Difference between revisions of "Test"
From Mech
Jump to navigationJump to search| Line 1: | Line 1: | ||
==Diodes== |
|||
Symbol for a diode: [[Image:diode_symbol.gif]] |
|||
We can take advantage of the properties of a p-n junction to make a '''diode''', which is an electrical component that only allows current flow in one direction. A diode made of silicon needs about 0.7V across it in the forward direction ('''forward biased''') in order to conduct. At a large negative voltage ('''reverse biased'''), the diode fails and the current forces its way though. This is known as '''diode breakdown'''. |
|||
[[Image:diode_N4002_profile.gif]] |
|||
===Applications of Diodes=== |
|||
====Half-Wave Rectifier==== |
|||
A half-wave rectifier cuts off half of a sine wave, leaving only the positive or negative side (although not perfectly, because of the diode voltage drop). |
|||
The schematic for a simple rectifier: |
|||
[[Image:half_wave_rectifier_schematic.gif]] |
|||
The graph for half-wave rectifier: |
|||
[[Image:half_wave_rectifier_graph.gif]] |
|||
====Flyback Diode==== |
|||
An approximate model of a DC motor is a resistor and inductor in series. (For more details on DC motors, see the section on [[Brushed_DC_Motor_Theory|brushed DC motors]].) If there is a current <math>i</math> through the motor, and we suddenly break the circuit by opening a switch, for instance, there will be a spike in the voltage across the inductor. This is because the current has to instantly go to zero (there is no longer a path for it), and the voltage across the inductor is proportional to the rate of change of current, |
|||
<math>v_L = L (di/dt)</math>. This may exhibit itself in a spark at the switch. A mechanical analog of instantly zeroing the current through an inductor is bringing the velocity of a moving mass to zero, instantly. |
|||
To alleviate this problem, we can put a '''flyback diode''' across the motor terminals, as shown below: |
|||
[[Image:diode_flyback_schematic.gif]] |
|||
This diode allows current to continue flowing through the motor, satisfying the inductor's "desire" to continue to allow current to flow. The energy in the inductor is dissipated over time by the resistance of the motor, and the current drops smoothly to zero. Note that no current flows through the diode when the switch is closed, as it is reverse biased. |
|||
The flyback diode must be sufficiently large to carry the potentially large currents of the motor. |
|||
'''Question''': If the voltage supply is +10V relative to ground and the diode drop is 0.7V, what is the voltage across the inductor immediately after the switch is opened? Make sure you have the sign right, where positive voltage means the left side has a higher potential than the right side. (Remember, the voltage across an inductor can change discontinuously, but not the current, if there is any possible path for the current.) |
|||
====Peak Detector==== |
|||
We can make a peak detector if we hook up our circuit like this: |
|||
[[Image:peak_detector_schematic.gif]] |
|||
The graph of the input and output voltages looks like this: |
|||
[[Image:peak_detector_graph.gif]] |
|||
Each time the input voltage dips, the diode prevents the capacitor from draining. |
|||
There are two problems with this circuit: First, the voltage at <math>V_{out}</math> is always one diode drop less than the maximum <math>V_{in}</math>. Second, the circuit has a very low impedance, and the capacitor drains a lot of current. A better peak detector can be built with op-amps. |
|||
====Voltage Clamp==== |
|||
A voltage clamp limits the output voltage, which is useful when we need to protect circuits from high voltages. We specify the maximum and minimum voltage we want by applying voltages across a diode. We must also take into account the voltage drop, so the voltages we apply should be <math>V_{max}-0.7</math> and <math>V_{min}+0.7</math> |
|||
[[Image:diode_voltage_clamp.gif]] |
|||
In this circuit, if <math>V_{in}</math> exceeds <math>V_{max}</math>, diode <math>D_1</math> becomes forward biased and conducts, thus forcing <math>V_{out}</math> to stay at <math>V_{max}</math>. If <math>V_{in}</math> drops lower than <math>V_{min}</math>, diode <math>D_2</math> conducts and prevents the <math>V_{out}</math> from dropping any lower. |
|||
==Zener Diodes== |
|||
symbol for a zener diode:[[Image:zener_diode_symbol.gif]] |
|||
Zener diodes have a low breakdown voltage, or zener voltage, and can allow reverse current to flow without being destroyed. A good zener diode can also maintain a fairly constant voltage over a large range of currents. The voltage-current graph of a zener diode looks like this: |
|||
[[Image:zener_graph.gif]] |
|||
===Voltage Regulator=== |
|||
We can use a zener diode to build a voltage regulator like this: |
|||
[[Image:zener_voltage_regulator.gif]] |
|||
<math>V_{out}</math> is the '''reverse breakdown voltage''' or '''zener voltage'''. In the circuit, ''V'' is greater than ''V<sub>Z</sub>''. This voltage regulator can only step down the voltage, not up. |
|||
Even if the current through the diode fluctuates, the voltage remains fairly stable. Choose a resistance R large enough such that the power specs of the zener diode are not exceeded, but not so large that it limits our current too much. |
|||
For example, if our diode has a zener voltage of 5V and can handle 1W, and our voltage source has a maximum voltage of 20V: |
|||
The largest amount of current our diode ever has to handle is the current through the diode when there is no external load. Therefore: |
|||
<math>I_{z(max)}=\frac{(20-5)V}{R}</math> |
|||
<math>P=IV=I(5V)=\left(\frac{(20-5)V}{R}\right)(5V)=1W</math> |
|||
<math>R=\frac{(15V)(5V)}{1W}=75 \Omega</math> |
|||
This means that we need at least 75Ω to protect our diode. Keep in mind that we must meet the power ratings for our resistor as well. |
|||
Because current is flowing through the resistor and diode even when there is no load, power is being wasted. We can do better with voltage regulator chips like a 7805 chip (78xx for xx volts). |
|||
==LEDs (Light Emitting Diode)== |
|||
symbol for an LED: [[Image:LED_symbol.gif]] |
|||
An LED is a type of diode that emits visible or infrared light proportional to the current passing through it (up to a maximum before it burns out). |
|||
A typical LED shines its brightest between 5mA and 30mA. Beyond its rated current, it may burn out. |
|||
[[Image:LED_photo.jpg|thumb|right|A photograph of an LED]] |
|||
Notice that on the LED pictured at the right, one of the legs is longer. The longer leg is the positive terminal. However, do not rely on the length of the leg to determine the polarity, because you will probably snip the legs short later on. Instead, look at the side of the plastic lens, and notice that it's not perfectly round; one side is flattened. The flattened side is the ground. |
|||
[[Image:LED_inside_photo.jpg]] |
|||
<br clear="all"> |
|||
==Photodiode== |
|||
Symbol for a photodiode:[[Image:photodiode_symbol.gif]] |
|||
Go to the [[Photodiodes_and_Phototransistors#Photodiodes|photodiode]] page for information on photodiodes. |
|||
==Bipolar Junction Transistors (BJT)== |
|||
The symbols for transistors:[[Image:transistor_symbol.gif]] |
|||
There are two types of of BJTs; n-p-n and p-n-p: |
|||
<b><u>npn</u></b> |
|||
[[Image:npn_transistor_diagram.gif]] |
|||
<b><u>pnp</u></b> |
|||
[[Image:pnp_transistor_diagram.gif]] |
|||
For the transistors, we have the following relationship for the voltages and curents: |
|||
<math>I_E=I_C+I_B</math> |
|||
<math>V_{BE}=V_B-V_E</math> |
|||
<math>V_{CE}=V_C-V_E</math> |
|||
Under normal operation, ''BE'' is forward biased, and ''CB'' is reverse biased. |
|||
BJTs have three regions of operation: |
|||
Cutoff (<math>V_{BE}<0.6V</math>): <math>I_C=0</math> (no collector current) |
|||
Active(<math>V_{BE}</math>≈<math>0.6V</math>): <math>I_C=</math>''β''<math>I_B</math> (collector current is proportional to base current by factor ''β'', which is typically about 100) |
|||
Saturated(<math>V_{BE}</math>≈<math>0.7V</math>): <math>V_{CE}</math> ≈ <math>0.2V</math> (for npn); collector current is mainly limited by its subsequent circuitry. |
|||
===Applications of BJTs=== |
|||
====Common Emitter Amplifier Circuit==== |
|||
In a common emitter circuit, the emitter of a transistor is grounded and an input voltage is applied to the resistor at the base. An example common emitter circuit with a transistor (''β''=100) is drawn below: |
|||
[[Image:transistor_common_emitter.gif]] |
|||
If <math>V_{in}</math> < 0.7V (same as the diode drop), the p-n junction is not forward biased, there is no base current, and <math>I_C</math> is equal to 0. |
|||
If <math>V_{in}</math> > 0.7V, then |
|||
<math>I_B=(V_{in}-V_B)/10000</math>''Ω''<math>=(V_{in}-0.7)/10000</math>''Ω'' |
|||
<math>I_C=</math>''β''<math>(V_{in}-0.7)/10000</math>''Ω'' |
|||
Now we calculate how large <math>V_{in}</math> has to be to saturate the transistor. When <math>V_{CE}=0.2V</math>, the transistor becomes saturated. If the drop across the collector and emitter of the transistor is equal to 0.2V, then the drop across the resistor at the collector must equal 9.8V. Thus: |
|||
<math>I_C=9.8/1000</math>''Ω''<math>=9.8mA</math> |
|||
<math>I_B=I_C/</math>''β''<math>=0.098mA</math> |
|||
Then we find <math>V_{in}</math>: |
|||
<math>V_{in}-0.7=(0.098mA)(10000</math>''Ω''<math>)</math> |
|||
<math>V_{in}=1.68V</math> or greater. |
|||
'''cutoff''':<math>V_{in}</math> < 0.7V |
|||
'''active''':0.7V < <math>V_{in}</math> <1.68V |
|||
'''saturated''':1.68V < <math>V_{in}</math> |
|||
If we are using the transistor as a switch, we should fully saturate it to save power and reduce heat. |
|||
====Emitter Follower==== |
|||
The emitter follower is a very simple current amplifier, taking advantage of the fact that <math>I_C=</math>''β''<math>I_B</math>. The high input impedance means that our imput signal does not need to "work" as hard. |
|||
[[Image:transistor_emitter_follower.gif]] |
|||
<math>V_E=V_{in}-0.7V</math> |
|||
====Push-Pull Follower==== |
|||
If we put two emitter followers together (npn transistor for positive voltages; pnp transistor for negative voltages), we get a push-pull follower. |
|||
[[Image:transistor_push_pull_follower.gif]] |
[[Image:transistor_push_pull_follower.gif]] |
||
Revision as of 16:18, 18 August 2009
Beware that there is a “dead band” when Vin is between 0.7V and -0.7V, because the voltage has not pass the cutoff voltage for either transistor. This is known as crossover distortion.
Push-Pull Follower Example: A (Poor) Motor Speed Controller
Suppose we want to control the speed of a motor by varying the input voltage between +5V and -5V. We attempt to do this by setting put a voltage divider with a potentiometer like in the circuit below.
