Transistor Switching Circuit Design
Update Time: 2022-12-08 11:28:13
Transistor switching circuits are ubiquitous in circuit design nowadays. The classic 74LS, 74ALS, and other integrated circuits use transistor switching circuits inside but only have the common driving capability. Transistor switching circuits are divided into two main categories, the classic TTL transistor switching circuits and MOS tube switching circuits.
Here we will introduce knowledge about switching circuits, including TTL transistor switching circuits, buzzer control circuits - passive buzzers, etc.
1. Transistor Switching Circuits
TTL transistor switching circuits are divided into small-signal switching circuits and power switching circuits according to driving capability. According to the triode connection, it is divided into emitter ground (PNP triode emitter connected to the power supply) and shot-following switching circuits.
a.Emitter ground switching circuit
The basic circuit in the above diagram is a bit far from the actual design circuit: due to the accumulation of charge at the base of the triode, there is a transition from on to off (when the triode is turned off, the base charge release rate becomes slower due to the presence of R1), so Ic does not immediately become zero). This means that the emitter ground switch circuit is related to the off time and cannot be directly applied to the high-frequency switching.
Working principle: When the transistor suddenly conducts (IN signal suddenly jumps), C1 instantly appears as a short circuit, quickly providing base current for the transistor, thus accelerating the transistor conduction. When the transistor suddenly shuts down (IN signal suddenly trips), C1 momentarily conducts, providing a low impedance path for releasing the base charge, thus accelerating the transistor shutdown.
The value of C is typically tens to hundreds of picofarads. r2 in the circuit ensures that the transistor stays off when there is no IN high input. r4 ensures that the transistor stays off when there is no IN low input. r1 and r3 are used for the base limit current.
Working principle: Because the TVS diode Vf is smaller than Vbe 0.2~0.4V, so most of the base current flows from the diode, then to the triode, and finally to ground when the triode is on, so the current flowing to the base transistor is small, and the accumulated charge is small. When the transistor is turned off (IN signal suddenly jumps), the discharge charge becomes less, and the turn-off action naturally becomes faster.
In the actual circuit design, the triode Vceo and Vcbo need to be considered to meet the voltage and the triode to meet the collector power consumption. Use the load current and hfe (take the minimum value of transistor hfe to calculate) to calculate the base resistance (leave a margin of 0.5 to 1 times the base current). Note the reverse withstand voltage of the special diode.
b. Emitter follower switching circuit
The emitter follower, also known as the emitter follower, is a typical negative feedback amplifier. From the transistor connection, it is a normal collector amplifier. The signal is input from the base and output from the emitter. The resistor Re connected to the emitter of the transistor plays an important role in the circuit. It acts as a mirror, reflecting the following characteristics of the output and input.
The input voltage use = ube + usc. Usually, Usc > Ube, ignoring Ube, usr ≈ usc. This means that the voltage amplification of the emitter limit follower is approximately equal to 1, i.e., the input voltage amplitude is approximately equal to the output voltage amplitude. When Usr increases, both ib and, i.e., increase, and the emitter voltage ue (usc) increases. Conversely, when Usr decreases, Usc also decreases. This indicates that the output voltage is in phase with the input voltage precisely because the output voltage is equal to the input voltage and also in phase. The output voltage follows the input voltage closely. We call this circuit a "radiation limit follower" with the following characteristics.
The emitter follower can obtain a large output current (i.e., = (1+β)ib) with a small input current. Therefore, it has the function of current amplification and power amplification. It is important to distinguish that the common multi-stage common emitter amplifier circuit does not amplify current and amplifies voltage, as opposed to emission.
2. Buzzer Control Circuit - Passive Buzzer
When BUZZ is a high voltage, transistor T1 (N-type triode) conducts, and the buzzer sounds.
3. IO control power switch on - using transistors and MOS tubes
MOS: a type of FET MOSFET tube can be made into enhanced or depletion type, P-channel or N-channel common type. However, the only practical applications are enhanced N-channel MOS tubes and P-channel MOS tubes, i.e., NMOS and PMOS.
ON condition: NMOS conducts when Vgs is greater than a certain value. When Vgs is less than a certain value, PMOS conducts.
Switching loss: Whether NMOS or PMOS, there is a conduction resistance after conduction, resulting in unavoidable losses. And now the MOS tube on-resistance is generally a few tens of milliohms.
MOS tube AO3401: P-channel enhanced field effect tube
On condition: generally, not more than -12V can be AO3401. The following is the impedance at different voltage drops.
The following is the switching control circuit in engineering applications.
a. Power control via IO pin - two 3401 MOS tubes
Here are two 3401 MOS tubes without switching control. Just after power up, VDD is equal to the input voltage. At this point, you can supply power in two ways. If J5 has no input voltage, it is powered by VBUS, and 5V is output through F1. The following circuit can use a switch instead of R10, Q201 is always on, and the internal diode voltage drop is about 0.5V.
Note: The two triodes are oriented differently, with Q200 having S on the left and D on the right. Q201 has D on the left and S on the right.
When J5 has voltage, Q200 conducts, and Q201 also meets the conduction conditions, the voltage is 0.1V.
Note: VBUS's right side is disconnected.
b. Voltage regulator and MOS tube voltage regulator circuit
Working principle: VCC can come from the left side VDD5V_Control, or the PC PS2 port supply Vpc_IN. VCC uses high voltage.
The left side Vpc_IN is powered by PS2, and the right side is powered by VCC. when PS2 is powered, the left side is 5V, and the right side is about 4.5V, which can meet the voltage requirements of the machine, and when the PS2 port is closed, the machine can work normally.
In order to reduce the voltage drop of PS2, the following circuit can be decided.
When the PS2 port is powered, the triple tube Q412 is on, Q411 is on, and VCC is close to Vpc_IN. At this time, the machine uses the PS2 port voltage (around 5V). When PS2 is not connected, current cannot flow from the machine to the PS2 port.
Test records using the above parameters.
The last two lines show that: the diode voltage drop inside the MOS diode is about 0.6V, and the Zener leakage current allows the transistor to conduct. The PN junction can conduct at about 0.6V.
Conclusion: The input voltage is at 3.3V, and the transistor conducts, indicating that the resistance of R436 is too large and needs to be reduced. Zener leakage current increases with the increase in input voltage, but when the voltage at both ends reaches 3.9V, the current should exceed 1mA.
In order to ensure that the input voltage can be stabilized at about 5V, the current must be increased and the resistance reduced, and when the input voltage is below 4.7V, the triode must be turned off.
Meet the PS2 input voltage in [4.6-5V] to achieve the effect of voltage regulation. Then connect the large keyboard to the machine. When the machine's power fails, the keyboard can work normally. The power tool also works normally when it is working.
Problems detected: The quality test showed that the terminal could not be shut down. It was found that there is still voltage at Vpc_In after the terminal is powered off. vcc (4.84V) passes through Q411 and generates 4.8V at Vpc_In. And the voltage drop of D405 is about 0.3V. When Vpc_IN is suddenly disconnected, the power supply VCC is on at the moment of disconnection, the triode is on, all the VCC is filled into the terminal, and the triode stays on.
The range of PS2 supply voltage is not easy to determine. When the terminal voltage is large, the circuit is forward-conducting. Also, the Vpc_IN voltage must be less than a certain value to prevent transistor Q412 from conducting.
For example, IRF530 characteristics: General VGS takes 12-15V, plus or minus 20V floating between.
But this is wrong because Vgs is too small.
The following issues need to be considered for microcontroller PWM drive high-voltage MOS (saturated conduction state VGS close to 10V).
Level conversion: High-level microcontroller output does not exceed 5V, generally 12-15V, so the driver circuit must have level conversion capability.
Phase conversion: MOS, as mentioned above, is used as an inverter, so the phase conversion is based on the phase of the load and the microcontroller output. If the MOS output MOS is required to conduct, the driver circuit is required to be in phase.
Switching frequency: different drive circuits have different frequency responses. For up to 1.5M switching frequency, with a simple triode, the simple self-riding circuit is difficult to meet the requirements, the basic need to choose a dedicated drive IC. Also, the general optocoupler cannot work in the switching state above the frequency of tens of K. If you want to isolate 6N137 better, there are special with optical isolation and drive optocoupler, 1.5M Still can not reach.
Drive current: MOS does not consume drive power at rest, but the input is capacitive. In order to conduct the switch as soon as possible to reduce switching losses, the need to charge Cgs at the fastest rate, so the drive circuit has a very important parameter Peak drive current, such as 200MA, 600MA, 1A, 2A, 4A, 6A.
The operating voltage of the driver circuit: generally, the maximum VGS can not exceed 20V, so the operating voltage of the driver circuit should not exceed 18V. The above circuit needs to add 15V; of course, you can step down from 40V.
DV/DT problems: electromagnetic interference will increase because MOS is easily damaged under high DV/DT. To solve these problems, increasing the rise/fall time of the driver circuit output is sometimes necessary. A simple way to do this is to add a small resistor between the driver output and the G-pole.
4. Signal Level Conversion
a. Improving the basic transistor switch of the circuit
Sometimes, the low level we set may not allow the transistor to turn off, especially when the input level is close to 0.6 V. To overcome this critical condition, corrective measures must be taken to ensure that the transistor must be turned off. The following figure gives an improved circuit designed for both cases.
The circuit on the left of the figure above A diode is connected in series between the base and the emitter so that 0.6 V. increases the input voltage value for the base current conduction. In this way, the transistor will not work conduction even if the value of Vin is close to 0.6 V due to a signal source fault. Therefore the switch can remain in the off state.
The circuit on the right side of the diagram above contains a secondary releasing resistor, R2, which is designed with the appropriate R1, R2, and Vin values to ensure that the switch turns off at the critical input voltage. As shown in the diagram above, R1 and R2 form a series voltage divider circuit before the base-emitter junction is on (IB0), so R1 must pass a fixed (varying with Vin) voltage. And the base voltage must be below the Vin value. Even if Vin is close to the threshold (Vin = 0.6 volts), the base voltage will still be pulled down below 0.6 volts by the auxiliary shutdown resistor connected to the negative supply. Due to the careful design of R1, R2 and VBB values, the base will still have enough voltage to turn on the transistor independent of the auxiliary shutdown resistor as long as Vin is in the high range.
b. Acceleration Capacitor
1) An RB resistor is connected in parallel to the acceleration capacitor
In applications where fast switching action is required, the switching speed of the transistor switch must be increased. The following figure shows a common method. This method is only in parallel with an RB resistor on the acceleration capacitor, so when Vin rises from zero voltage and sends current to the base, the capacitor can not be instantaneously charged, so the same short circuit. However, at this time, there is a transient high current flowing from the capacitor to the base, thus speeding up the conduction of the switch tube. Later, until the charging is complete, the capacitor is the same as an open circuit, which does not affect the normal operation of the triode.
Once the input voltage drops from a high level to a zero voltage level, the capacitor turns the base-emitter junction into reverse bias in a very short period, causing the transistor switch to quickly turn off charging to a positive voltage due to the role of the left end of the capacitor, so the moment the input voltage drops, the voltage across the capacitor does not change instantaneously and will remain at a fixed value, so the input voltage immediately drops so that the base-emitter junction becomes reverse biased, quickly shutting down the transistor. Correctly selected acceleration capacitors can reduce the switching time of the transistor switch to less than a few microseconds, and most acceleration capacitors are on the order of hundreds of PF.
2) It is very close to a small signal amplifier circuit, only with one less output coupling capacitor.
Sometimes the load of the triode switch is not added directly between the collector and the power supply, but the connection is shown in the figure below. This connection is very close to the small-signal amplifier circuit, except for one less output coupling capacitor. This connection is the opposite of the normal connection. When the transistor is off, the load is enabled. When the transistor is on, the load is cut off. These two forms of circuits are common, and we must be able to resolve them.
One of the most common applications of transistor switches is to drive indicators that indicate the operating condition of a specific point of the circuit, whether the motor's controller is energized, or whether a certain limit switch is passed or a digital circuit is energized. in a high state.
3) Output state of a digital flip-flop using transistor switches
The following figure shows the output state of the digital flip-flop using a transistor switch. If the output of the flip-flop is high (usually 5 volts), the transistor switch conducts and leaves an indicator light, so the operator can know the current operation of the flip-flop by looking at the light without the need to detect it with a meter.
Sometimes the output current capacity of the signal source (such as the flip-flop) is too small to drive the transistor switch. At this time, to avoid overloading the signal source and misoperation, the improved circuit shown below must be used. When the output is high, first drive the emitter with transistor Q1 to amplify current, and then Q2 conduct to drive the light. Because the input impedance of the emitter and input stage is quite high, the flip-flop should provide a small amount of input current. The digital trigger can work satisfactorily.
Analysis: If FREOF is 5V high, the output FREOUT should be a square wave of about 1.3K Hz.
The waveform is as follows: the left side of C39 and the right side of C41 is a square wave around 1.3K, one high and one low.
In the figure below, when a 1Hz square wave signal is input, the intercept on the left side of waveform C3 is as follows. Theoretical calculation: The same for charging and discharging. First, calculate the charge/discharge constant TC=RC in ohms and F.
The following circuit TC=1K*1uf=1ms 3TC can usually reach 0.95E and 4.75V, so that 3ms can reach 4.75V, consistent with the waveform.
The following figure shows a simple control circuit: when KSEL is high, KCLK1 and KCLK0 pass, and KDAT1 and KDAT0 pass.
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