Level conversion and Driving Circuit Analysis of Transistor

3.3V-5 V level conversion circuit

For example, if the left side is connected to a 3.3 v cmos level, it can be an I/O port of STM32 or FPGA, and the right side output is 5 V to realize conversion from 3 V to 5 V.

Analyze the functions of each resistor (the core idea is to hold the transistor's Vbe on at a constant value of about 0.7V ):

1. If there is no R87, then when the US_CH0 high level is directly added to the BE of the transistor, where is the voltage of 0.7V?

2. Suppose there is no R91. If the US_CH0 level is unknown, does Trig output a high level or a low level by default? Therefore, R91 serves as a fixed level. At the same time, if there is no R91, as long as the input is greater than V, the transistor will be turned on. The threshold voltage is too low, and R91 will improve the threshold voltage (see the analysis of the buzzer in section 2 ).

   However, when R91 is added, you must note that if R91 is too small, the base voltage is similar.

   Only when Vb> 0.7V can US_CH0 be turned on at high power levels. The Vb = 1.36 V

3. If there is no R83, when the input US_CH0 is high (when the transistor is turned on), The D5V0 (5 V High) is directly added to the CE level of the transistor, And the CE level of the transistor is easily damaged.

Further analyze its working mechanism:

1. When the input is high, the transistor is turned on, and the output is clamped at the Vce of the transistor, the circuit test result is only 0.1 V.

2. When the input is low and the transistor is not turned on, the output is equivalent to pulling up the input of the next level circuit with a 10 k resistor. The actual test result is 5.0 V (no-load)

   Note:

   For heavy current loads, the features of the above circuit will not be so good, so it has been emphasized that this circuit is only applicable to the level conversion of load from 10 mA to dozens of mA.

Buzzer drive circuit

The above is the circuit uploaded from Zhou licong's iMX283 Development Board. It can be either an active source or a passive buzzer. To analyze:

1. Calculate the current (β = 120 of S9013, set the buzzer current to 15mA ):

   The input threshold voltage is calculated as follows:

   R1 plays a role in providing threshold voltage.

2. The main difference between the active buzzer and the passive buzzer is that the passive buzzer is essentially a inductive component, and its current cannot be transient. Therefore, there must be a continuous stream diode D1 to provide continuous streams. Otherwise, the two ends of the buzzer will have a back-induction potential, resulting in dozens of volts of peak voltage, may damage the drive transistor, and interfere with other parts of the entire circuit system. If the operating voltage in the circuit is large, a diode with a high voltage value should be used. If the operating frequency of the circuit is high, a high-speed diode should be used.

3. The basic way to design this circuit is to determine the load (Buzzer 10mA ~ 80mA) Current and input threshold voltage. Calculate the values of R1 and R2.

ULN2x03 driver circuit

For the above driver circuit:

1. The load is connected to an infrared diode, and its series resistance is a throttling resistor to control the infrared emission intensity.
2. The input is connected to the common I/O port of the PWM function of STM32 (set the push-pull output), and The comport is connected to the output voltage of 5 V.

For the above circuit test (Power = 5.0 V ):

1. 3.3 V for input and 0.6 V for output
2. Input 0 V and output 5.0 V
3. Input is not connected, output is 5.0 V

Therefore, ULN2003/2803 can also be used for level conversion. Why? What is the relationship between ULN2803/2003 and the transistor? Its internal implementation is two transistors.

Its structure has three features:

1. Output Collector leakage, so you can connect your own pull resistance, pull the signal to the corresponding level, ULN2803 manual shows that the maximum voltage can withstand is 50 V

2. In the data manual, the input threshold voltage at Ic = 250mA is VI (on) = 2.7 V.

3. The COM end is connected with a reverse diode: it is connected to the output power supply, which is used to drive the load inductor such as the motor to provide the current loop protection circuit when the power is on and off; the output voltage is higher than the COM end voltage, the voltage will be clamped around VCOM + 0.4V (the diode voltage drop here is small)

The difference between ULN2003 and ULN2803 is that ULN2003 only has eight channels, while ULN2803 has nine channels.

Compared with the self-built transistor circuit, it has better current driving characteristics. Therefore, the self-built transistor circuit is suitable for level switching and small current driving, ULN2803 and ULN2003 are suitable for larger current drives (Datasheet says the maximum drive current can reach about 500mA ). Therefore, ULN2803 and ULN2003 (other such as 75452, MC1413, and L293D) are commonly used to improve the load carrying capacity of the system (motor, large LEDs, relays, etc ).

Note: The transistor can also be used to drive the mos tube. The relevant content will be tested and verified in the future. We will discuss it later!