xunke

Constant current source circuit, LED current = 0.7/R162 = 0.7/120 = 5.83mA

zhuyebb

Starts with VDD/

chunyang

15818739085 Published on 2019-1-14 22:46 Hello, why does the current flowing through the LED rise slowly?

I don't know what you mean? If you have other questions, it's best to ask them in another thread. There is no such thing as "slowly rising" in the original post.

xunke

The simulation result shows that the LED current is about 5.1mA The original file of LTSpice simulation 三极管恒流源仿真电路.zip (537 Bytes, downloads: 2)

2019-1-15 18:51 上传

点击文件名下载附件

maychang

xunke posted on 2019-1-15 18:52 The simulation result shows that the LED current is about 5.1mA. The original file of LTSpice simulation

Using simulation to verify the calculation on the 21st floor is good and worthy of encouragement.

15818739085

chunyang posted on 2018-12-10 16:43 When VDD is powered on and the gate voltage of Q27 reaches the turn-on voltage, Q27 is turned on, the LED lights up, and R162 in the circuit provides the base voltage to Q28. When the current flowing through the LED...

When the current flowing through the LED reaches a certain level, I don't understand this part. Please help me analyze it. Thank you.

lingking

xunke posted on 2019-1-15 18:52 The simulation result shows that the LED current is about 5.1mA. The original file of LTSpice simulation

I used multisium simulation and it is also about 5mA

lingking

xunke posted on 2019-1-15 18:52 The simulation result shows that the LED current is about 5.1mA. The original file of LTSpice simulation

How is the simulation performance of LTspice?

15818739085

chunyang posted on 2018-12-12 15:12 As for the value of R161, this resistor determines the collector current of the transistor, and the gate input impedance of the MOS tube is very high and can be considered as an open circuit (that is, it does not affect...

How to calculate the base current of this transistor? Are the base current and collector current in a magnification relationship?

maychang

15818739085 Published on 2019-1-15 20:59 When the current flowing through the LED reaches a certain level, I don’t understand this part. Please help me to analyze it. Thank you.

See the first post picture. Before the power is turned on, there is obviously no current in the LED, and the gate-source voltage of the MOS tube is zero. After the power is turned on, the capacitor between the gate and source of the MOS tube is charged by the 47k resistor. During the charging process, the voltage between the gate and source gradually rises, and the drain current of the MOS tube gradually increases. When the current flowing through the LED reaches a certain level, causing the base voltage of Q28 to rise high enough to cause Q28 to start conducting, the gate potential of Q27 will be pulled down, thereby reducing the current flowing through the LED.

chunyang

15818739085 Published on 2019-1-15 20:59 When the current flowing through the LED reaches a certain level, I don’t understand this part. Please help me to analyze it. Thank you.

The current flowing through the LED forms a loop through Q27 and R162. The larger the current, the higher the voltage drop on R162. Once the voltage drop on R162 reaches about 0.7V, Q28 starts to conduct. It is because the conduction of Q28 lowers the gate potential of Q27, thereby reducing the conduction degree of Q27, thereby limiting the current.

chunyang

15818739085 Published on 2019-1-15 21:39 How to calculate the base current of this transistor? Is the base current and collector current in a magnification relationship?

The base current of Q28 is difficult to calculate specifically because of the discreteness of device manufacturing. For example, what is the base potential of Q28 to make it conduct? I don’t know, I can only say it is about 0.7V, and Q27 and LED are also non-linear components. The manufacturing discreteness will affect their VA characteristics, and even change with temperature. To calculate, it must be based on specific devices, and after rigorous testing, the specific parameters of LED, Q27 and Q28 must be obtained under constant temperature conditions.

chunyang

15818739085 Published on 2019-1-15 21:39 How to calculate the base current of this transistor? Is the base current and collector current in a magnification relationship?

The calculation of the original poster's circuit does not need to be highly accurate. It only needs to calculate the value of R162 according to the required current limiting threshold. For example, the current limiting threshold is 5mA, and the emitter junction voltage drop of Q28 is 0.7V, so 0.7/0.005=140 ohms. However, there is no such value in the existing resistor series. The actual value needs to be approximated in the optional series. In the original poster's figure, 120 ohms is used. The threshold under this value will be slightly higher than 5mA. The error does not matter and will not cause any adverse effects. A netizen on the 24th floor did a simulation. In the simulation software, the Q1 emitter junction voltage drop corresponding to Q28 is defined as 0.6165V in the software, which may not be the case in reality. However, since it does not need to be precise, you don't need to care about the specific value.

xunke

lingking posted on 2019-1-15 21:27 How is the simulation performance of LTspice

It’s okay. It mainly depends on the component model. LTspice is free and not big. Some great people in the xmo forum like to use it. I usually use it to simulate analog circuits. It’s pretty good. I don’t like multisium very much. The waveform on the oscilloscope keeps flickering.

lingking

lingking posted on 2019-1-15 21:27 How is the simulation performance of LTspice

I also feel that the simulation capability of multisium is relatively poor. I have also heard university teachers say the same thing. I will switch to LTSPICE later, mainly for the simulation of some auxiliary transistor circuits.

15818739085

maychang posted on 2019-1-16 08:18 See the first post. Before the power is turned on, there is obviously no current in the LED, and the gate-source voltage of the MOS tube is zero. After the power is turned on, the gate-source voltage of the MOS tube...

I have understood it now. Thank you for the teacher's explanation.

15818739085

chunyang posted on 2019-1-16 11:42 The current flowing through the LED forms a loop through Q27 and R162. The larger the current, the higher the voltage drop on R162. Once the voltage drop on R162 reaches about 0.7V...

I have understood it now. Thank you for the teacher's explanation.

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A story about a triode