Analysis of Capacitance and Inductance Charge and Discharge of Dynamic Circuit

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1. Charge pump charging and discharging scheme
   1. Principle introduction

 

       

As shown in the figure above, the working steps of the dynamic circuit scheme are as follows:

1. PWM1 and PWM2 signals control the MOS tube ⑤ ⑥ to charge the capacitor ⑧ (PMW1 and PWM2 MCU frequency lock).
2. Capacitor ⑧ and diode ⑨⑩ form a unidirectional charge pump, which charges ⑪ to a negative voltage.
3. Resistor ④⑦ limits the charge and discharge current.
4. The fuse ③ protects the operating current of the system.
5. The MOS tube ① converts the monolithic drive signal, and then drives the MOS tube ⑤.
   1. Work process analysis

First, simulate the circuit of this level of PWM, the simulation circuit is as follows:

 

       

When the input waveform is high (green), the left side of capacitor C3 is the power supply voltage V1, and the right side charge is released through D5, and finally the right side voltage is maintained at 0.7V.

         When the input waveform becomes low, the voltage difference on both sides of the capacitor will not change suddenly. Because the voltage on the left side becomes 0, the voltage on the right side changes to -(V1-0.7V) (that is, the voltage difference before the capacitor is maintained). When the capacitor C1 is charged, the current flows from C1 to C3, which causes the voltage of C3 to rise. Finally, if diode D1 is ignored, C1 and C2 will equally divide the voltage-(V1-0.7V), but due to the existence of the diode, the level of C3 will change Will be 0.7V higher than C1.

         When the output waveform becomes high here, the left side of capacitor C becomes V1, and the level on the right side is partially neutralized by C1, so the level on the right side will be greater than 0.7V. At this time, D5 discharges and the right side is discharged. The level is discharged again to 0.7V.

         Repeat the above steps several times, and finally the voltage at C1 will become -(V1-0.7V-0.7V). Due to the above simulation, V1 is 3.3V, so the level at C1 will finally stabilize at about 1.9V.

1. 1. Scheme simulation
      1. Parameter calculation
2. The relationship between capacitor voltage and charging time:

Vt-capacitor voltage, V0-initial voltage, Vmax-supply voltage, R-charging impedance, C-capacitance, t-time, E-supply voltage.

When V0=0:

 

At this time:

The specific charging curve is as follows:

 

1. The relationship between capacitor voltage and discharge time:

Vt-capacitor voltage, E-capacitor initial voltage, R-discharge impedance, C-capacitance, t-time.

1. Parameter calculation

For the current design, there are the following parameters, capacitance 100uF load 600R, power supply voltage 24V, if the power supply voltage is allowed

Swing at 20V~22V, there is discharge time:

For charging from 20V to 22V, that is, from 0.83E to 0.91E, compared with the above, there are roughly:

          If the charging rate is 50%, the charging time is the same as the discharging time, substituting t charge = 2.85mS, C = 100uF, you can get:

         That is, the charging resistance is about 28.5R.

         In summary, when the discharge resistance is 600R, the capacitance is 100uF, and the output voltage changes in the range of 20V~22V,

The circuit should be configured as follows:

1. The charging resistance is about 28.5R
2. PWM cycle is around 175Hz
   1. 1. Waveform simulation

The simulation based on the above parameters is as follows:

 

 

         As shown in the figure above, the simulation results are basically consistent with the settlement.

         Then introduce a dynamic circuit, the diode is an ideal diode, and the voltage drop is set to 0. The simulation results are as follows:

 

   The simulation results are as above. Due to the increased capacitive reactance and delay of the double capacitors, the voltage drops slightly, which is basically around -20V.

Use dual MOS to control power on and off, the simulation results are as follows:

 

         Pull back the diode to 0.7V voltage drop, the simulation results are as follows:

         Adjust the charging resistance to 10R and increase the frequency to 10KHz. The simulation results are as follows:

        

         In summary, when the capacitance is 100uF, the charging resistance is 28.5R, and the operating frequency is 175Hz, the load can be basically guaranteed

600R works at a level of about 20V.

         If you need to improve the performance of the charge pump, you can adjust it from both the charging resistance and the operating frequency.

inductance

1. Inductance charging and discharging scheme (switching power supply scheme)
   1. An Introduction

         This scheme can be regarded as a deformation of the transformer isolation scheme. The transformer in the transformer isolation scheme is replaced by an inductor.

Using the inductance's freewheeling function to work, the workflow of this solution is as follows:

1. PWM1 controls the optocoupler isolation①, periodically on and off.
2. When the inductor ② is disconnected, it enters the freewheeling mode, and the capacitor ④ is charged to a negative voltage through the diode ③.
3. V1+ and V1- supply power to the second-level circuit, and the second-level circuit works on the same principle as the first-level circuit.
4. Generate voltages V2+ and V2- at OUT to drive external relays.
   1. Scheme simulation

As shown in the figure above, use R3 and 4R to limit the system current to 6A, and use 22uH common inductors for inductance sampling.

The working frequency is 25KHz, and the first-stage working waveform is as follows:

         As shown in the figure above, when the MOS tube is turned off (when the blue is low), the current flowing through the inductor will not disappear immediately (the red part), but

Turn to diode freewheeling. At this time, the V2 voltage is instantly pulled down to a negative voltage. As the freewheeling current decreases, the V2 page gradually approaches 0.

During this process, V3 is charged, and the voltage gradually increases due to no release path.

         When the MOS is turned on (when the blue is high), the current increases sharply. Due to the existence of the inductance, it provides an instantaneous

The induced voltage, as the current becomes stable, the V2 voltage is gradually pulled down, approaching zero.

         Through the above analysis, when the first stage is working, it will continuously charge V3 or C1, so the second stage circuit can be used as a power supply.

         The working status of the second level is as follows:

         Similar to the working principle of the first stage, V6 will be continuously charged and provided to the analog load R6 using 100R internal resistance.

         Lengthen the entire sampling time, and get the simulation waveform as follows:

         As shown in the figure above, the system can finally provide a voltage drop of about 20V.

         Adjust the operating frequency of the system to 1KHz, and the simulation waveform is as follows:

         As shown in the figure above, the system cannot work normally under low frequency conditions.

1. Transformer isolation scheme
   1. Introduction

The functional diagram of the program is as follows:

         As shown above, the working flow of this circuit is as follows:

1. PWM1 drives the optocoupler isolator①, and then drives the relay ② to be turned on periodically.
2. The relay ② is turned on periodically, and an AC signal will enter the rectifier bridge ③.
3. The AC signal passes through the rectifier bridge ③ and becomes a DC signal and enters the capacitor ④. The capacitor ④ stores the energy of the DC signal.
4. PWM2 drives the optocoupler isolator ⑤, and uses the energy stored in the capacitor ④ to convert it into a voltage signal through the resistor ⑥, and then drives the MOS tube ⑦.
5. MOS tube ⑦ is turned on periodically, the steps are the same as ①~③, and finally the capacitor ⑩ stores energy.
6. The capacitor ⑩ drives an external relay.
   1. Simulation and analysis

1. Working principle of transformer

 

 

As shown above:

u1-power supply voltage; i1-primary coil working current; r1-primary coil internal resistance; e1-primary coil reverse electromotive force;

N1- primary coil turns; L1- primary coil inductance.

u2-output voltage; i2-output current; r2-secondary coil internal resistance; e2-secondary coil induced electromotive force; N2-secondary line

Number of turns; L2-Secondary coil electromotive force.

         Φ-magnetic flux, I-total current of primary coil, R-load of secondary coil.

         Its working process is as follows:

 

         1: Add power U1 to the primary coil;

         2: U1 generates an increasing current i1;

         3: The magnetic flux F produced by the gradually increasing i1;

         4: The changing magnetic flux F generates an induced voltage U2 in the secondary coil;

         5: U2 generates induced current i2, and i2 generates magnetic flux superimposed on F;

         6: Newly added magnetic flux induces current in the primary coil;

1. Transformer operating frequency problem

         According to the above working process, if the transformer wants to work normally, the primary coil must produce a changing current. For the beginning

Level coil is a classic RL charging and discharging process:

         In this process, there are: τ=L/R

Circuit charging, i=Io[1-e^(-t/τ)], Io is the final stable current.
The circuit is discharged, i=Io×e^(-t/τ)], Io is the current in L before the short circuit.

For the primary coil, Io=U/R, U is the power supply voltage, R is the internal resistance of the coil, so the charge and discharge time t can be calculated, then

For the transformer, the operating frequency should be less than 1/t, otherwise the transformer will work in DC mode and lose its effect.

 

         Build a simulation circuit, U=100V, L=10H, R=100R, t=L/R=0.1, final current Io=1A, current=0.5A

Then there are:

0.5=1 [1-e^(-t/0.1)]

t=0.06931s      

1. Transformer working current problem

         For the above transformer, we assume N1=N2, L1=L2, and without considering the saturation problem, the equivalent circuit is as follows:

 

        For the right side of the secondary coil, e2=u1 when it is just powered on. As the rate of change of φ becomes slower, e2 gradually decreases, so:

i2max=U1/(r1+r2+R)

         For the primary coil, I=i1+i2, the process is as follows:

1. Just powered on, L is equivalent to open circuit, I1=0, I=i2
2. In the current change, i2 decreases, i1 increases, I=i1+i2
3. When DC, L is equivalent to short circuit, i1 is the largest, i2=0, I=i1=U/R1

 

 

 

        

In the above simulation, set r1=r2=R=115, N1=N2, L=3.8H, the initial state is:

I=i2=24/(115+115+115)=0.069A

         Then i2 gradually decreases, i1 gradually increases, i1 can be calculated with the above charging equation, or the secondary coil is disconnected, that is

i2=0 simulation, this process satisfies:

I=i1+i2

         When the maximum reaches DC, u2 becomes 0, i2=0, and there is

I=i1=24/115=0.208A

1. Transformer performance characteristics 1

Ignoring problems such as magnetic leakage, for transformers:

Φ = U / (4.44fN)

         It can be seen that the magnetic flux is inversely proportional to the frequency f and the number of turns N. The higher the frequency, the less the magnetic flux, the less the magnetic flux, the less output power, so

The problem of insufficient output power for high-frequency signals in low-frequency transformers.

1. Transformer performance characteristics 2

Since the transformer works in the coil charging and discharging state, the longer the charging and discharging time, the greater the current (see the above formula), so the low frequency

The signal in the high-frequency transformer (generally the inductance L is small) is prone to excessive current and even burnout.

1. Transformer performance characteristics 3

For the transformer secondary coil:

i2=U/(r1+r2+R)

         It can be seen that the larger the internal resistance of the coil will directly affect the load capacity of the transformer, and the excessive internal resistance will also cause the transformer to heat up seriously, so the internal resistance of the transformer is generally made small.