I was stumped by the switch control of the transistor

tongshaoqiang

I was stumped by the switch control of the transistor [Copy link]

 

As shown in the circuit above: I want to use the output of the microcontroller to control the on and off of the backlight. When the microcontroller is at a high level of 3.3v, 050 is turned on, and the 5V voltage is added to the backlight LED through 8050.

Question: After theoretical calculations, and then through experimental tests, I adjusted the resistance values of R7 and R6. The voltage between Vce is always around 2.2v (the saturation voltage drop should be around 0.2v). Can anyone tell me where the problem lies?

maychang

This NPN transistor Q6 cannot be saturated in this way. Even without R7, the base can only reach 3.3V to ground (MCU pin high level), while the collector is 5V, which is not lower than the base.

maychang

Your transistor is emitter output, that is, common collector connection. Change to common emitter connection to control LED. To change to common emitter, the emitter should be grounded, the original ground end of the LED should be connected to the transistor collector, and the upper end of R6 should be changed to 5V.

Also, how much current do your backlight LEDs need? Why is R6 so large?

tongshaoqiang

maychang posted on 2024-8-8 18:15 Your transistor is emitter output, that is, common collector connection. Change to common emitter connection to control the LED. Change to common emitter, the emitter should be...

Thanks for the guidance! The backlight wants to pass 10mA current, and the voltage drop on the backlight is 3v.

tongshaoqiang

maychang posted on 2024-8-8 18:10 This NPN transistor Q6 cannot be saturated according to this connection method. Even without R7, the base can only reach 3.3V to ground (MCU pin high level) at most...

I see. After the transistor is turned on, the PN junction between the base and the emitter is turned on, and the emitter voltage will not be higher than the base voltage.

maychang

tongshaoqiang posted on 2024-8-8 18:41 Thanks for the guidance! The backlight wants to pass 10MA current, and the voltage drop on the backlight is 3V.

[The backlight wants to pass 10MA current, and the voltage drop on the backlight is 3v]

It should be 10mA and 3V, and uppercase and lowercase letters must be distinguished.

The power supply voltage is 5V, the LED drops 3V, and the voltage on R6 is 2V (ignore the saturation voltage drop of the transistor). 10mA passes through R6, so it is correct to use 200 ohms.

maychang

tongshaoqiang posted on 2024-8-8 18:41 Thanks for the guidance! The backlight wants to pass 10MA current, and the voltage drop on the backlight is 3V.

However, 100 kilo-ohms is too large for R7. The minimum current amplification factor of Q6 is 100. The collector current is 10mA, and the base current is required to be 0.1mA. The maximum output voltage of the microcontroller is 3.3V, and it is impossible to output 0.1mA current through a 100 kilo-ohm resistor.

patrick_huang11

100 kilo-ohms is too large for R7. The minimum current amplification factor of Q6 is 100. The collector current is 10mA, so the base current is required to be 0.1mA. The maximum output voltage of the microcontroller is 3.3V, and it cannot output 0.1mA current through a 100 kilo-ohm resistor.

patrick_huang11

tongshaoqiang posted on 2024-8-8 18:41 Thanks for the guidance! The backlight wants to pass 10MA current, and the voltage drop on the backlight is 3V.

xiexie I understand. After the transistor is turned on, the PN junction between the base and the emitter is turned on, and the emitter voltage will not be higher than the base voltage.

tongshaoqiang

maychang posted on 2024-8-8 18:46 [The backlight wants to pass 10MA current, and the voltage drop on the backlight is 3v] It should be 10mA and 3V, and the upper and lower case must be distinguished. The power supply voltage is 5V, LED ...

"It should be 10mA and 3V, uppercase and lowercase letters must be distinguished."

Once again, I am impressed by the rigor of the older generation of technical people, and I will continue to demand the same of myself in the future!

tongshaoqiang

maychang posted on 2024-8-8 18:49 But 100 kilo-ohms is too large for R7. The minimum current amplification factor of Q6 may be 100, the collector current is 10mA, and the base current is required to be 0.1mA. The microcontroller...

Thanks again to maychang for his guidance.

When adjusting the base resistance, measure the voltage at points AB in the figure above: when the base terminal is connected to a high level, the backlight D lights up, and the voltage at point A is 2.9V, as expected;

When the base input is low, Q6 is cut off, backlight D is off, and the voltage at point B to ground is about 3.4V, the voltage at point A to ground is 5.3V (same as the input voltage at R6), and the voltage at points AB is 0V. In my opinion, the voltage at point B should also be 5.3V. I don't know why it is 3.4V?

maychang

tongshaoqiang posted on 2024-8-9 09:52 Thanks again to Mr. Maychang for his guidance. When adjusting the base resistance, measure the voltage at points AB in the figure above: the base terminal is connected to a high level...

[According to my thinking, the voltage at point B should also be 5.3V. Why is it 3.4V? ]

One reason is that the voltage-current characteristic of LED is not linear, but similar to the forward characteristic of ordinary diode, or even more nonlinear. When the current is relatively small, the voltage drop is relatively large. The second reason is that the internal resistance of your multimeter may be relatively small. When the voltage range of the multimeter is connected between point B and the ground, a little current needs to pass through. This little current is enough for the LED to produce a voltage drop of nearly 2V.

tongshaoqiang

maychang posted on 2024-8-9 10:18 [According to my idea, the voltage at point B should also be 5.3V. I don’t know why it is 3.4V? ] One of the reasons is that the voltage-current characteristic of LED is not linear...

niuyite

There is some very good simulation software now, you can simulate first and then make the circuit!

振动试验仪器

The second reason is that the internal resistance of your multimeter may be relatively small. When the voltage range of the multimeter is connected between point B and ground, a little current needs to pass through. This little current is enough for the LED to produce a voltage drop of nearly 2V.

That is to say, the voltage to ground measured by the multimeter is not the true value.

The current can be calculated by measuring the collector current when the transistor is cut off with an ammeter in series, or by measuring the voltage drop on a 200 ohm resistor with a voltage range. The voltage drop of the diode can be calculated based on the volt-ampere curve of the light-emitting diode.

振动试验仪器

When the transistor is cut off, the potential of points A and B is measured with the ground as the common point. If the potential of points A and B is measured with the positive pole of the power supply as the common point, will there be different results?