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Hello everyone, in this lesson, let's take a look at the power supply method of op amp plus MOS

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The core of this circuit is an operational amplifier plus a MOS tube

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The working principle of this circuit is a bit complicated, but its workflow is relatively simple

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First of all, there must be a power supply input,

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which is not only the power supply of the operational amplifier,

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but also the power supply of the D pole of the MOS tube.

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With the power supply, the operational amplifier will work to control the conduction and cut-off of the MOS tube

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After both the operational amplifier and the MOS tube start working, a power supply will be output

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This power supply is the power supply after step-down

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Let's see how this circuit actually works

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Let's first look at the working logic of the operational amplifier

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The operational amplifier is relatively simple to work, it has two input pins and one output pin

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One of these two input pins identifies + and the other identifies -

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We also call it the same input and reverse input

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When the voltage of the + pin is greater than that of the - pin, the output pin will output a high level

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This high level is generally the power supply of the operational amplifier.

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When the voltage of the + pin is lower than the - pin,

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the operational amplifier will directly ground the output pin,

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that is, it will become a low level

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After understanding the working logic of the operational amplifier, let's take a look at this circuit

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First of all, both the operational amplifier and the MOS tube must be powered

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The operational amplifier gets 12V power supply, and the MOS tube gets 1.5V power supply

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The positive pole of the operational amplifier is connected to a series voltage divider circuit

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Let's take a look at the voltage divided by this series voltage divider circuit.

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The calculated voltage should be 1.054V

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Let's round up and that's 1.05V

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Then the positive terminal of the op amp is 1.05V

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When the circuit has not started working, the VCCSA pin is not powered, that is, 0V

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At this time, the positive of the operational amplifier is greater than the negative,

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then it will output a high level

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This high level is 12V

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After 12V is given to the N-channel MOS tube, the MOS tube can be fully turned on

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After it is fully turned on, the 1.5V will be fully turned on

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After 1.5V is fully turned on, it will come back

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At this time, its negative pole becomes 1.5V

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At this time, the positive is smaller than the negative, and the operational amplifier will output a low level

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When the low level is output, the MOS tube is turned off, and the output voltage becomes 0V again.

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After the output voltage becomes 0V, the positive of the operational amplifier is greater than the negative

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Then at this time it will output high level again, and the MOS tube will be turned on again.

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Cycle like this

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So the output waveform of this circuit should be

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0V 1.5V 0V 1.5V 0V 1.5V

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But we want to output a direct current

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This waveform is obviously not a direct current and cannot be used as a stable power supply

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If we look closely at the circuit, we can see that there is a capacitor

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This capacitor is also the core of the op amp plus MOS circuit

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We didn't analyze this capacitor just now

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In the beginning its positive pole is 1.05V and its negative pole is 0V

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Then it outputs a high level to turn on the MOS tube

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After the 1.5V of the MOS tube is turned on, it will slowly charge the capacitor

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When the capacitor is charged to 1.05V, its positive pole is not greater than the negative pole

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At this time, it will output a low level, and the MOS tube will be cut off.

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After the MOS tube is cut off, the capacitor is discharged to the outside

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When the electricity in the capacitor is discharged a little, suppose it becomes 1.04V

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At this time, the positive pole is greater than the negative pole again,

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it will output a high level and continue to charge the capacitor

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In this way, the reciprocating cycle outputs a 1.05V DC

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This is how the op amp plus MOS power supply works