| 4. Sensing Speed |
The operation of a train at the correct speed is fundamental to rail safety. There have been many accidents that have been the result of excessive speeds. Old locomotives used to detect the speed mechanically. Modern machines do it electronically.
A toothed wheel on the end of one of the axles passes through the poles of an electromagnet, and minute variations in the current are detected. They are cleaned up and passed to be processed electronically. This arrangement is shown below:
The signals are combined with a measured pulse using a logic gate called an AND gate. A pulse counter works out how many pulses there are per second; a calibrated read out is given to the driver. The schematic shows the idea:
At the heart of every timing circuit there is a capacitor, including the one that produces the pulse exactly one second long that is in the schematic above. A capacitor is a short-term charge store. At its simplest it is two parallel plates separated by air.
The diagram above shows how the capacitor works. In most capacitors a slice of insulating material called a dielectric is put between the plates. This increases the capacitance.
The symbol for a capacitor is shown below:
High value capacitors are electrolytic. Electrolytic capacitors need to be connected with the correct polarity. If not, they can explode. Polarity is not important with non-electrolytic capacitors.
When charge is separated on a capacitor, a potential difference results. The potential difference is proportional to the charge, so we can write:
Q µ V
We can show this on a graph:
The graph is a straight line. We find that the gradient is the capacitance, which is defined as:
the charge per unit potential difference across the plates.
Capacitance is measured in Farads (F), where 1 F = 1 C V-1
Now that we know the constant of proportionality, we can write the equation as:
Q = CV
in which the physics codes are Q - Charge (C), C - Capacitance (F), and V - potential difference (V).
Be
careful not to confuse "C" meaning coulombs with "C" the physics code for
capacitance.
A 1 F capacitor is a very large capacitor (they do exist). Electronic engineers use microfarads (mF), nanofarads (nF) and picofarads (pF).
1 mF = 1 × 10-6 F
1 nF = 1 × 10-9 F
1pF = 1 × 10-12 F
Capacitors have a working voltage. If the voltage is bigger than this, the dielectric can break down and the capacitor conducts. The capacitor is wrecked.
When a capacitor is discharged, the rate of discharge is proportional to the charge left on it. The means that the discharge is exponential. The graph below shows the voltage plotted against time. We use the voltage to represent the charge since Q = CV.
We can see that it takes 31 seconds for the voltage to fall to 50 % of its original value. It takes a further 31 seconds for the voltage to fall to 25 % of its original value. So we can say that the voltage falls by an equal fraction in equal time.
Using a derivation that uses the mathematical trick of calculus, we get the equation:
V = V0 e-t/RC
The physics codes are as follows:
V - voltage at any time;
V0 - the voltage at the start;
t - the time (s);
R - the resistance of the resistor through which the capacitor discharges (W);
C - the capacitance of the capacitance (F);
e - the exponential number, 2.718...
The charge follows the same pattern:
Q = Q0 e-t/RC
The current follows the same pattern:
I = I0 e-t/RC
The product between resistance and capacitance (RC) that appears in the formula is called the time constant and is measured in seconds. It may seem strange that ohms × farads = seconds, but it's true. If we multiply the base units of ohms and farads, we end up with seconds.
To manipulate the formula you need to understand about natural logarithms. If you are not sure, refer to the maths pages in the book, ask your teacher, or another student. If you are doing Mathematics A level, you will be able to do it standing on your head!
The RC time constant is used by electronics engineers. It is the time taken for the voltage to fall to 37 % of its original value. We will see how. Let t = RC.
Basic equation first:
V = V0 e-t/RC
Rearrange and substitute:
V = V0 e-RC/RC
So:
V = V0 e-1
Since e-1 = 0.3678 = 0.37, we can write:
V = 0.37 V0
We can also look at the "half-life", the time taken for the charge to fall to half its original value:
V = V0 e-t/RC
Rearranging:
V = e-t/RC = 0.5
V0
To get rid of the e we need to take natural logarithms (written ln or loge):
ln(0.5) = -t/RC
Therefore:
-0.693 = -t/RC
The minus signs cancel so we are left with:
t = 0.693 RC
So the "half life" is about 69 % of the time constant.
This is an important electronic timing circuit. It is an unspectacular 8-pin chip which can be bought for a few pence. Its pin-out arrangement is this:
We will consider the 555-timer in monostable mode, the circuit for this being:
The circuit will give a single output pulse like this:
The components R and C determine the time period T of the output pulse.
T ≈ 1.1 RC
Once triggered, the circuit cannot be re-triggered to extend the period T.
With suitable values for components, we can get the 555-timer to give a pulse exactly 1 second long that we wanted at the start of this section.
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