This article mainly introduces you to the 3V-5V level conversion circuit diagram, children's shoes that are of interest or have not been understood before can be observed!
Now let's analyze the role of each resistor (the core idea of ​​grabbing is that the Vbe of the triode is constant at around 0.7V):
Assuming no R87, when the high level of US_CH0 is directly added to the BE of the triode, where is the voltage of 0.7V going?
Suppose there is no R91. When the state of US_CH0 is uncertain, the default is to output high or low level of Trig? Therefore, R91 functions as a fixed level. At the same time, if there is no R91, just turn on the triode by inputting “0.7Vâ€, the threshold voltage is too low, and R91 has the function of raising the threshold voltage (refer to the second section for the analysis of the buzzer).
Only when Vb"0.7V can make US_CH0 turn on when it is high. Vb=1.36V in the above figure
Suppose there is no R83. When the input US_CH0 is high (when the transistor is turned on), D5V0 (5V high level) is directly added to the CE level of the triode, and the CE and triode of the triode are easily damaged.
Further analysis of its working mechanism:
When the input is high, the transistor is turned on, and the output is clamped to the Vce of the triode. The test result for the circuit is only 0.1V.
When the input is low, the transistor is not conducting, and the output is equivalent to pulling up the input of the next-stage circuit using a 10K resistor. The actual test result is 5.0V (no-load).
Please note:
For high current loads, the characteristics of the above circuit will not perform as well, so it has been emphasized here - this circuit is only suitable for level shifting of loads from 10 mA to tens of mA.
This article mainly talks about the 3V-5V level conversion circuit diagram, let's learn together:
The left end of the figure is connected to the 3.3V CMOS level, which can be an IO port of STM32, FPGA, etc., and the output of the right end is 5V level, achieving conversion from 3.3V to 5V.Now let's analyze the role of each resistor (the core idea of ​​grabbing is that the Vbe of the triode is constant at around 0.7V):
Assuming no R87, when the high level of US_CH0 is directly added to the BE of the triode, where is the voltage of 0.7V going?
Suppose there is no R91. When the state of US_CH0 is uncertain, the default is to output high or low level of Trig? Therefore, R91 functions as a fixed level. At the same time, if there is no R91, just turn on the triode by inputting “0.7Vâ€, the threshold voltage is too low, and R91 has the function of raising the threshold voltage (refer to the second section for the analysis of the buzzer).
3V-5V level conversion circuit diagram
However, adding R91 should pay attention to it: if R91 is too small, the base voltage is similar.Only when Vb"0.7V can make US_CH0 turn on when it is high. Vb=1.36V in the above figure
Suppose there is no R83. When the input US_CH0 is high (when the transistor is turned on), D5V0 (5V high level) is directly added to the CE level of the triode, and the CE and triode of the triode are easily damaged.
Further analysis of its working mechanism:
When the input is high, the transistor is turned on, and the output is clamped to the Vce of the triode. The test result for the circuit is only 0.1V.
When the input is low, the transistor is not conducting, and the output is equivalent to pulling up the input of the next-stage circuit using a 10K resistor. The actual test result is 5.0V (no-load).
Please note:
For high current loads, the characteristics of the above circuit will not perform as well, so it has been emphasized here - this circuit is only suitable for level shifting of loads from 10 mA to tens of mA.
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