What does it do?
| The Darlington driver subsystem is an electronic switch that provides an output signal powerful enough to drive output subsystems requiring high current. |
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How does it operate?
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Inside a Darlington Driver
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The subsystem acts as an inverter; the output signal is the inverse of the input signal.
The darlington driver circuit consists of two transistors connected together to form a darlington pair.
The first transistor boosts the current to the second transistor, giving much more current gain than an ordinary transistor. |
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Darlington Driver circuit
Click on the circuit diagram to download a Livewire file of the circuit that you can investigate and add to your own circuit. |
To turn on a Darlington Driver the base voltage (the input voltage) needs to be 1.4V (twice the value needed for a single transistor.
While it is possible to construct a Darlington Driver using a pair of transistors, it is simpler and cheaper to use manufactured Darlington Drivers (with two transistors already combined inside them), such as the BCX38B, the more modern BCX38C and the MPSA13.
These are shown in a circuit diagram using the same symbol as a single transistor.
The BCX38C can provide up to 0.8A for a load. The MPSA13 can provide up to 0.4A. |
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The resistor R1 is used to limit the current from the previous subsystem. For use with a CMOS IC R1 should be about 22k. For use with a PIC R1 should be about 1k2. Full details on selecting this resistor are given in a section of this web site.
The output subsystem is connect between the supply rail (+Vs) and the output signal. The output subsystem is sometimes called the load resistance.
The resistor R2 is included to help with testing – it pulls the output signal up to Vs when the transistor is off. |
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ULN2803A Darlington Driver IC
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ICs are available that contain several Darlington Drivers. These are useful when more than one load is being driven. This is quite common when using a PIC.
As an example, the ULN2803A IC contains eight Darlington Drivers and each of these can provide up to 0.5A.
All the Darlington Drivers in the IC include a base resistor, so it is not necessary to include one. They also include a protective diode, so it is not necessary to add one to an output device that could generate a 'back emf', such as a motor, solenoid or relay. |
Possible applications
- Providing enough current to drive an output device that needs a relatively high current, such as a motor or solenoid.
Making
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Pins of the BCX38B, BCX38C and MPSA13
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How part of the PCB might look |
When using a Darlington Driver it is important to be clear about which pin is the collector (C), which is the base (B) and which is the emitter (E).
Manufacturer’s data sheets usually show the pins viewed from underneath (unlike IC pins, where the view shown is from above).
The PCB shows the basic circuit.
Build and test the unit that will provide the input signal before building the Darlington Driver.
Testing
Make sure that the signal going out (on the green PCB track) is the inverse of the input signal (on the blue PCB track).
Fault finding
If there is a fault, check that the Darlington Driver is connected the right way round. Then check the resistor values. Check the tracks and solder joints.
Alternatives
- Transistor – cheaper but provides less current
- MOSFET (Transducer driver) – can provide even more current but is more expensive
- The L293D driver can drive high current output devices and reverse two motors but is more expensive.
Web links
Return to list of datasheets
