There are several equations used to convert one network to the other:
Δ and Y networks are seen frequently in 3-phase AC power systems (a topic covered in volume II of this book series), but even then they’re usually balanced networks (all resistors equal in value) and conversion from one to the other need not involve such complex calculations. When would the average technician ever need to use these equations?
Application of Δ and Y Conversion
A prime application for Δ-Y conversion is in the solution of unbalanced bridge circuits, such as the one below:
The solution of this circuit with Branch Current or Mesh Current analysis is fairly involved, and neither the Millman nor Superposition Theorems are of any help since there’s only one source of power. We could use Thevenin’s or Norton’s Theorem, treating R3 as our load, but what fun would that be?
If we were to treat resistors R1, R2, and R3 as being connected in a Δ configuration (Rab, Rac, and Rbc, respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit:
After the Δ-Y conversion . . .
If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration.
Resistors R4 and R5, of course, remain the same at 18 Ω and 12 Ω, respectively. Analyzing the circuit now as a series/parallel combination, we arrive at the following figures:
We must use the voltage drops figures from the table above to determine the voltages between points A, B, and C, seeing how they add up (or subtract, as is the case with the voltage between points B and C):
Now that we know these voltages, we can transfer them to the same points A, B, and C in the original bridge circuit:
Voltage drops across R4 and R5, of course, are exactly the same as they were in the converter circuit.
At this point, we could take these voltages and determine resistor currents through the repeated use of Ohm’s Law (I=E/R):
Nodal Voltage Analysis complements the previous mesh analysis in that it is equally powerful and based on the same concepts of matrix analysis. As its name implies, Nodal Voltage Analysis uses the “Nodal” equations of Kirchhoff’s first law to find the voltage potentials around the circuit.
So by adding together all these nodal voltages the net result will be equal to zero. Then, if there are “n” nodes in the circuit there will be “n-1” independent nodal equations and these alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchhoff’s first law equation, that is: “the currents entering a node are exactly equal in value to the currents leaving the node” then express each current in terms of the voltage across the branch. For “n” nodes, one node will be used as the reference node and all the other voltages will be referenced or measured with respect to this common node.
For example, consider the circuit from the previous section.
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three nodes are assumed to have voltages, Va, Vb and Vc with respect to node D. For example;
As Va = 10v and Vc = 20v , Vb can be easily found by:
again is the same value of 0.286 amps, we found using Kirchhoff’s Circuit Law in the previous tutorial.
From both Mesh and Nodal Analysis methods we have looked at so far, this is the simplest method of solving this particular circuit. Generally, nodal voltage analysis is more appropriate when there are a larger number of current sources around. The network is then defined as: [ I ] = [ Y ] [ V ] where [ I ] are the driving current sources, [ V ] are the nodal voltages to be found and [ Y ] is the admittance matrix of the network which operates on [ V ] to give [ I ].
Nodal Voltage Analysis Summary.
The basic procedure for solving Nodal Analysis equations is as follows:
1. Write down the current vectors, assuming currents into a node are positive. ie, a (N x 1) matrices for “N” independent nodes.
2. Write the admittance matrix [Y] of the network where:
Y11 = the total admittance of the first node.
Y22 = the total admittance of the second node.
RJK = the total admittance joining node J to node K.
3. For a network with “N” independent nodes, [Y] will be an (N x N) matrix and that Ynn will be positive and Yjk will be negative or zero value.
4. The voltage vector will be (N x L) and will list the “N” voltages to be found.
We have now seen that a number of theorems exist that simplify the analysis of linear circuits. In the next tutorial we will look at Thevenins Theorem which allows a network consisting of linear resistors and sources to be represented by an equivalent circuit with a single voltage source and a series resistance.
The above LR series circuit is connected across a constant voltage source, (the battery) and a switch. Assume that the switch, S is open until it is closed at a time t = 0, and then remains permanently closed producing a “step response” type voltage input. The current, i begins to flow through the circuit but does not rise rapidly to its maximum value of Imax as determined by the ratio of V / R (Ohms Law).
This limiting factor is due to the presence of the self induced emf within the inductor as a result of the growth of magnetic flux, (Lenz’s Law). After a time the voltage source neutralizes the effect of the self induced emf, the current flow becomes constant and the induced current and field are reduced to zero.
We can use Kirchhoff’s Voltage Law, (KVL) to define the individual voltage drops that exist around the circuit and then hopefully use it to give us an expression for the flow of current.
Kirchhoff’s voltage law (KVL) gives us:
The voltage drop across the resistor, R is I*R (Ohms Law).
The voltage drop across the inductor, L is by now our familiar expression L(di/dt)
Then the final expression for the individual voltage drops around the LR series circuit can be given as:
We can see that the voltage drop across the resistor depends upon the current, i, while the voltage drop across the inductor depends upon the rate of change of the current, di/dt. When the current is equal to zero, ( i = 0 ) at time t = 0 the above expression, which is also a first order differential equation, can be rewritten to give the value of the current at any instant of time as:
Expression for the Current in an LR Series Circuit
Where:
V is in Volts
R is in Ohms
L is in Henries
t is in Seconds
e is the base of the Natural Logarithm = 2.71828
The Time Constant, ( τ ) of the LR series circuit is given as L/R and in which V/R represents the final steady state current value after five time constant values. Once the current reaches this maximum steady state value at 5τ, the inductance of the coil has reduced to zero acting more like a short circuit and effectively removing it from the circuit.
Therefore the current flowing through the coil is limited only by the resistive element in Ohms of the coils windings. A graphical representation of the current growth representing the voltage/time characteristics of the circuit can be presented as.
Transient Curves for an LR Series Circuit
Since the voltage drop across the resistor, VR is equal to I*R (Ohms Law), it will have the same exponential growth and shape as the current. However, the voltage drop across the inductor, VL will have a value equal to: Ve(-Rt/L). Then the voltage across the inductor, VL will have an initial value equal to the battery voltage at time t = 0 or when the switch is first closed and then decays exponentially to zero as represented in the above curves.
The time required for the current flowing in the LR series circuit to reach its maximum steady state value is equivalent to about 5 time constants or 5τ. This time constant τ, is measured by τ = L/R, in seconds, where R is the value of the resistor in ohms and L is the value of the inductor in Henries. This then forms the basis of an RL charging circuit were 5τ can also be thought of as “5*(L/R)” or the transient time of the circuit.
The transient time of any inductive circuit is determined by the relationship between the inductance and the resistance. For example, for a fixed value resistance the larger the inductance the slower will be the transient time and therefore a longer time constant for the LR series circuit. Likewise, for a fixed value inductance the smaller the resistance value the longer the transient time.
However, for a fixed value inductance, by increasing the resistance value the transient time and therefore the time constant of the circuit becomes shorter. This is because as the resistance increases the circuit becomes more and more resistive as the value of the inductance becomes negligible compared to the resistance. If the value of the resistance is increased sufficiently large compared to the inductance the transient time would effectively be reduced to almost zero.
LR Series Circuit Example No1
A coil which has an inductance of 40mH and a resistance of 2Ω is connected together to form a LR series circuit. If they are connected to a 20V DC supply.
a). What will be the final steady state value of the current.
b) What will be the time constant of the RL series circuit.
c) What will be the transient time of the RL series circuit.
d) What will be the value of the induced emf after 10ms.
e) What will be the value of the circuit current one time constant after the switch is closed.
The Time Constant, τ of the circuit was calculated in question b) as being 20ms. Then the circuit current at this time is given as:
You may have noticed that the answer for question (e) which gives a value of 6.32 Amps at one time constant, is equal to 63.2% of the final steady state current value of 10 Amps we calculated in question (a). This value of 63.2% or 0.632 x IMAX also corresponds with the transient curves shown above.
Power in an LR Series Circuit
Then from above, the instantaneous rate at which the voltage source delivers power to the circuit is given as:
The instantaneous rate at which power is dissipated by the resistor in the form of heat is given as:
The rate at which energy is stored in the inductor in the form of magnetic potential energy is given as:
Then we can find the total power in a RL series circuit by multiplying by i and is therefore:
Where the first I2R term represents the power dissipated by the resistor in heat, and the second term represents the power absorbed by the inductor, its magnetic energy.
AC Inductor Phasor Diagram
These voltage and current waveforms show that for a purely inductive circuit the current lags the voltage by 90o. Likewise, we can also say that the voltage leads the current by 90o. Either way the general expression is that the current lags as shown in the vector diagram. Here the current vector and the voltage vector are shown displaced by 90o. The current lags the voltage.
We can also write this statement as, VL = 0o and IL = -90o with respect to the voltage, VL. If the voltage waveform is classed as a sine wave then the current, IL can be classed as a negative cosine and we can define the value of the current at any point in time as being:
Where: ω is in radians per second and t is in seconds.
Since the current always lags the voltage by 90o in a purely inductive circuit, we can find the phase of the current by knowing the phase of the voltage or vice versa. So if we know the value of VL, then IL must lag by 90o. Likewise, if we know the value of IL then VL must therefore lead by 90o. Then this ratio of voltage to current in an inductive circuit will produce an equation that defines the Inductive Reactance, XL of the coil.
Inductive Reactance
We can rewrite the above equation for inductive reactance into a more familiar form that uses the ordinary frequency of the supply instead of the angular frequency in radians, ω and this is given as:
Where: ƒ is the Frequency and L is the Inductance of the Coil and 2πƒ = ω.
From the above equation for inductive reactance, it can be seen that if either of the Frequency or Inductance was increased the overall inductive reactance value would also increase. As the frequency approaches infinity the inductors reactance would also increase to infinity acting like an open circuit.
However, as the frequency approaches zero or DC, the inductors reactance would decrease to zero, acting like a short circuit. This means then that inductive reactance is “proportional” to frequency.
In other words, inductive reactance increases with frequency resulting in XL being small at low frequencies and XL being high at high frequencies and this demonstrated in the following graph:
Inductive Reactance against Frequency
The slope shows that the “Inductive Reactance” of an inductor increases as the supply frequency across it increases.Therefore Inductive Reactance is proportional to frequency giving: ( XL α ƒ )
Then we can see that at DC an inductor has zero reactance (short-circuit), at high frequencies an inductor has infinite reactance (open-circuit).
A Flyback voltage or an Inductive Flyback is a voltage spike created by an inductor when its power supply is removed abruptly. The reason for this voltage spike is the fact that there cannot be an instant change to the current flowing through an Inductor.
Time Constant of the Inductor determines the rate at which the current can change through an inductor. This is similar to the time constant of a Capacitor, which determines the rate at which its voltage can change.
Time Constant of Inductor τ = L/R, where L is Inductance in Henries and R is series resistance in Ohms.
Similar to a capacitor, it takes nearly 5 time constants (5τ) to dissipate the current in an inductor.
Assume, in the above circuit, the inductor is 10H and the series resistor is 10Ω. So, when the switch is closed, maximum current flows through the inductor.
Now, let us see what happens when the switch is opened suddenly.
First, let us calculate the time constant. Using the formula of time constant and substituting the above assumed values, it is clear that the time constant is 1 second.
So, it will take approximately 5 seconds from the moment the switch is open, to completely stop the flow of current. This means that current flows in the circuit even after the switch is open (assuming it would take a few milli seconds to completely open the switch). How is this possible?
This can be understood from the inductors point of view. The switch gap, which is essentially air, is viewed as a huge resistor by the inductor and the resistance is in the order of few mega Ohms. This means that the circuit is still closed from inductor’s point of view with a huge resistor filling in the air gap.
Now that it is confirmed that the circuit is still closed, the inductor will try to dissipate the current and in order to do so, the inductor will drop voltage across the air gap resistance by reversing its polarity by using the energy stored in it in the form of magnetic field.
Now, the inductor tries to flow current as per its current dissipation curve. This could be problematic according to Ohm’s Law, V = I x R.
Even for a small current, when it is multiplied by the huge air resistance (few hundreds of Mega Ohms) will lead to a very high voltage across the air resistor. This is the origin of the Flyback Voltage or Voltage Spike.
Impact of Flyback Voltage on Switches
As there is no physical resistor when the switch is opened, sparks / arcs will occur between the switch and the other terminal, if a mechanical switch is used. All the energy from the arc is usually discharged across the contacts of the switch in the form of heat.
This could potentially damage the switches permanently or drastically reduce the lifetime of the switches. when speaking of switches, they can be mechanical switches or semiconductor switches like Transistors.
How Flyback Diode can prevent Voltage Spikes?
To protect the switch from being damaged due voltage spikes or inductive Flyback, a Flyback Diode or a Freewheeling Diode is used. The basic idea behind the use of a Flyback Diode is to provide an alternative path for the inductor for its current to flow.
The above image shows the same inductor circuit but with additionally a Flyback Diode. It is important to note that the diode is connected in reverse bias when the switch is closed.
As a result, the diode doesn’t impact the operation of the rest of the circuit when the switch is closed and maximum current flows through the inductor.
But when the switch is opened, the change in polarity of the inductor will make the diode to be placed in forward bias. Hence, the diode will allow current to flow at a rate determined by the time constant of the inductor.
The resistance of the diode, when it is forward biased, is very less and hence the voltage drop across the diode will be significantly less for the current to flow. This prevents arc at the switching device and as a result protects the switching device from damage.
eries Diode – By far the cheapest and easiest to install is a diode in series with the positive power lead. Anode toward the power supply and Cathode (banded end) going to the radio. For a few pennies a 1N400x series diode will handle many QRP rigs up to 1A and a 1N540x series will work up to 3A. Although simple and cheap it comes at a cost. The diode will drop about 0.6-0.9V which could result in lower transmit power and if the equipment uses voltage regulators this lower voltage may be under the regulator level.
Bridge Rectifier – The most foolproof for reverse polarity protection. For about a dollar or two, the bridge will correct the polarity no matter how you connect the power supply. With a bridge there are always 2 diodes conducting which can result in greater losses of round .8-1.5V. This works good only when a single power supply is used with a single radio but should be avoided when multiple pieces of equipment are connected together via the grounds.
Fuse with crowbar diode – With most protection circuits there is usually some form of loss in the process, as seen in the diode examples above. However, the fuse with crowbar diode has no voltage drop loss. The crowbar works by putting a diode across the voltage lines with the Cathode toward the positive side. This in effect leaves the path through the diode open unless the polarity is reversed. When that happens the diode now becomes an almost short circuit causing the fuse to blow. The cost of course if that a hefty diode needs to be used to handle the large current flow until the fuse opens up, not to mention a new fuse each time it’s tested.
MOSFET1 – Not only can MOFSET’s do a great job at switching, a P-channel MOSFET does an excellent job of reverse polarity protection. Depending on your needs the cost is from a few pennies to a dollar or two. Under normal circumstances the MOSFET is turned on with a small voltage drop across it, typically fractions less than the diode example above. When the polarity is reversed the MOSFET is forced off and creates an open circuit. See the sidebar on MOSFET selection for more information in selecting a P-channel MOSFET.
MOSFET RPP Rules:
VDS – Voltage Drain to Source – must be greater than the power supply voltage.
RDS(on) – Resistance of the Drain to Source when the MOSFET is turned on. This must be as low as possible. Usually the lower the Rds(on) the more expensive.
Vgss – Voltage from Gate to Source – The maximum voltage allowed from Gate to Source. This should also be greater than the power supply voltage. Many MOSFET’s will have a lower Vgss. In these cases the addition of a simple zener diode and resistor are required as follows:
Resistor is not critical, almost any value between 10K and 100K is good.
Zener voltage must be less than the MOSFET Vgss. Additionally it must be greater than the MOSFET Vgs(th) – Threshold Voltage from Gate to Source – This is the minimum voltage required to fully turn on the MOSFET.
There is a small capacitor in the Drain to Gate circuitry. This capacitor is optional but recommended and only used for static discharge protection.2
