For the boost circuit, we still have a lot in practical applications, for example, our product is powered by lithium battery, but we need to drive a 12V fan, we need to boost the lithium battery voltage to 12V to supply power to the fan;
Let's start with the topology:
Principle:
- Charge phase
The switch S1 is closed, Vin passes through the inductor L1, and passes S1 to the ground, this process charges the inductor, we know that the inductor is an energy storage element, the diagram is as follows:
At this point, we can conclude that the voltage at both ends of the inductor during the charging phase, Uc = Vin
我们知道电感公式U = L*di/dt,即Vin = L*di/Ton,这里Ton代表充能时间;
Some people may have doubts here, isn't this Vin directly to the ground short-circuited, so that the circuit will be prone to failure, let's explain: the characteristics of the first inductor are to hinder the change of current, that is, the current at both ends of the inductor can not change abruptly, so in the charging stage, the current is gradually rising; The second is that the S1 switch is not always closed, and if it is always closed, the inductor will burn out at a certain stage. The closing time of S1 varies according to the load, that is, the current on the inductor rises to a certain point and then releases its own energy;
This brings us to the parameters of the inductive device: rated current and saturation current
The rated current is the maximum current that works continuously, and the saturation current is the maximum current that reaches magnetic saturation inside the inductor. That is, the maximum current through the inductor cannot exceed the saturation current of the inductor
If the magnetic saturation current is exceeded, the inductance value will decrease, the ability to hinder the current will decrease, and the current through the inductor will rise in a straight line, resulting in inductance damage; As illustrated
Therefore, when designing the circuit, we should pay special attention to the charging time here, and ensure that the peak current through the inductor during the charging stage cannot exceed its saturation current;
2. Release the energy phase
Disconnect S1, supply Vin and inductor energy through diode D1 to Vout, and charge capacitor C1 at the same time.
At this point, we are able to derive the voltage at both ends of the inductor during the energy release phase, Uf = Vout + 0.3 - Vin
我们知道电感公式U = L*di/dt，即Vout + 0.3 - Vin = L*di/Toff，这里Toff表示释放能量时间；
In the process of releasing energy, S1 is disconnected, and the right end of the inductor L1 will generate a large induced electromotive force, and due to the clamping effect of the diode, the voltage on the left side of the diode in the above figure will be clamped at about Vout+0.3 (voltage drop at both ends of the diode). Therefore, there is no risk of high pressure here;
In practical application, the above S1 is replaced by an NMOS tube, and we control the charge and discharge energy of the inductor by controlling the switch of the NMOS tube;
In summary, we know that the current waveform of the inductor over the entire cycle is as follows:
The current change of the inductor is the same during the Ton period and the Toff period, and the Imax cannot exceed the saturation current of the inductor;
由上文中公式:vin = l*di/from,得l *di= vin*from
Vout + 0.3 - Vin = L*di/Toff，得L *di=Vin*Ton=（Vout + 0.3 - Vin）*Toff≈（Vout - Vin）*Toff
Then Vin*Ton = (Vout - Vin)*Toff, which is the formula for volt-second equilibrium;
又Ton+Toff = T =1/f，占空比D = Ton/T;
The f here is the switching frequency, which is generally given in the boost chip we use;
我们化简Vin*Ton = (Vout - Vin)*Toff
It can be concluded that the duty cycle D = 1- Vin/Vout, and Vin and Vout are generally known conditions;
Let's talk about the concept of ripple rate, we use r to denote, ripple rate is the change of current / average current, generally we will according to the design requirements, the ripple rate is generally set between 0.2~0.4; We can set the parameters according to our requirements for ripple;
Let's look at it again, we know that Vin = L*di/Ton, then the inductance L = Vin*Ton/di, and di is the current change, so di = Io*r, where Io is the average current of the inductor
或者根据公式Vout + 0.3 - Vin = L*di/Toff
The above two formulas can find the inductance value, but it should be noted that there is still a gap between the theoretical calculation and the actual situation, and the inductance value still needs to be further confirmed according to the actual debugging results.