Binary Outputs: Transistors Used as a Switch
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1 Binary Outputs: Transistors Used as a Switch Background A is able to output 5V at 25mA. Some devices require too much power for a to drive these directly. These include: A stepper motor which draws 1A at 12V An LED array (i.e. a tail light), which draws 12V and 200mA A heater, which draws 2.4A at 25V. If the device you're trying to drive requires more voltage or more current, you need a buffer circuit. This lecture covers three types of buffers: relays, transistors, and MOSFETs. Relays: A simple way to allow a to turn on and off a device is to use a relay. For example, the following is available from Marlin P. Jones ( Input: 3 32V < 1mA Output: 40 10A max If you connect the input to a, the can turn on and off anything you power with up to 480VDC. With a similar module, you can turn on and off anything powered by 120VAC as well. Solid State Relay from AC versions capable of driving up to 480VAC at 10A are also available. Transistors Transistors can also be used as a switch only they're much cheaper and smaller than a relay. Transistors are threeterminal devices and can be of type NPN or PNP. The main one we will use in the NPN transistor. The three terminals of an NPN transistor are called the collector, base, and emitter. Between the emitter and the collector is a reverse biased PN junction. Normally, this prevents current flow from the collector to the emitter (electrons, being negatively charged, flow the opposite direction). The switch is open. If you apply current to the base to the emitter, however (which is a forward biased PN junction), holes flow from the Ptype base and electrons flow from the Ntype emitter. JSG 1 September 16, 2016
2 Collector electrons N P N holes Base Emitter NPN Transistor: The base current controls the current from collector to emitter If the base is very thin, most of the electrons pass through the base and appear at the collector, creating current flow from the emitter to the collector. (The switch is closed). If the doping on the emitter side is times higher than the doping in the Ptype base, you get times more electrons than holes. This creates current amplification. The model for a transistor reflects this: Between the base and the emitter is a diode. It takes 0.7V to turn on a Silicon diode. Between the collector and the emitter is a currentcontrolled current source. The current flow is times the base to emitter current. C C B 1 B 0.7V B*Ibe E E Symbol and Circuit Model for an NPN Transistor: The arrow indicates a diode from base to emitter. The transistors we use in lab have parameters given in the following table: Zetex 1051A Zetex 968 Type NPN PNP NPN PNP Current Gain max(ic) 200mA 200mA 4A 4.5A max(vce) 40V 40V 40V 12V Vce(sat) 300mV 400mV 210mV 300mV price $0.037 $0.037 $0.73 $0.71 The current gain tells you what is. max(ic): When used as an "on" switch, this is the maximum current this transistor can conduct. max(vce): When used as an "off" switch, this is the maximum voltage the transistor can block. Vce(sat): When used as an "on" switch, the voltage drop across the transistor. Price: The reason we use the 3904 and 3906 transistors in lab. We go through a lot of them. JSG 2 September 16, 2016
3 Note that Vce depends upon the base current and the collector current. Normally, we assume this is 0.2V for all transistors as a ballpark estimate. Using a Transistor as a Switch To use an NPN transistor as a switch First, design a circuit which turns on your device including an extra 0.2V drop to model the losses in the transistor Next, break the path from power to ground, placing the NPN transistor in series with the emitter tied to ground. Finally, add a resistor to the base chosen so that I b I c Example: Design a circuit so that a processor can turn on and off a 1W LED at 100mA. Assume the LED has the following specification: Vf = 350mA mA 1W Prolight 1W High Power LED: $1.14 ea. Solution: First, design a circuit that drives the LED at 100mA along with a 0.2V drop. Assuming a 5V power supply, R c 5V 3.5V 0.2V 100mA 13 Next, add an NPN transistor (3904) in series with the emitter tied to ground. Finally, add so that meaning I c 100mA I b Ic 1mA Let Ib = 2mA. is then. When on, I b I c R b 5V 0.7V 2mA 2150 The exact value of isn't critical so long as I b I c. Round to 2.2k JSG 3 September 16, 2016
4 5V 5V 100mA Rc 13 Ic 100mA Rc V 0.2V 2.2k Ib 2mA 3.5V 0.2V Design of a Circuit to Drive 100mA Through the LED (left) and Final Circuit which Allows a to Turn On and Off the LED (right) One way to look at what's going on with this circuit is to look at the loadline relative to the transistor: When Ic = 0, Vce = 5V When Vce = 0V, Ic = 115mA The base current, Ib, controls the collector current, Ic as I c I b When Ib = 0, Ic = 0 and the transistor is off. This is easily done by outputting 0V from the processor. If I b is more than 100mA, the current Ic clips at 100mA. This is called the "on" state or the transistor is saturated (it can't draw any more current). Ice (ma) Saturated "On" beta*ib Active Region ( avoid ) Off Vce (Volts) Load Line for the NPN Transistor: Ideally you want to operate in the off or on (saturated) state. Both of these states are what you want for an on/off switch: Off: Ic = 0 JSG 4 September 16, 2016
5 On: Vce = 0 (ideally) or 0.2V (in practice) It also has the advantage that the power the transistor has to dissipate is zero at both of these states: The current is zero when off, meaning P = VI = 0 The voltage is zero when of, meaning P = VI = 0 (approximately) What you want to avoid is operating inbetween these two points where the power dissipated by the transistor is a maximum. This is why you design for I b I c : you want to make sure the transistor is saturated. Choosing I b I c gives you a safety margin. Example 2: Design a circuit so that the can drive an 8 Ohm speaker at 5V. Solution: 8 Ohms at 5V means you're trying to drive 625mA through the speaker I c 5V 0.2V 600mA. 8 To do this, you'll need to use a Zetex 1051 transistor. To turn on the speaker, the base current needs to be I b I c I b 600mA 300 2mA Let Ib = 3mA. Then R b 5V 0.7V 3mA 1433 A 1k resistor would work (it makes Ib slightly larger which still saturates the transistor). 5V Ic 600mA 8 Ohm speaker 1k Ib 3mA 0.2V NPN Zetex 1051 Circuit to allow a to drive an 8 Ohm speaker at 200mA (0.32W) JSG 5 September 16, 2016
6 Note when driving motors (inductive loads) If you have an inductive load, you need to add a flyback diode to the circuit. The problem is that when the transistor is turned on, energy is store in the magnetic field of the inductor as E 1 2 LI2 When the transistor turns off, the current goes to zero meaning the energy has to go somewhere. What happens is the voltage shoots up as V L di dt until it finds a path to ground. This is how sparkplugs work: the alternator is an inductor which stores energy. When the current is brought to zero, the magnetic field collapses and the voltage shoots up until it finds a path to ground with the spark plug being that path. For our transistor circuits, the path to ground is the transistor meaning you're going to fry the transistor when you turn it off. To limit the voltage, a flyback diode is used. This can be a diode tied to the power supply (limiting Vc to 12.7V for the circuit to the left) or a Zener diode can be used (limiting the voltage to 15V for the circuit on the right). 12V 12V Inductive Load L Flyback Diode Inductive Load L 1k Vc 1k Vc Flyback Diode 15V When you have an inductive load, you need to add a flyback diode to prevent frying the transistor when the transistor turns off. JSG 6 September 16, 2016
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