A transformer can only convert alternating current. How can I convert direct current?

How to handle 12 Volt to 5V Well ？ It can be realized through transformer , But the transformer can only convert alternating current , How do you convert DC power ？

I once naively thought it was such a way to reduce blood pressure , For example, the load is 5 o , Then you have to get 5 Volt drop , According to the series partial pressure principle , It needs to be connected in series 7 Ohm resistance , The load will get a voltage drop of five volts , This method can really be implemented 12 V to v , But what if the load resistance changes ？

When it becomes 10 Ou Shi , At this time, the voltage on the load becomes 7V, To ensure that the voltage applied to both ends of the load is 5 v , We need to take R1 Replace with 14 Ohm resistance , So we can get five volts , But the problem is that the resistance of the load changes all the time , How can we ensure that the voltage at both ends of the load is maintained at when the load resistance changes 5V Well ？

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The method is to adjust all the time R1 Value , This is a little difficult for people , Because we have to measure the voltage of the load , One side to adjust R1 Value , If the work is left to manual work , Obviously, it can't be done well , In response to this question , Intelligent man invented an intelligence R1.

intelligence R1 It's actually a bunch of circuits , It can help us measure the output voltage Vout, Then adjust the resistance for yourself , Keep the output at 5 v , Its core is the in-phase amplifier , In the mold , Amplifiers are very difficult to learn , But it is also very simple if it is only applied .

In this circuit , We just need to know when VS Greater than VF When , Its output voltage decreases , When VS Less than VF When , Its output voltage increases .

3.3V Is the reference voltage of the amplifier ,D It's a regulator , Even if the input voltage changes , It can also stabilize the voltage at 3.3 v , So the reference voltage of the amplifier VF Namely 3.3 v .

This is the sampling voltage , It is related to the output voltage , When Vout At five volts , Amplifier input VS be equal to 3.3 v , If Vout Greater than five , be responsible for VS Greater than 3.3 v , When Vout be equal to 5V When , The rest 7V Applied to the triode , At this time, the triode works in the linear amplification region .

It doesn't matter if you don't know what a linear amplification region is , We only need to know that when the output voltage of the amplifier increases , The output current of the triode also increases , When the output voltage of the amplifier decreases , The output current of the triode is also reduced , Its voltage stabilizing principle is as follows .

If the input voltage or load change is Vout Greater than 5 v , be VS Greater than VF, At this time, the output of the amplifier UB smaller , At the same time, the output current of the triode becomes smaller , This causes the load voltage to drop again 5 v , Again , When Vout Less than 5 When you fall ,VS Less than VF, At this time, the output of the amplifier UB Bigger , This causes the load voltage to rise again 5V, This is the principle of voltage stabilization .

Simply put, it is always detecting the output voltage , And compare the output voltage with the reference voltage , Then the output voltage is controlled by triode .

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This stack of circuits is essentially the above R1, But this resistor is smart , It can adjust its own resistance , The reason why this amplifier is the core , Because sampling control is done by it , It is equivalent to the circuit CPU, From the schematic view , It is a circuit symbol .

But in fact, it is also a chip , Because it has many transistors , Resistance and capacitance components constitute such a thing , Very complicated , Even if we don't know how it works , You can also use it easily , Not just amplifiers , This whole circuit has long been designed as a chip , So we can use this kind of chip directly , Without knowing how it works .

Using the chip can also make the circuit more simple and stable ,, This circuit constitutes a linear regulated power supply , Its advantage is that the output ripple is small , Less peripheral electronic components are required , The disadvantage is low efficiency , such as 12V drop 5V, Then we should put the other 7V Sacrifice in the form of heat , If a large current is output , The chip will be obviously hot .

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