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Hello everyone, we found the input power supply of 5700XT in the last class,

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and then we will look at the next step according to the sequence

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According to the timing, the next step should be to generate a 5V power supply

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Let's look for a 5V power supply in the catalog

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As you can see, there is no page for 5V power supply in the catalog

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Then we can temporarily skip the 5V power supply

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When the 5V power supply is used in the later sequence, go back and find the 5V power supply

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If the 5V supply is skipped, then the next supply is 1.8V

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Let's look for 1.8V in the catalog

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You can see that 1.8V is very easy to find him on page 25

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Jump to page 25

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Observe the circuit after finding the 1.8V power supply

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It can be seen that this is a PWM circuit

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This is its PWM power supply chip

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Let's get to know the pins

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VCC, chip power supply pin

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CSP and CSN, this is a pair of current detection pins

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COMP is a feedback regulator, the feedback compensation pin

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FB, voltage detection voltage adjustment pin

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FBG is the ground of FB

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EN, enable signal

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This is a setting pin, it's grounded, don't worry about it

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Then there are 5 ground pins

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LG, down tube driver

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Phase pin

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High tube drive pin

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Bootstrap presser pin

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VCCP is a power supply pin

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PGOOD, power good signal

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Although this power supply chip has more pins, its workflow is also the most common PWM chip workflow.

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The first step is to have a chip power supply

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The second step is to have an enable signal

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With power supply and the enable signal, it will output power supply

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After the output power supply, it will come back to adjust the voltage through the FB pin

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When the voltage is stable, it will send a PG signal through the PG pin

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Then, the external conditions that this chip needs are the power supply in the first step

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and the turn-on signal in the second step.

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Let's look at the power supply first

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Its power supply comes from +12V_BUS_B

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Now with +12V_BUS, but without +12V_BUS_B

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Next, let's look for this power supply and see where it comes from

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After searching, it can be found that it is directly renamed from the gold finger 12V power supply 12V_BUS

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through an inductor, and will not be controlled by any

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So at this time, 12V_BUS has been generated

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Let's draw this power supply

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After the 12V_BUS_B is available, the power supply of the 1.8V power supply chip will be available

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Let's draw this step as well

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After the chip has power supply, let's look at the open signal

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It can be seen that the turn-on signal is controlled by two MOS tubes

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The turn-on signal must be high level, then the MOS tube will be cut off

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It is an N-channel MOS transistor

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The N-channel MOS transistor is turned on at high level and cut off at low level

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So its G pole should be low

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Its G pole must be low level, Q904B will be turned on, then its B level must be high level

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It should be noted that the EN signal of the 1.8V power supply chip is not connected to the pull-up resistor

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So this signal is not a 3.3V high level during normal operation

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In normal operation, as long as this EN signal is not grounded, it is normal

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In actual maintenance, the EN signal cannot be directly pulled up to high level

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If you want to create an EN signal, you can only remove this tube

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We just concluded that this 1.8V_EN should be high

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Let's see how it can be high

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It can be seen that it comes from a circuit like this

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1.8V_EN signal should be high level, then here is also high level

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The level below should also be high, let's look at the top first

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If it is high level here, then Q1000A will be cut off

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Q1000A is an NPN transistor

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The NPN transistor is high-level conduction and low-level cut-off

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Then it will be cut off, and its second pin will be low level

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If the two pins are low, the two transistors are all turned on.

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After the two transistors are all turned on, their second pins will be connected to the power interface

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When the power interface is plugged into the power line,

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this signal will be pulled down to the ground, that is, low level

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Then the second pin here will be low level

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If Q1001A and 1001B are to be turned on, their B-levels must be high

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Because they are all NPN transistors

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The second pin of Q1001A is connected to the 12V golden finger power supply, and the voltage is divided in series

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As long as there is 12V gold finger power supply, its second pin will be high level

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And the triode below is connected to a 12V series voltage divider

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It's just that this 12V is the 12V of the power interface

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Let's look at the following one

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The following part is also the same

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If it is high level here, then pin 5 must be grounded as low level

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If it wants to be low level, then both tubes must be turned on.

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The second pin of the above tube is connected to the 3V series voltage divider

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The tube below is connected to the series voltage divider of the 12V B power interface.

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If the 1.8V_EN signal is at a high level,

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it is actually necessary to check whether the 12V gold finger power supply,

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the 12V power supply interface, and the 3V power supply are normal.

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That is, to detect whether the input voltage is normal

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Let's draw this step briefly

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So far, the power supply and start signal of the 1.8V power supply chip are available.

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With the power supply and enable signal, the chip will work,

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and it will generate 1.8V power supply and a PG signal

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Well, so far, 1.8V has been generated normally