A.C. circuits Part 2
2E3 12 Recall that current lags potential difference by 90° in an inductor and that current leads by 90° in a capacitor.
Re-writing the syllabus makes it easier to understand
Recall that potential difference leads current by 90°in an inductor and that current leads potential difference by 90° in a capacitor.
However the voltage across the capacitor will lag the current by 900 and the voltage can be determined by V = I x Xc
So you cannot just add the voltages but they have to be added using a vector diagram.
Memory aid: This section is quite difficult to grasp for some students so an old hand at Amateur Radio, at a recent BRATS club meeting, suggested the following was added to the course notes.
C I V I L
C I V......V I L
with a Capacitor the current leads the voltage the so I leads the V thus this gives us Capacitor IV
and with an Inductor voltage leads the current so V leads the I thus this gives VI L (Inductor)
With that in mine continue with the topic.
Capacitor - Current leads Potential Difference (C I V )
If the black curve now represents the current through a capacitor then the blue curve represents the potential difference across a capacitor. The current leads the potential difference by 900.
This capacitive reactance can be calculated by the formula XC= 1/2fC
f is in Hertz with C in Farads
Inductor - Potential difference leads the Current ( V I L )
If the black curve represents the potential difference across an inductor then the blue curve represents the current across an inductor. The potential difference leads the current by 900.
So let us explain further. When the Black curve is at position 0 the blue curve has not yet reached the same 0 position thus the black cure is said to lead the blue curve.
This inductive reactance can be calculated by the formula XL=2fL
f is in Hertz with L in Henrys
PD (voltage Wave forms)
Taking this further at the point of supply of AC the wave forms of PD potential difference (voltage) and Current are in phase but when it encounters a capacitor then from above the current leads the PD (voltage) CIV thus in fact across the capacitor the PD or voltage wave will lag behind the supply waveform.
Had it been an Inductor then the PD (voltage) leads the current VIL thus in fact across the Inductor the PD or voltage wave will lead the supply waveform.
Can you draw the appropriate wave forms ??
Recall that the term 'reactance' describes the opposition to current flow in a purely inductive or capacitive circuit where the phase difference between V and I is 90°.
2E3 12 Understand the formulae for the reactance of a capacitor or inductor in terms of frequency and component value. Calculate the unknown term given the other two.
The equations are :-
Reactance is the opposition to current flow in an AC circuit and can be loosely thought of as AC resistance and has values in ohms. This "AC resistance" is not actually resistance and it must be remembered that a capacitor stores energy electrostatically and an inductor stores energy magnetically.
Taking the case of the capacitor it can be loosely thought of as a very small rechargeable battery which accepts a charge and can then accept no more. On the next half cycle the capacitor then gives up that charge back to the supply and charges in the opposite direction. For this reason a "perfect capacitor" (one with no losses) does not get hot. It will be appreciated that the capacitor will charge and once charged can pass no more electricity hence it then apparently acts as an open circuit.
To differentiate from "normal" resistance we call that caused by the capacitor Capacitance Reactance and that by an inductor Inductive Reactance and both have values in ohms.
, in ohms, represents capacitive reactance in an AC circuit.
, in ohms, represents inductive reactance in an AC circuit.
In the formulae that follow they use certain symbols :-
= frequency in Hz = inductance in henries = capacitance in Farads
If we draw a graph of inductive reactance , against Frequency, we get :-
If we draw a similar graph for capacitive reactance ,against frequency we get:-
Now, if we have an LC circuit and draw both graphs on a common frequency line we get :-
We have spoken about the reactance of the coil (inductor) which is like resistance but the coil still has "normal" resistance and for these calculations we call this and identify it in the circuit diagram as a resistor.
Let == 10k and = 10R
10mV RMS at the resonant frequency.
as and cancel each other out, so =
Thus = = 1 mA
Voltage across or = x or
Thus Voltage across or =
Thus Voltage across or = 10 volts
The "gain" between the input and the output is known as magnification factor which in this case = 1000 (10mV rising to 10V = 1000 times) and thus the
magnification factor at resonance =
Assuming that we loosely couple a signal generator into the inductor of a parallel LC circuit and measure the voltage across with an RF voltmeter
is the resonant frequency giving a peak voltage, .
The points either side are generally accepted as limits of usable frequency - This range to is the bandwidth [eg. 12kHz for speech] for a parallel LC circuit, which could be the load in a power valve RF amplifier.
This measurement, requiring only a voltmeter, is used in a common form of Q meter. Q meters give a measure of "Goodness" of a component or circuit.
In this case =
ie the Bandwidth compared to the resonant frequency = SELECTIVITY
In the exam the sheet of equations shows this formula as
fC = centre frequency fU = the upper frequency and fL = the lower frequency
There is also something called Carson's rule which is related to the above
This defines the approximate modulation bandwidth required for a carrier signal that is frequency-modulated by a spectrum of frequencies rather than a single frequency.
The Carson bandwidth rule is expressed by the relation Bw = 2(Af+Δf )
Where Bw is the bandwidth requirement,
Af is the highest modulating frequency
Δf is the carrier peak deviation frequency
For example, Carson's rule would say that the bandwidth or speech (300Hz to 3kHz) with a peak deviation of 5kHz would be.
Bw = 2(3kHz+5kHz)
Longwave Coils (for interest only)
If we build a receiver for say 120kHz, the RF stage (if we still require 12kHz speech bandwidth) will have a measured "MAX"
= = =
NOTE: a of is very low !!
Providing is greater than a single figure the following basic statements are true with small and negligible errors.
= , the MAGNIFICATION FACTOR is =
The parallel resistance of a tuned circuit at resonance is x . This is known as the DYNAMIC resistance
The general expression for Dynamic Resistance is :-
= which does not include any frequency component.
It is obvious from this that the larger compared to , the higher will be the Dynamic Resistance and .
2E4 12 Understand the use of capacitors for AC coupling (DC blocking) and decoupling AC signals (including RF bypass) to ground.
It should be clear to you that from the construction of a capacitor (two separate plates with no link between them) provided no path for DC to pass. Thus it can be said that capacitors block DC and thus can be used for blocking DC in a circuit.
On the other hand in an AC circuit current appears to pass because of the build up and decay of the charge on one plate and then the other AC changes direction of its flow of electrons.
If therefore there is the possibility of an AC signal (and this include RF which is also AC) in a DC circuit then this can be channelled to ground via a capacitor and as such this is called decoupling hence the term "decoupling capacitor".
a look at page 35 Fig 5.19 and C3 is acting as
a "decoupling capacitor" as the circuit is amplifying AC
signals but any that might flow from the emitter of the
transistor will be taken to ground via the Capacitor C3 but
any DC signal will not be passed in stead it will travel to
ground through R4.
2E5 12 Understand the use of inductors for DC decoupling (AC blocking)
An inductor is a passive electronic component which is capable of storing electrical energy in the form of magnetic energy. Basically, it uses a conductor that is wound into a coil, and when electricity flows into the coil it will generate a magnetic field.
If we assume that an AC is flowing through the inductor. When current is about to flow to the inductor, the magnetic field generated by that current cuts across the other windings, giving rise to an induced voltage and thus preventing any changes in the current level.
These effects of the induced voltage
are produced even when the direction in which the current
is flowing is reversed. Before overcoming the induced
voltage that is attempting to block the current, the
direction of the current is reversed so that there is no
flow of current.