Transistor switch circuit design

The following mainly through the use of NPN transistor switching circuit design, PNP transistor switching circuit similar to NPN.
First, the transistor switch circuit design feasibility and necessity

Feasibility: The person who used the transistor is clear, transistor has a characteristic, is saturated and cut off state, precisely because of these two states, so that it can be applied to the switching circuit is possible.

Necessity: Suppose we in the design of a system circuit, some voltage, signal, etc. need to be cut off during the operation of the system, but can not be cut by mechanical way, at this time can only be processed by software, which requires the transistor switch circuit as the basis.

Second, transistor basic switch circuit overview

The following (Fig. 1) is the most basic transistor switching circuit, the NPN base is connected to a base resistor (R2), and a load resistor (R1) is connected to the collector.

First of all, we need to understand that when the base of the transistor has no current, the collector also has no current, the transistor is in the cut-off state, that is, when the base current will cause the collector to flow through a larger amplification current, that is, into saturation, equivalent to close. Of course the base must have a voltage input that meets the requirements to ensure that the transistor enters the cutoff zone and saturation zone.

Fig. 1 NPN Basic switching circuit
Three, transistor switch circuit design and analysis
(1) Conditions of the cutoff zone and saturation zone
1, enter the cut-off zone conditions: the above mentioned to make the three-level tube into the cutoff zone condition is when the base has no current, but under what circumstances can achieve this requirement. In the case of a silicon transistor, the base pole is connected to the emitter with a forward bias of approximately 0.6V, so to make the transistor cutoff, the base input voltage (Vin) must be less than 0.6V, so that the base current of the transistor is zero. Usually at design time, in order to make the transistor must be in the cut-off state, often make VIN value less than 0.3V. Of course, the closer the input voltage of the base to 0V will ensure that the transistor must be in the cutoff state.
2, enter the saturation zone condition: First collector to be connected to a load resistor R1, the base to be connected to a base resistor R2, as shown in figure. 1. To transmit the current to the load, the transistor collector and emitter must be shorted. It is therefore necessary to have VIN reach a sufficiently high potential to drive the transistor into the saturated working area. When the transistor is saturated, the collector current is quite large, almost so that the entire supply voltage VCC is across the load resistor, so VCE is close to 0, and the transistor collector and emitter are almost short-circuited. Ideally, according to Ohm's law, when the transistor is saturated,
1) The collector saturation current should be:
Ic (saturated) =vcc/r1------------------(equation 1) collector saturation current
2) The minimum base current should be:
Ib (saturated) =ic (Sat)/β=vcc/(β*r1)--------(equation 2) Base saturation current

The above expression shows the basic relationship between the IC and IB, the β value in the formula represents the DC current gain of the transistor, and for some transistors, there is a significant difference between the AC β value and the DC Beta value. To close the switch, the VIN value must be high enough to send the lowest base current that is greater than or equal to (equation 2). Since the base circuit is just a series circuit of a base resistor, base and emitter interface, VIN can be solved by the following formula:

3) base input voltage vin should be at least:

Vin=ib (saturated) *r2 + 0.6v===== "vin= (+0.6V*VCC*R2)/(Β*R1)----(equation 3) base saturation input voltage

Once the base voltage is greater than or equal to (equation 3) the calculated value, transistor conduction, that is, into the saturation zone, so that all the supply voltage across the load resistance R1, and completed the closed action of the switch.

(2) example analysis of the transistor as a lamp switch
The following circuit diagram. 2, the internal resistance of the bulb is 16 ohms, the base string connection resistance is 1 K, the transistor DC current gain of 150, now we want to determine the voltage of VIN is how much time can be the transistor in the cutoff, saturation, that can make the lamp lit or extinguished.

Fig. 2
1, lamp off
As long as the VIN is less than 0.3V, at this time the transistor into the cutoff area, collector No current flow, the light bulb naturally extinguished.

2. Light bulb
To make the lamp light, the transistor collector must have current flow, that is, to enter the saturation zone. According to the formula, we can calculate:

The saturation current of the collector is (according to Equation 1): Ic (saturated) =24v/16r=1.5a

The base saturation current is (according to Equation 2): Ib (saturated) =24v/(150*16) =10ma

The base input voltage is (according to Equation 3): vin=10ma*1k+0.6v=10.6v

Therefore, when VIN is greater than or equal to 10.6V, the light bulb will light up, whereas when VIN is less than or equal to 0.3V, the bulb will turn off.

As can be seen from this example, to use the transistor switch to control the load current of up to 1.5A of the open and close operation, only need to use very small control voltage and current can be. In addition, although the transistor flows through the large current, but does not need to install the heatsink, because when the load current flows, the transistor is saturated, its VCE toward zero, so its current and voltage multiplied by the power of very small, there is no need for heat sinks.

(2) example analysis of the transistor as a voltage output switch

1, the supply voltage Vcc=9v;vin using the MCU's Gpio port control, the output voltage is: 0V and 3.3V, the output voltage of the requirements Vout is 4v/10ma.

2, 9014 of the technical parameters:

Collector Max dissipation Power pcm=0.4w (Tamb=25℃)

Collector Maximum allowable current icm=0.1a collector base Breakdown voltage bvcbo=50v Collector Emitter Breakdown voltage bvceo=45v emitter Base Breakdown voltage bvebo=5v

Base emitter saturation pressure drop VBE (Sat) =1v (ic=100ma; IB=5MA)

β=150

 
 
Fig. 3
3. The value of the resistor (R1) on the computed collector
    the maximum allowable current of the collector is icm=0.1a, so r1=vcc/0.1a=9v/0.1a= 90R, so the minimum collector resistor is 90R, we might as well set the R1 resistance to 10K. So we take r1=10k. Since Vout has a maximum current output of 10mA, it is set to 20mA or 30mA in order to retain enough headroom.
We now set the power of 20MA,R1 to PR1=20MA*4V=0.08W<1/8. Finally, we can set the R1 as 10K SMD resistor (1/8w).
4, Calculate load resistance (value of R3)
    when vin=0v, the transistor cutoff, 9014 of the Collector No current flow, the value of Vout is R1, R2 these two resistors. We can calculate the resistance value of R3 according to the voltage divider:
r3= (r1*vout)/(Vcc-vout) = (10k*4v)/(9v-4v) =8k
Because the 8K resistor is more difficult to buy, so we set a more common 8.2K, so r3= 8.2K SMD Resistor (1/8w).
5, calculating the base resistance (value of R2)
    We already know that the upper limit of Vin is 3.3V, according to Formula 1, 2, 3 can calculate the value of R2:
r2= (VIN-VBE) *β*r1/vcc= (3.3V-1V) *150*10k/9v=383k, final set r2=370k/SMD resistor

Parameters determined: r1=10k/0603
r2=370k/0603
r3=8.2k/0603
Test results:
vin=3.3v time: Test vbe=0.567v close to 0.6V, transistor has entered the saturation zone.
The universal watch shows that Vout is 0.1V, which is actually vce=0.1v<<4v.
VIN=0V: The universal meter shows Vout at 4.06V, which is the 4V voltage output that was originally conceived.

Fig. 4 (vin=3.3v) transistor enters saturation zone

Fig. 4 (vin=0v) transistor entering the cutoff area