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Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off state
When used as an AC signal amplifier, the base biasing voltage of a bipolar transistor is applied in such a way that it will always operate within its “active” region. That is the linear part of its output characteristic curves are used. However, both the NPN & PNP type bipolar transistors can also be made to operate as solid state “ON/OFF” switches simply by biasing the transistors base terminal differently allowing us to operating the transistor as a switch.
Solid state switches are one of the main applications for the use of transistors to switch a DC output “ON” or “OFF”. Some output devices, such as LED’s only require a few milliamps at logic level DC voltages and can therefore be driven directly by the output of a logic gate. However, high power devices such as motors, solenoids or lamps, often require more power than that supplied by an ordinary logic gate so transistor switches are used.
If the circuit uses the Bipolar Transistor as a Switch, then the biasing of the transistor, either NPN or PNP is arranged to operate the transistor at both sides of the ” I-V ” characteristics curves we have seen previously.
The areas of operation for a transistor switch are known as the Saturation Region and the Cut-off Region. This means then that we can ignore the operating Q-point biasing and voltage divider circuitry required for amplification, and use the transistor as a switch by driving it back and forth between its “fully-OFF” (cut-off) and “fully-ON” (saturation) regions as shown below.

The pink shaded area at the bottom of the curves represents the “Cut-off” region while the blue area to the left represents the “Saturation” region of the transistor. Both these transistor regions are defined as:
Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.
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Then we can define the “cut-off region” or “OFF state” when using a bipolar transistor as a switch as being, both junctions reverse biased, VB < 0.7v and IC = 0. For a PNP transistor, the Emitter potential must be negative with respect to the Base.
Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.
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Then we can define the “saturation region” or “ON mode” when using a bipolar transistor as a switch as being, both junctions forward biased, VB > 0.7v and IC = Maximum. For a PNP transistor, the Emitter potential must be positive with respect to the Base.
The bipolar transistor operates as a “single-pole single-throw” (SPST) solid state switch. With a zero signal applied to the Base of the transistor it turns “OFF” acting like an open switch and zero collector current flows. With a positive signal applied to the Base of the transistor it turns “ON” acting like a closed switch and maximum circuit current flows through the device.
The simplest way to switch moderate to high amounts of power is to use the transistor with an open-collector output and the transistors Emitter terminal connected directly to ground. When used in this way, the transistors open collector output can thus “sink” an externally supplied voltage to ground thereby controlling any connected load.
An example of an NPN Transistor as a switch being used to operate a relay is given below. With inductive loads such as relays or solenoids a flywheel diode is placed across the load to dissipate the back EMF generated by the inductive load when the transistor switches “OFF” and so protect the transistor from damage. If the load is of a very high current or voltage nature, such as motors, heaters etc, then the load current can be controlled via a suitable relay as shown.

The circuit resembles that of the Common Emitter circuit we looked at in the previous tutorials. The difference this time is that to operate the transistor as a switch the transistor needs to be turned either fully “OFF” (cut-off) or fully “ON” (saturated).
An ideal transistor switch would have infinite circuit resistance between the collector and emitter terminals when turned “fully-OFF” resulting in zero current flow. Zero resistance between the collector and emitter when turned “fully-ON”, results in maximum current flow between them.
In practice when the transistor is turned “OFF”, small leakage currents flow through the transistor and when fully “ON” the device has a low resistance value causing a small saturation voltage (VCE) across it. Even though the transistor is not a perfect switch, in both the cut-off and saturation regions the power dissipated by the transistor is at its minimum.
In order for the base current to flow, the Base input terminal must be made more positive than the Emitter by increasing it above the 0.7 volts needed for a silicon device. By varying this base-emitter voltage VBE, the base current is also altered and which in turn controls the amount of collector current flowing through the transistor as previously discussed.
When maximum collector current flows the transistor is said to be Saturated. The value of the Base resistor determines how much input voltage is required and corresponding base current to switch the transistor fully “ON”.
Using the transistor values from the previous tutorials of: β = 200, Ic = 4mA and Ib = 20uA, find the value of the Base resistor (Rb) required to switch the load fully “ON” when the input terminal voltage exceeds +2.5 volts.

The next lowest preferred value is: 82kΩ, this guarantees the transistor switch is always saturated.
Again using the same values, find the minimum Base current required to turn the transistor “fully-ON” (saturated) for a load that requires 200mA of current when the input voltage is increased to 5.0V. Also calculate the new value of Rb.
Transistor Base current:

Transistor Base resistance:

Transistor switches are used for a wide variety of applications such as interfacing large current or high voltage devices like motors, relays or lamps to low voltage digital IC’s or logic gates like AND gates or OR gates.
Here, the output from a digital logic gate is only +5v but the device to be controlled may require a 12 or even 24 volts supply. Or the load such as a DC Motor may need to have its speed controlled using a series of pulses (Pulse Width Modulation). transistor switches will allow us to do this faster and more easily than with conventional mechanical switches.

The base resistor, Rb is required to limit the output current from the logic gate.
We can also use the PNP Transistors as a switch, the difference this time is that the load is connected to ground (0v) and the PNP transistor switches the power to it. To turn the PNP transistor operating as a switch “ON”, the Base terminal is connected to ground or zero volts (LOW) as shown.

The equations for calculating the Base resistance, collector current and voltages are exactly the same as for the previous NPN transistor switch. The difference this time is that we are switching power with a PNP Transistor (sourcing current) instead of switching ground with an NPN Transistor (sinking current).
Sometimes the DC current gain of the bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors are used. Here, one small input transistor is used to switch “ON” or “OFF” a much larger current handling output transistor.
To maximise the signal gain, the two transistors are connected in a “Complementary Gain Compounding Configuration” or what is more commonly known as a “Darlington Configuration” were the amplification factor is the product of the two individual transistors.
Darlington Transistors simply contain two individual bipolar NPN or PNP type transistors connected together so that the current gain of the first transistor is multiplied with that of the current gain of the second transistor to produce a device which acts like a single transistor with a very high current gain for a much smaller Base current.
The overall current gain Beta (β) or hfe value of a Darlington device is the product of the two individual gains of the transistors and is given as:
So Darlington Transistors with very high β values and high Collector currents are possible compared to a single transistor switch. For example, if the first input transistor has a current gain of 100 and the second switching transistor has a current gain of 50 then the total current gain will be 100 * 50 = 5000.
So for example, if our load current from above is 200mA, then the darlington base current is only 200mA/5000 = 40uA. A huge reduction from the previous 1mA for a single transistor.
An example of the two basic types of Darlington transistor configurations are given below.

The above NPN Darlington transistor switch configuration shows the collector terminals of the two transistors connected together with the emitter of the first transistor connected to the base terminal of the second transistor. Therefore, the emitter current of the first transistor becomes the base current of the second transistor switching it “ON”.
The first or “input” transistor receives the input signal to its base. This transistor amplifies it in the usual way and uses it to drive the second larger “output” transistors. The second transistor amplifies the signal again resulting in a very high current gain. One of the main characteristics of Darlington Transistors is their high current gains compared to single bipolar transistors.
The Sziklai configuration is similar to the Darlington pair except that it consists of two types of transistors, an NPN and a PNP. The Sziklai configuration is also known as a complementary Darlington pair. Current gain is about the same as in the standard NPN Darlington pair.
The difference is that the base current of transitor TR2 is the collector current of TR1 instead of the previous emitter current, as in the NPN Darlington arrangement. Nevertheless, the advantage of the Sziklai pair, compared to the standard Dalington pair, is that it takes less drive voltage to turn it “ON”. This is because it only has one base-emitter barrier potential to overcome, since it is effectively an equivalent to a single NPN transistor.
As well as its high increased current and voltage switching capabilities, another advantage of a “Darlington Transistor Switch” is in its high switching speeds making them ideal for use in inverter circuits, lighting circuits and DC motor or stepper motor control applications.
One difference to consider when using Darlington transistors over the conventional single bipolar types when using the transistor as a switch is that the Base-Emitter input voltage ( VBE ) needs to be higher at approx 1.4v for silicon devices, due to the series connection of the two PN junctions.
Then to summarise the tutorial when using a Transistor as a Switch the following conditions apply:
In the next tutorial about Transistors, we will look at the operation of the junction field effect transistor known commonly as an JFET. We will also plot the output characteristics curves commonly associated with JFET amplifier circuits as a function of source voltage to gate voltage.
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⁹excellent
Thank you for the great material. At the end of this section, you have a brief description of the Darlington pair and provide schematics for both the Darlington pair (NPN+NPN) and the Sziklai pair (NPN+PNP). However, the difference between these circuit configurations is not explained. You have a separate section about these circuits. It would be great to add a link here to that separate section. This would show the reader that a more detailed explanation is available.
This is the stupidest fake electronics website I ever seen so you say you don’t need to use a voltage divider to make a transistor turn on and off where are you getting your information from Pinocchio
I want to connect a transistor as an automatic switch
good
Very important note
Hey! Thanks for all the hard work you did to complile all these tutorials they are very helpful.
You guys know nothing about how to use a transistor as a switch I don’t know why you pretend to
This is exactly how you would use a transistor as a switch. Other tutorials teach similar methods.
Hi,
I’m new to this. I did some electronics in my physics subject long long time ago. I’m trying to revisit this area. Thanks you very much for all these tutorials.
I have a question regarding the overall collector current gain in the NPN Darlington configuration. I thought the overall collector current gain is β1+ β2+β1β2 instead of just β1β2.
Here is my calculation.
Assume the manufactured voltage drop between the base and emitter is 0.7V of the two silicon transistors. And β1 and β2 are their respective current gain. Therefore,
1. VBE1=0.7+0.7=1.4V as two forward biased diodes joined in series.
2. VBE2=0.7V
3. IB1=(Vin – VBE1)/RB, and IC1= IB1β1=β1(Vin – VBE1)/RB
4. IB2=IE1=IC1+ IB1=(β1+1)(Vin – VBE1)/RB
5. IC2=β2IB2=(β1+1)β2(Vin – VBE1)/RB
6. Total current flowing through RL=IC1+IC2. Hence, IC (total)=(β1+ β2+β1β2)(Vin – VBE1)/RB and the overall β gain is (IC1+IC2)/IB1=β1+ β2+β1β2
Appreciate if you can help clarifying this.
Best regards
As stated in our tutorials about Darlington Transistors, the total current gain is aproximately equal to: β1*β2. If the current gain of the two transistors is the same, then the total current gain of the Darlington pair is: β2
Oh, sorry. I haven’t gone that far to the tutorial from your link. Thanks for pointing that out. Cheers
This is very special knowledge for me.
Thanks for your good article , I think something new hits in my head so I could go forward in my project.
Interested
HI – GSVD-5884/ND28
When multiplexing of two CA 7 segment display, how to use only one pin of 8051 as switching between MSB and LSB display??
Use an Inverter (NOT Gate). Connect the output pin to the LSB display and input to inverter. Connect the output to the inverter to the MSB display. A logic 0 or 1 on the output pin will switch between the two displays.
Transistor proons .
It is very useful. …we can collect alot of information with alot of depth….I heart ful thank them for sharing the detailed information which is important for the learners….
Very interesting, you have break it in such away some one can understand it.thanks
nothing
PLEASE, HOW CAN YOU CALCULATE THE R RESISTANCE FOR THE FIRST EXAMPLE?
YOU DIDN´T MENTION IT IN YOUR COMENTS…
THANKS!
R = V/I, then if you know the voltage, and you can estimate the current, you can calculate the resistance
Quite educative…. It helped alot