First, I'm going to look at some of basics of transistors.
Types of Transistors:
There are two types of standard transistors, NPN and PNP, with different circuit symbols. The letters refer to the layers of the semiconductor material used to make the transistor.
A Darlington pair is two transistors connected together to give a very high current gain.
Transistor currents:
The diagram shows the two current paths through a transistor.The small base current controls the larger collector current.
When the switch is closed a small current flows into that base (B) of the transistor. It is just enough to make the LED B glow dimly. The transistor amplifies this small current to allow a larger current to flow through its collector (C) to its emitter (E). This collector current is large enough to make the LED C light brightly.
When the switch is open no base current flows, so the transistor switches off the collector current. Both LEDs are off.
A transistor amplifies current and can be used as a switch.
- The base-emitter junction behaves like a diode.
- A base current I.B flows only when the voltage V.BE across the base-emitter junction is 0.7V or more.
- The small base current I.B controls the large collector current Ic.
- Ic = h.FE * I.B (unless the transistor is full on and saturated) h.FE is the current gain (strictly the DC current gain), a typical value for h.FE is 100 (it has no units because it is a ratio).
- The collector-emitter resistance R.CE is controlled by the base current I.B:
I.B small R.CE reduced transistor partly on
I.B increased R.CE = 0 transistor full on (saturated)
- A resistor is often needed in series with the base connection to limit the base current I.B and prevent the transistor being damaged.
- Transistors have a maximum collector current Ic rating.
- A transistor that is full on (with R.CE = 0) is said to be saturated.
- When a transistor is saturated the collector-emitter voltage V.CE is reduced to almost 0V.
- The emitter current I.E = I.c + I.B, but I.c is much larger than I.B, so roughly I.E = I.c
This is two transistors connected together so that the current amplified by the first is amplified further by the second transistor. The overall current gain is equal to the individual gains multiplied together:
Darlington Pair current gain - h.FE = h.FE1 * h.FE2
(h.FE1 and h.FE2 are the gains of the individual transistors)
This gives the Darlington pair a very high current gain, such as 10000, so that only a tiny base current is required to make the pair switch on.
A Darlington pair behaves like a single transistor with a very high current gain. It has three leads (B, C and E) which are equivalent to the leads of a standard individual transistor. To turn on there must be 0.7V across both the base-emitter junctions which are connected in series inside the Darlington pair, therefore it requires 1,4V to turn on.
A Darlington pair is sufficiently sensitive to respond to the small current passed by your skin and it can be used to make a touch-switch as shown in the diagram. For this circuit which just lights a LED the two transistors can be any general purpose low power transistors. The 100k Ohm resistor protects the transistors if the contacts are linked with a piece of wire.
Using a transistor as a switch:
When a transistor is used as a switch it must be either OFF or fully ON. In the fully ON state the voltage V.CE across the transistor is almost zero and the transistor is said to be saturated because it cannot pass anymore collector current Ic. The output device switched by the transistor is usually called the 'load'. The power development in a switching transistor is very small:
- In the OFF state: power = Ic * V.CE, but Ic = 0, so the power is zero.
- In the full ON state: power = Ic * V.CE, but V.CE = (almost), so the power is very small.
Protection diode:
If the load is a motor, relay or solenoid (or any other device with a coil) a diode must be connected across the load to protect the transistor from the brief high voltage produced when the load is switched off.
The picture shows how a protection diode is connected 'backwards' across the load.
Current flowing through a coil creates a magnetic field which collapses suddenly when the current is switched off. The sudden collapse of the magnetic field induces a brief high voltage across the coil which is very likely to damage transistors or IC's. The protection diode allows the induced voltage to drive a grief current through the coil (and diode) so the magnetic field dies away quickly rather than instantly. This prevents the induced voltage becoming high enough to cause damage to transistors or IC's.
Advantages of relays:
- Relays can switch AC and DC, transistors can only switch DC.
- Relays can switch high voltages, transistors cannot.
- Relays are a better choice for switching large currents (>5A).
- Relay can switch many contacts at once.
- Relays are bulkier than transistors for switching small currents.
- Relays cannot switch rapidly, transistors can switch many times per second.
- Relays use more power due to the current flowing through the coil.
- Relays require more current than many IC's can provide, so a low power transistor may be needed to switch the current for the relay's coil.
Most ICs cannot supply large output currents so it may be neccessery to use a transistor to switch the larger current required for output devisec such as lamps, motors and relays. The 555 timer IC is unusual because it can supply a relatively large current of up to 200mA which is sufficient for some output devisec such as low current lamps, buzzers and many realy coils without needing to use a transistor.
A resistor R.B is required to limit the current flowing into the base of the transistor and prevent it being damaged. However, R.B must be sufficiently low to ensure that the transistor is thoroughly saturated to prevent it overheating. A safe rule is to make the base current I.B about five times larger than the value which should just saturate the transistor.
Choosing a suitable NPN transister:
The circuit diagram shows how to connect an NPN transister, this will switch on the load when the IC output is high.- The transistors maximum collector current Ic (max) must be greater than the load current Ic. load current Ic = supply voltage Vs / load resistance R.L
- The transisters minimum current gain h.FE (min) must be atleat five times the load current Ic devided by the maximim output current from the IC. h.FE (min) > 5 * (load current Ic / max. IC current)
- Choose a transistor which meets these requirements and make a note of its properties: Ic (max) and h.FE (min).
- Calculate an appropiate value for the base resister: R.B = (V.c * h.FE) / 5 * Ic (where Vc = IC supply voltage).
- Then choose the nearest standard value for the base resister.
The circuit diagram shows how to connect a PNP transistor, this will switch on the load when the IC output is low.
A transistor inverter (NOT gate):
Inverters (NOT gate) are availabkle on logic ICs but if you only require one inverter it is usually better to use this circuit. The output (voltage) is the inverse of the input signal:
- When the input is high the output is low.
- When the input is low the output is high.
Heres the link were I found all the information on transistors -
http://www.kpsec.freeuk.com/trancirc.htm
Well done Rich, I remember learning about these! Your one step ahead already!
ReplyDeleteThanks Mark, I just have to lesrn about H-bridge circuits and I'll be able to start hackibg :).
ReplyDelete