Homework Answers

Here are the answers to the cover work questions.  If you don't understand anything, scroll down the blog to get more information, or come to ask me at lunchtime or after school.

First of all the simulation task you should have completed in class:

The input of the system is made using a voltage divider with a thermistor and a variable resistor.  Because this is a heat sensor the thermistor must go at the top because as the temperature increases the resistance decreases. As the resistance decreases, less voltage is dropped over the thermistor and therefore a larger share of the supply voltage is dropped over the variable resistor and therefore the signal voltage/voltage out of the voltage divider increases. The base resistor protects the transistor by limiting the current. When the voltage between the base and the emitter of the transistor (VBE) reaches 0.7V the transistor saturates, allowing current to flow from the collector to the emitter and therefore through the lamp, lighting it. The variable resistor is used to adjust sensitivity. Its resistance value could be changed to make the lamp light at different temperatures.  In this case if was adjusted to 1.8k to ensure the lamp would be on above 25.  This could be increased to make the lamp come on at a lower temperature, or decreased to make the lamp light at a higher temperature.

The answers to the theory questions are below:

HOMEWORK DEADLINE

You should complete the 4 past paper questions by:

Monday 27th June

Please be aware that in question 7 d) iii) (tumble drier) it asks for the value of a type 256-051.  This is from an older paper.  Use the type 1 thermistor.

HOMEWORK DEADLINE

You should be working to complete the cover sheets I left you on Friday by:

Tuesday 21st June

This includes a print out of your circuit with a description of how it works, and the answers to the theory questions on the double sided sheet. 

The content of this blog (keep scrolling down!) will help you answer these questions if you are unsure, or come back at lunchtime/after school to ask any questions.

Transistor Circuits - Inputs and Outputs

As mentioned before transistors are used as transducer drivers.

Lets consider some systems.

An automatic street lamp will come on at night when the light level gets too low.
We can draw a systems diagram:

We know how to make a light sensor using a voltage divider with a LDR.
As the light level decreases the resistance of an LDR increases, and so the voltage dropped over it will increase.  This means that there will be a higher voltage signal in the dark if the LDR is at the bottom of the voltage divider.

Using a variable resistor as the top resistor will give you flexibility in the light level required to switch on the lamp.

The signal voltage produced by the light sensor will depend on the light level - this is the signal to the transducer driver (as in the system diagram) so next we have to add the transistor.

We can then lastly add the lamp.

This is how to take a system diagram to a circuit diagram.

Some output transducers need a bigger voltage or current than the transistor/sensing circuit can provide.  For this we need to use a relay - just as we did in the telescope project to make the motor turn because the Alphaboard circuit provided 5V and the motor required 12V.

This page in your notes (P39) shows the relay as a magnetic switch. This means that there is no physical connection between circuits with different power supplies, a magnet pulls a switch closed to complete a higher power circuit.

When a relay switches it creates a large ElectoMagnetic Force (EMF) which creates a very large current.  If this flows back into the transistor it will destroy it and your circuit wouldn't work. Therefore we have to add a diode over the relay to stop the back EMF destroying the transistor.

So our street lamp may require a bigger supply voltage than the sensing circuit.  So we can add a relay in to ensure the lamp lights brightly.

Note the diode.  It protects the transistor from back EMF as the relay switches.

Motors often need greater current/voltage to turn them, so we can use a relay:

This circuit uses a light sensor as its input made using a voltage divider with a LDR at the top and a variable resistor at the bottom.  So as the light level increases the resistance of the LDR will decrease, decreasing the voltage dropped over it, therefore increasing the voltage dropped over the variable resistor and the voltage out of the voltage divider.  When the voltage between the base and emitter (VBE) is 0.7v the transistor will saturate, allowing current to flow from the collector to the emitter, energizing the relay.  This will pull the switch closed in the motor circuit and the motor will turn.

Note that the motor circuit is connected using only one leg of the relay switch, so that when the relay is not energized nothing will happen.

However using the Single Pole double throw relay above only allows the motor to turn in one direction.  We can use a Double Pole Double Throw relay to allow the motor to turn in both directions: one when it is dark and one when it is light:

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.

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

Power in Electrical Circuits

Power is the rate at which an electrical circuit transfers electrical energy to another form - i.e. if we had a bulb it would be the rate at which electrical energy is transferred to heat and light energy.  It is dependent on the voltage and current flowing in the circuit.

Power is measured in watts.

Joule's law states that

P = IV 

where P = Power (W) I = Current (A) and V = Voltage (V)

We may not know current in our circuit, but instead we might know the resistance.  So we can combine this law, and Ohm's law to make different permutations of the same formula:

P = IV,  and V = IR  so P = I x I x R   so we can use the formula P = I2R

P = IV and I = V/R so P = V/R x V   so we can use the formula P = V2/R

Lets consider these circuits:

Combined Circuits - Series and Parallel

We have looked at series circuits and we have looked at parallel circuits, so now we must consider circuits which have elements of both.

You need to remember all the rules for each type of circuit:

In a series circuit the total of all the voltages dropped in the circuit must equal the supply.

In a parallel circuit the voltage dropped in each branch is the same.

In a series circuit the current is the same at all points.

In a parallel circuit the currents in each branch are added to give the total circuit current OR from Kirchoff's law, the current entering a node must equal the currents exiting a node.

In a series circuits all the resistances are added to give the total resistance.

In a parallel circuit the reciprocal  (1/Ω) are added, or if there are only two resistors in parallel the special total resistance formula can be used: Rt = Product/Sum.

Lets consider this circuit.

To work out where the series and parallel parts are follow your finger around the circuit.  When you reach a node (junction/join) you know that you have come to a parallel branch.

First you should work out the equivalent resistance of the parallel resistors.  This, put simply, is finding out what resistance value these two have created, or working out the single value of resistor which could replace these two.

This makes it easier to see that these two parts of the circuit are in series with each other, so we can then work out the total resistance:

Knowing the total resistance we can find the total circuit current:

So going back to the circuit as it was, we can see that the circuit current flows through R1, and it then splits.

So we can work out the voltage dropped over R1 and from that work out the voltage dropped over the parallel branches (remember that this will be the same voltage in each branch):

Knowing the voltage dropped over each of the parallel resistors and the resistance, we can work out the current using ohm's law.

And finally we can check if our current calculations are correct using Kirchoff's Law.