Transistor as a switch to learn

Transistor as a switch, is the transistor working in the saturation zone, the NPN transistor as an example (all of the following)

As shown in the figure: when the vin=0, transistor b-e There is no forward bias, and cutoff, equivalent to the state of disconnection, at this time vce=vcc,c pole current ic=0;

When the transistor is closed (in the saturation zone), the load RC and transistor equivalent in series, through the RC current, through Ohm's Law, ic= (Vcc-vce)/RC, (at this time the Ic is saturation current, that is, the C pole can reach the maximum current, Vce is the C-pole and e-pole saturation voltage, Generally take 0.2v), according to the figure shown in the parameters, at this time can be calculated to get ic=9.8ma, and according to the HFE (DC gain, general transistor manual will give, this inside value of =ic/ib), so you can get Ib=196ua, Also because the transistor conduction when the B-pole and e-pole will have a 0.7v pressure drop, when the voltage on the RB is VIN-VBE (the voltage between the B-pole and the e-pole), assuming vin=5v, can be calculated rb= (vin-vbe)/ib=22kω;

The above calculation procedure can be described, when the vin=5v,rb=22kω,vcc=10v, transistor closed, and can be like RC to provide 9.8mA of current; When vin=0, the transistor disconnects.

According to these can be designed to transistor as a switch circuit, for example, with transistors for the buzzer design a switching circuit, assuming the buzzer forward pressure drop of 1v, working current of 10mA, transistor vcc=vin=5v;

First can determine the transistor closed vce=0.2v, the general load is connected to the C-pole transistor, transistor closure when the load is equivalent to the transistor in series, thus vcc=5v, vce=0.2v, Can be determined that the load on the voltage of 4.8v, and the normal operation of the buzzer voltage from the condition can be known as long as 1v larger than a little more, 4.8v far more than its operating voltage, in order to prevent its burnout, the need to add a sub-pressure module, to get the figure circuit (the figure with light-emitting two-ray replacement buzzer)

First determine the transistor, according to the requirements of the transistor C-pole need to allow the current to be greater than 10mA, the general small power transistor 2n3904 can (specific transistor selection to see the operating current of the load), if vin=0v, transistor disconnect buzzer does not work Buzzer normal work requires transistor into the saturation zone and the buzzer end voltage of 1v, the operating current of 10mA, because the transistor into the saturation zone, so vce=0.2v, you can calculate the rc= (Vcc-1v-vce)/10ma;

And according to the HFE (DC gain) =ic/ib, the IB=IC/HFE can be obtained;

After the Ib, and because the transistor is in the conduction state, so the pressure drop between the B-pole and e-pole is 0.7v, that is, vbe=0.7, so can be obtained, rb= (VIN-VBE)/ib;

This will determine the value of RB and RC, when the vin=5 buzzer is working normally;

If the value of VCC has changed, use the same method to calculate, now use transistor as switch control light emitting diode, buzzer, motor, relay ...

The above is the NPN transistor, in fact, PNP transistor is often used when the switch is used, but the PNP transistor is Vin is the high-level transistor is disconnected, vin low-level transistor is conduction;

The NPN transistor current flows from the C-pole to the e-pole outflow, while the PNP type is the inflow from the e-pole and the C-pole outflow;

NPN transistor is the C-pole load and the VCC,PNP type is the e-pole VCC, but the load is still connected to the C-pole;

As shown in the figure is the PNP type transistor control light emitting diode

This is not a detailed derivation.

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