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Charge pump boost principle - this article will give you a simple understanding
2022-07-05 08:09:00 【Learning notes of hardware Xiaobai】
Charge pump , A noun that sounds very tall , Its circuit design is actually applied to many fields . It can not only achieve the purpose of boosting , It can also be used to generate negative pressure . According to its principle , Let's understand it by ourselves , There is no need for Xiaobian to copy and paste online articles !
Take the field in which Xiaobian is engaged , It is mainly integrated in the charging chip .
Design of charge pump circuit , It's not hard to say , Because if you want to design , It can be designed simply . Let's make it simple , It can't be simple , Because Xiaobian has seen the use of very complex charge pumps , But anyway , The truth is the same after all .
So , Based on the situation encountered at work , Today, I still want to talk about the circuit design that uses the charge pump to boost the voltage .
The diagram shows the design of the simplest boost circuit .
XFG1 The peak value provided is 5V The frequency is 10KHz The square wave . So the circuit can be from XFG1 High and low levels are analyzed .
Without considering the diode voltage drop
(a) When XFG1 Input is 0V when , here VCC adopt D1 And D2 by C3 C4 Charge , here A2 B2 The point voltage is 5V.
(b) When XFG1 Input is 5V when , It can be equivalent to two 5V The power supply is connected in series , here A2 The instantaneous voltage at point is 10V, because A2 And B2 There is a differential pressure , So the charge will change from A2 flow B2, As time goes by , Final C3 Voltage and C4 The average is equal , Are all 7.5V. here C3 The pressure difference between positive and negative electrodes is 2.5V.
When XFG1 Input again 0V when , Since the capacitance voltage cannot be changed suddenly , So at this time A2 The point voltage is 2.5V. because D2 The existence of C4 The charge of cannot flow to C3,VCC Charge flow C3,C3 It will soon be charged to 5V.
(4) When XFG1 Input is 5V when ,C3 The state of is back (b) The state of , The upper voltage is 10V, The charge continues to flow C4, Final A2 B2 Both voltages are 8.75V.
(5)XFG1 Such repeated high-low level conversion .C3 The voltage of is also XFG1 For the high time , The instantaneous voltage at the upper end becomes 10V, It keeps giving C4 Charge , therefore C4 The upper voltage keeps rising , To the end C4 The upper voltage is 10V when , No longer transfer charge , At this time, the output voltage is 10V, by VCC Double of , It plays the role of boosting .
Due to the existence of diode, about 0.4V Pressure drop of , So the final output voltage is a little different from the ideal value .
Such a simple charge pump boost circuit is designed , The principle is also roughly the same . Understand the circuit design , We can also extend by ourselves , Design triple boost , Quadruple boost, etc , This depends on everyone to play !
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