conduction of Q2 will also turn on Q3, and the conduction of Q3 will increase the base potential of Q2, maintaining the conduction of Q2. Even if the current through the 0.12 ohm resistor disappears, Q2Q3 will still remain on.

Q3 is turned on, the collector potential rises, and U2 is turned off through D4.

maychang

If we talk about the function, then the 0.12 ohm resistor is used to sample the load current. When the load current exceeds a certain value (about 5.8A), U2 will be shut down to protect the power supply. The function of Q1Q2Q3 can be said to be overcurrent protection.

maychang

The Q2Q3 circuit is a bistable circuit. After power-on, when there is no overcurrent fault, Q2Q3 is in the off state. Once an overcurrent occurs, Q2Q3 flips to another stable state where both Q2Q3 are on.

Therefore, the overvoltage protection formed by Q4 is different from the overcurrent protection formed by Q1Q2Q3. The overvoltage protection of Q4 will automatically recover after the overvoltage disappears, but Q1Q2Q3 cannot automatically recover after the overcurrent disappears because of the bistable state. U2 will remain off unless the power at J1 is turned off and then on again.

maychang

Q2Q3 and its associated circuits are in the off state after J1 is powered on. This is achieved by the voltage across capacitors C4 and C5 being zero before power-on. Because the voltage across C4C5 cannot change suddenly, Q2Q3 is in the off state at the moment of power-on. After the power-on process is completed, Q2Q3 is still in the off state until overcurrent occurs.

maychang

As for the functions of R1, D2, C1, and D3 in the circuit, it is not clear yet.

maychang

The so-called DC-DC circuit refers to a device that converts one DC voltage into another DC voltage, which can be either step-up or step-down. This circuit does not have the function of stepping up or stepping down, so it is not a DC-DC circuit. As mentioned above, this circuit has the functions of reverse connection protection, overvoltage protection, and overcurrent protection.

qwqwqw2088

U1 anti-reverse is no problem

R2, Q1 part is over-current detection

chunyang

The picture you drew is not a DCDC circuit. The first red box from the left is for reverse connection protection; the second is for overcurrent protection; the third is part of the self-locking circuit, that is, when overcurrent occurs, the circuit is self-locked to disable output; the fourth is for overvoltage protection.

kal9623287

mingplus posted on 2022-12-22 10:56 1. Q1: OCP 2. Q3: Latch, if OCP, Q1 ON =》Q2 ON =》Q3 ON => U1&U2 OFF. Once latched, release O ...

When Q3 is ON, Q3 should burn out

Why does it burn?