Misunderstanding of switching triode

    In digital circuit design , It is often necessary to drive buzzer by digital signal through switching current spreading device 、LED、 Relay and other devices that need high current , Most commonly used is the triode . However, in use , If not designed properly , The triode will not work in the normal switching state , Unable to achieve the desired effect .

        Pictured （a） Shown , use NPN triode , The buzzer is connected to the collector of the triode , Drive signals are common 3.3V perhaps 5V   TTL level , High level conduction , The resistance is taken as the empirical value 4.7KΩ, When the triode is turned on, it is assumed that the high level is 5v, The base current is ：

        Ib=（5-0.7）V ÷4.7KΩ = 0.9mA

It can make The triode is completely saturated .

Pictured （b） Shown , use NPN triode , Also connect the buzzer to the triode collector , The difference is Yes, the driving signal is 5V Of TTL level .

Both of the above circuits can work normally , as long as PWM The drive signal operates at a suitable frequency , Buzzer （ Active power ） Will make the loudest sound .

chart 2 Sum graph 1 contrast , The biggest difference is the transmitter connected to the triode by the driver .

Pictured （c） Shown , When the triode is turned on, it is assumed that the high level is 5V, The base current is

        Ib=（5-0.7-UL）V ÷4.7KΩ 

among ,UL Is the pressure drop across the driven device . It can be seen that , Also take the positive resistance as 4.7KΩ , The current flowing through the base will be larger than the figure 1 Medium （a） The circuit current should be small , How small is it UL How much? ： If UL more , So the corresponding Ib It will be very small , It is very likely that the triode will not work in the saturated state , Make the drive device unable to operate . Some people think that it is better to turn down the base resistance , However, it is difficult to know the voltage drop of the driven device , The voltage drop of some driven devices is variable , thus , It is difficult to choose a suitable base resistance ： The resistance value is too large , Will cause the drive to fail ; Resistance selection is too small , The loss becomes larger . therefore , Not as a last resort , Drawing is not recommended 2 Two kinds of circuits .

Pictured 3, The driving signal is 3.3V level , And the turn-on voltage of the driven device needs 5V. stay 3.3V Single chip microcomputer circuit , If you're not careful , It is easy to design these two circuits .

Pictured （e） Shown , This is typical “ Emitter positive deviation , Collector reverse bias ” Amplifier circuit , Or emitter emitters . When PWM The signal is 3.3V when , The emitter voltage of triode is 3.3V-0.7V=2.6V, Unable to meet expectations 5V.

Pictured （f） Shown , This is a failed circuit . First , This circuit cannot be disconnected , When the drive signal PWM by 3.3V High level is ,Ube=5V-3.3V = 1.7V The triode can still be turned on , So the circuit cannot be disconnected . ad locum , Some people will say that they have used this circuit , He has no problem , And the voltage of single chip microcomputer is also 3.3V. I personally believe that this person uses OD（ Open a leak ） Driving mode , And it's real ＯＤ Or is it ５Ｖ Tolerable ＯＤ, such as ＳＴＭ３２ Many of the IO Can be set to OD Door drive mode , Output high level , The signal becomes a high resistance state , The current flowing through the base is 0, Triode can effectively cut off , Time chart （f） Is still valid .

To sum up, several circuits , obtain The two optimal circuits in the figure above . And graph （1） Different , chart （4） Add a between the base and the transmitter 100KΩ The resistance of , This resistor has a certain function , The triode can have a known default state . When the input signal is removed , The triode is still in the cut-off state . In terms of security , It is necessary to add this resistance , Or it can make the triode work in a better switching state .