Applied Electronics - Outcome 2 - Operational Amplifiers

Operational Amplifiers are components which can be configured to have many uses.  You need to know about 6 of these configurations.

General Information about Op-Amps:

• They have an infinite gain (open loop)
• They have an infinite input impedance (resistance) and so no current flows into the Op-Amp
• They have no output impedance
• There is no difference in the potential between the two inputs

• The Op-Amp has two terminals: A non-inverting terminal (labelled +) and an inverting terminal (labelled -)
• It normally has a positive and a negative supply so all voltages are measured relative to ground.
• Like transistors and MOSFETs if the Op-Amp is fully switched on, it is saturated.

These pictures (from this website) show the pin out diagram of the 741 Op-Amp chip.

Op-Amps can be open loop or closed loop.  If open loop they have no feedback and therefore the gain is infinite.  The first open loop Op-Amp to consider is the comparator.

This compares the two inputs named here as Vi and Vref. Vi goes to the inverting terminal and Vref to the non-inverting terminal. 

• If Vi > Vref then the Op-Amp will saturate negatively so Vo = (85%) -Vcc
• If Vi < Vref then the Op-Amp will saturate positively so Vo = (85%) +Vcc

Due to the characteristics of the Op-Amp in reality, it will saturate at 85% of the supply.

The other open loop configuration you need to know about is the voltage follower.

The two simplest, closed loop Op-Amps are the non-inverting and inverting amplifiers.

Inverting:

The inverting amplifier has a feedback loop with a feed back resistor Rf.  The configuration of this and the input resistor, Ri gives the Op-Amp a gain.  This means that the input voltage is amplified to give the output voltage. 
The gain is found using the equation: Av = -(Rf/Ri)
The output voltage is found by multiplying the Vi by the gain: Vo = Av.Vi, so Vo = -(Rf/Ri) . Vi


Non-Inverting:

The non-inverting amplifier also has a feedback resistor to create a gain.
The gain is found using the equation Av = 1+Rf/Ri
The output voltage is therefore found using the equation Vo = 1+Rf/Ri . Vi

The most important thing to remember about the non-inverting amplifier is to remember about the 1!  This can sometimes make the non-inverting amplifier unsuitable as it always has a gain of more than 1 and you may need a decimal gain.  In this situation you may need to use two inverting amplifiers, one to give the gain and one with a gain of -1 to invert the signal back to a positive one.

Difference Amplifier:

The difference amplifier magnifies the difference between the the two inputs.

Summing Amplifier:

The summing amplifier is used to combine more than one input voltage to give an output voltage.  This is done by multiplying each input by a gain (Rf/Ri) and adding them together.  The inputs can be given different gains depending on their importance/input voltage.  i.e. if a input produces a small voltage, you can make the gain for that input bigger in relation to the others to give it equal importance.

In this way, a summing amplifier can be used as a digital to analogue converter.  Because a microprocessor can only give a high or a low signal, the output will always be either 5v or 0v.  So if you have a signal of 1001, decimal 9, the output voltage should reflect this.  By 'weighting' the input resistors you can create a DAC.

Tachogenerator investigation

This is my experiment to see if I can upload a video onto the blog from the iPad! Hopefully you can see the investigation we did today to see if the tachogenerator generated a linear output proportional to motor speed.

Case Study for Today

Today in our case study period you should have started looking at the Analysis and Description section of the case study.  This should be between 500 and 800 words.

Start by researching how your product works.  You should be able to draw (or find) a system diagram which helps describe the parts of your product and how they work together to produce your output.  Once you have a very general description, and a general system diagram, you should break your system down further to describe it in more detail.  A more detailed system diagram should then be described.

To help you check through this section you may find these questions useful:

• Are all my systems diagrams clear?
• Is there an overall, simple diagram with inputs, outputs and feedback?
• Are all the components clearly identified and analysed in a paragraph after the diagram?
• Has the system been fully analysed and explained?
• Has an evaluation of the technology involved been achieved?

Make sure that you keep a note of all the websites you have used to get information and/or pictures from.  You need to reference these properly at the end of your case study report.

Applied Electronics - Outcome 1 - MOSFETs

MOSFETs at a glance:

N-channel MOSFET

• MOSFETs are similar to transistors, but they are voltage operated devices. 
• This means that (ideally) it draws no current. (This means that you can use ohm's law in your calculations to find the Vin unlike a transistor where you must use the voltage divider equation because the transistor will draw current from the supply/voltage divider)
• They can be N-channel, or P-channel
• It has a very high input resistance, and no (low) current, therefore has a low power consumption
• It has a slower switching speed than a transistor
• It is very sensitive to static!  Beware when building with them!

•  MOSFETs operate above a threshold voltage VT.  When the voltage between the gate and the source (VGS) is above the threshold voltage, the MOSFET will saturate and ID will be its maximum and will not increase past this point.
• Until full saturation the ID will increase proportionally.  It is then possible to calculate the transconductance which is the ratio between ID and VGS as demonstrated below:

•  MOSFET circuit calculations involve mostly ohm's law.  This diagram shows the voltages dropped and the currents in the circuits:

 

So, a similar question to the homework - Use the diagram above to help you understand the labels and relationships between the voltages and currents.

This MOSFET has a Threshold voltage of 5v. When saturated the ID is 3mA and the source resistor is 2k (The supply voltage is 12v).  The current flowing through the voltage divider is 5uA.

Find the values of R1 and R2 (in the voltage divider) to ensure saturation.

So if ID = 3mA then IS will equal 3mA.  If the source resistance is known then the VS can be found.

VS = ISRS

        = 3x10-3 x 2x103 

        = 6v 

If we know VS and we know VGS (remember that if the MOSFET is saturated then VGS = VT) so we can find VG which is the voltage dropped over R2.

VG = VGS + VS

       = 3 + 6

       = 9v

If we know VG and we know that the current flowing through the voltage divider, then we can work out the two resistances.  Remember that we can use Ohm's law because no current goes into a MOSFET so it is all flowing through the voltage divider.

 VG = V2 = IR2

           9  = 5 x10-6 x R2  

         R2 = 9/5 x10-6

               = 1.8M Ω

V1 = VCC - V2              R1 = 3/5 x10-6

      = 12 - 9                 = 600kΩ

      = 3v 

Find the VDS if the Drain resistor is 500Ω.

 VD = ISRS                                      VDS = VCC - VD - VS

          = 3x10-3 x 500                 = 12 - 1.5 - 6

          = 1.5v                            = 4.5v

Applied Electronics - Outcome 1 - Transistors

In Higher we need to know about two configurations of Transistors, one to create a greater amplification of current, and one to allow two different things to happen depending on whether the input is positive or negative.

Darlington Pair:

The Darlington Pair uses two transistors with the emitter from one giving the base of the other (two in a row).  This gives double the current gain, but also needs double the VBE to saturate the pair.

Note: The easiest thing for you to forget is to carry out your transistor calculations with 1.4v instead of 0.7v.  Often this is the value that people most readily forget even in a single transistor!

Push Pull Driver:

The push pull driver uses the two different types of transistor so that when VBE is positive, the NPN transistor will saturate and when VBE is negative, the PNP transistor will saturate.

This allows the current to flow either from positive to 0v or from 0v to negative. If we connect a motor to the 0v (the emitters of each transistor) we can make it turn in both directions.  If we connect to different outputs to the collectors of each transistor, we can get a different output from a positive input signal than an negative one.

Applied Electronics - Outcome 1 - Revision of last year

The first part of Applied Electronics was a revision of Intermediate 2 and a reminder of the fundamental rules.

Also important is your ability to use Prefixes and symbols.  They are based on scientific notation where the powers are multiples of 3.  Your calculator should have a ENG button to put the numbers into this format for you.

It is also necessary to be able to use voltage dividers and to calculate values using the equation:

                                                         R2
Signal Voltage / Output Voltage = R1 +R2    x Vcc

You may need to combine the above with readings from the LDR or Thermistor (or thermcouple) graphs. These are logarithmic and so the axis aren't linear.  You have to take care when reading them.

To find a value, read along the axis of the value you know (you could be given either the temperature and asked to find the resistance, or the resistance and be asked to find the temperature)  This graph shows that at 25°c the resistance of a type 4 resistor is 50kΩ.

 


We also looked at transistors:

Transistors are current operated devices which perform two functions:

• An electronic switch
• Amplify current - so are used as transducer drivers

They are made up of layers of semi-conductive material which means that they will allow current to pass through them under certain conditions - in this case if a current is present at the base terminal. Depending on these layers transistors could be either NPN, or PNP but in Intermediate 2 we only need to look at NPN transistors.

Transistors have 3 legs:

The base leg is the "trigger" leg - if there is a current here, the transistor will saturate and allow current to flow between the collector and the emitter (in the direction of the arrow on the emitter).

The Transistor as a Switch

If there is no current at the base leg, the transistor material will act as an insulator and no current will flow.

If there is a current at the base the transistor material will act as a conductor and current will flow from the collector to the emitter.  (This is just like you pushing a switch with your finger, allowing current to flow through a switch.)

It only takes a small base current to switch on the transistor.  When it is fully switched on we say that the transistor is saturated.  The voltage required to produce a current which will saturate a transistor is 0.7V.

We need to be able to use the above fundamental principles and the following transistor notation to carry out calculations:

The Transistor as an Amplifier

The nature of the transistor will see that a small base/input current, will be amplified and the collector current will be a lot greater.  Technically the base current and the collector current join at the emitter but because IB is so small in comparison to IC we say that IC ~ IE

Consider this circuit.

Because the lamp is on, the transistor must be saturated and VBE = 0.7V.

The gain of the transistor is found using the equation HFE = IC / IB

So if IB was 8mA and the IC was 800mA, the gain of the transistor would be 100.  That means that it has amplified the input current by 100x.

Transistor Calculations:

Calculate the current gain of the transistor in this circuit:

First we can work out what we know:

VBE = 0.7V because the lamp is on, the transistor must be saturated.
VR = Vsignal - VBE
      = 5.3V

We can then work out the base current:

IB = VR/R
    = 5.3/1000
    = 5.3mA

And the collector current:

IC = VL/RL
    = 6/150
    = 40mA

Now that we know both the base current and the collector current we can work out the transistor gain:

HFE = IC/IB
        = 40 / 5.3
        = 7.55

Note that there is no unit for current gain.

Calculate the voltage dropped over the lamp in this circuit:

First we can work out what we know:

VBE = 0.7V because the lamp is on, the transistor must be saturated.
VR = Vsignal - VBE
      = 4.3V

 If we know the voltage dropped over the base resistor we can work out the base current:

IB = VR/R
    = 4.3/1000
    = 4.3mA

If we know the base current and the current gain we can work out the collector current:

IC = HFE x IB
     = 10 x 4.3x10-3
     = 43mA

If we know the current flowing through the lamp and its resistance we can work out the voltage dropped over it:

V = ICR
   = 43x10-3 x 150
   = 6.45V