Switching Circuit Voltage Calculation

lhjsol

Switching Circuit Voltage Calculation [Copy link]

 

As shown in the figure, how to calculate the G pole voltage of Q2 when Q1 is turned on and off?

maychang

How to calculate the G pole voltage of Q2 when Q1 is turned on and off?

When Q1 is fully turned on, it is equivalent to a very small resistor, so the gate voltage of Q2 is the voltage divided by R1 and R5, which is 2.5V.

When Q1 is cut off, it is equivalent to an extremely high resistance and is regarded as an open circuit, so there is no current in R1 and the gate voltage of Q2 is VCC, which is 5V.

maychang

Q1 and Q2 are both marked 100u, 100u, I don’t know what this marking means.

qwqwqw2088

N tube controls P tube circuit

Since it is a simulation, put a voltage needle at the G pole and you will understand

The parameter identification of the two Qs is wrong

maychang

There is a problem in the simulation circuit in the first post: there is no load on the drain of Q2 to the ground. In this case, the input capacitor of the oscilloscope probe is charged to VCC and has nowhere to discharge. Therefore, even if Q2 is turned off for a short time, the voltage across the input capacitor of the oscilloscope probe will not drop, and the waveform displayed by the oscilloscope is wrong.

maychang

The description on the second floor does not take into account the gate input capacitance of Q2, let alone the Miller capacitance of Q2, so this description is only valid when the switching frequency is very low (capacitor charging and discharging can be ignored).

lhjsol

maychang published on 2019-12-1 10:23 The first post has a problem with the simulation circuit: Q2 drain has no load to ground. In this case, the oscilloscope probe input capacitor is charged to VCC and has nowhere to discharge, so Q ...

How should the load and oscilloscope be connected to the circuit correctly? 100u is the default parameter value of the simulation software, and I don’t know how to modify the parameters of the mos tube

maychang

lhjsol posted on 2019-12-1 21:51 How should the load and oscilloscope be connected to the circuit correctly? 100u is the default parameter value of the simulation software, and I don’t know how to modify the parameters of the MOS tube

"So how should the load and oscilloscope be connected to the circuit correctly?"

There is no load at all in the simulation diagram, so there is no way to talk about how to connect it correctly.

The easiest way to add a load is to use a resistor of appropriate size from the drain of Q2 to ground.

lhjsol

maychang published on 2019-12-2 08:18 "How should the load and oscilloscope be connected to the circuit correctly?" There is no load in the simulation diagram, so there is no way to talk about how to connect them correctly...

When the G pole of Q2 is turned on, the actual measured voltage value will fluctuate up and down. I don't know why

maychang

lhjsol posted on 2019-12-3 13:51 The actual voltage value measured when the G pole of Q2 is turned on will fluctuate up and down. I don’t know why

"The actual measured voltage value will fluctuate"

That is most likely a load current change, causing the gate voltage to change due to the Miller effect.

lhjsol

maychang posted on 2019-12-3 14:16 "The actual measured voltage value will fluctuate up and down" It is likely that the load current changes, and the gate voltage changes due to the Miller effect. ...

Is there any way to solve it? In addition, I saw some capacitors directly connected to GS or GD. Why?

maychang

lhjsol posted on 2019-12-3 19:32 Is there any way to solve it? In addition, I saw some capacitors are directly connected to GS or GD. Why?

“Is there any solution?”

If the voltage fluctuation across the load meets your requirements (not greater than a certain value), there is no need to "solve" it.

maychang

lhjsol posted on 2019-12-3 19:32 Is there any way to solve it? In addition, I saw some capacitors are directly connected to GS or GD. Why?

"Also, I saw some people connect capacitors directly to GS or GD. Why is that?"

The capacitor between the gate and source is used to reduce the fluctuation of the gate-to-source voltage. At this time, the capacitor and the previous stage output impedance form a low-pass filter circuit.

As for the capacitor between the drain and the gate, this is rare. Please post the original picture you "saw".

lhjsol

maychang posted on 2019-12-3 19:46 "In addition, I saw some capacitors directly connected to GS or GD. Why?" The capacitor is connected between the gate and the source to reduce the gate-to-source...

C3 is not connected by default, I don't know what it does

lhjsol

Also, can you help me figure out how to calculate the softstart time here?

maychang

lhjsol posted on 2019-12-4 12:43 C3 is not connected by default, I don't know what it does

"C3 is not connected by default, I don't know what it does"

If C3 is connected, the voltage at the VOLED terminal will rise slowly. After 12VEN gives a signal, it will take several seconds for D1 to reach full brightness.

You might as well build a circuit and give it a try.

lhjsol

maychang posted on 2019-12-4 15:26 "C3 is not connected by default, and I don't know what it does." If C3 is connected, the voltage at the VOLED end will rise slowly. After 12VEN gives a signal, D1 ...

Please help answer the question on the 15th floor, thank you

maychang

lhjsol posted on 2019-12-4 17:40 Please help answer the question on the 15th floor, thank you

The circuit on the 15th floor lacks necessary parameters and cannot be calculated.

Miller capacitance is the capacitance between the gate and drain multiplied by the voltage gain of the tube. Now the voltage gain of Q503 cannot be calculated because we don't know how big the drain load is. The drain load will never be R522 plus the light-emitting diode, and there must be output current from LCD_VDD.

maychang

lhjsol posted on 2019-12-4 17:40 Please help answer the question on the 15th floor, thank you

However, the Miller effect can still be estimated. For current MOS tubes, the transconductance is always above 1S, so the voltage amplification factor is estimated to be above 100 times, and it is temporarily estimated to be 200 times. Then the Miller capacitance generated by the Miller effect is 200 times that of C0402, that is, 20uF. The time constant of the Miller capacitance and R524 is 20ms.

This estimate does not take into account R521, nor the capacitance between the drain and gate of the MOS itself. It is only accurate in order of magnitude.

afymzr

Very good, worth learning.