Power
Energy is the capacity to do something. In physics, the law of conservation of energy states that energy cannot be created or destroyed (by natural means). However, energy can be converted from one form to another. The rate which energy is converted is called power, measured in Joules per second which we also call Watts (W).
Power is calculated by multiplying the current (coulombs per seconds) by the voltage (joules per coulomb):
Electronic circuits need a source of power. We group these power sources into two main groups based on how they function.
Voltage Source
Voltage sources maintain a constant potential difference (voltage) between its two terminals (connection points). The symbol for a voltage source is:
Voltage sources supply enough current to the circuit so that the voltage can be maintained. Ideally, voltage sources would maintain this voltage no matter what, but practically speaking this is not possible. In reality, voltage sources are made from more complicated electronic circuits (LDO, switching converters), or they are simply batteries. Either way, the circuit will exhibit an effect where the voltage drops as more current is drawn from the source. Mathematically, this can be stated as the ideal voltage (y-intercept) and a line with a decreasing slope:
This can be represented in the circuit as a voltage source and resistor:
The effect of the voltage V becoming lower as more current is draw us called the output resistance (more about this in a future blog). Ideally this resistance would be zero.
Current Source
Similarly, current sources maintain a constant current through itself. The symbol for a current source is:
It maintains a constant current by adjusting the voltage over itself. Remember it is always the voltage that causes charge to flow. A current source just controls the voltage so that it maintains the desired current. Practically speaking, there are semiconductor effects which will cause the current to depend slightly on the voltage of the circuit that it is connected to. Is represents the ideal current and Ip represents the non-ideal effects of the semiconductor that it is made of:
The can be represented by a current source which supplies the ideal current, and a resistor (which is dependant on the voltage over it):
Dependant Sources
Sources can either be fixed or depend on another value in the circuit. There are 4 types of controlled sources:
- Voltage Controlled Voltage Source (VCVS)
- Current Controlled Voltage Source (CCVS)
- Voltage Controlled Current Source (VCCS)
- Current Controlled Current Source (CCCS)
The most common controlled source is voltaged controlled current source, as it is the basis for how transistors operate. In the circuit below, we have a circuit with a fixed valued voltage source, and a voltage depantant current source. Find V3. All resistors are 100 Ω.
This is acomplished by solving for V2 by first finding the total current in the loop using Ohm's law. No current will flow in the wire between the two loops; all the current though the 10V sourse goes through R1 and R2 and back to the 10V source. Using Ohm's law, this gives:
Next, we use Ohm's law again to find V2:
Now, we know the current in the dependant current source is:
Finally, using Ohm's law we can find the voltage V3:
In the future, we will see that this is the basic principle on how transistor amplifiers work.
Power Dissipation
Resistors are components that absorb power. When this power is absorbed, it is converted into heat.
The power that is absorbed by a resistor is calculated using the power equation. We can substitute Ohm's law in two of its forms (2) and (3) into the power equation (1) and get two useful equation for calculating power.
In the previous example, find the power dissipated in the resistor R3.
Since we know the voltage, current, and resistance we can use any three of the power equations.
Maximum Power Rating
Resistors are designed with maximum power ratings. This is the maximum amount of power that can be absorbed and dissiapted as heat by the resistor without causing any issues. If this limit is exceeded, then the ressitor may burn up, causing a short or open circuit to form.
Short Circuits
A short circuit is a path of no (or very little) resistance between an element that is supplying power like a voltage source will and the ground node or between two elements that are supplying power.
This will result in a theoretically infinite current, which will cause the circuit to burn up.
Open Circuits
An open circuit is a path that is broken or intentionally does not exist so that no current can flow. This can be a good thing or a bad thing depending on what is desired.
Example 1
A 9V battery (represented by a voltage source and series resistor) has an internal resistance of 2Ω and is connected directly to a 10Ω resistor. How much power will be dissipated by the battery alone?
We can find the current using Ohm's law as usual (see previous post):
Now, we can use the current to find the amount of power dissipated by the battery.
Note: If you want to use V²/R to find the power or I*V, do not not use 9V as the voltage, instead calculate the voltage over the 2Ω resistor.
Example 2
A non-ideal current source is connected to the resistor network below. Find the power dissipated by resistor R3. Assume all resistors are 100Ω.
What we need to do is to find the current through R3. We can do this by first simplifying the circuit. R3 and R5 are in series, this simplifies the circuit to:
Now, combining the new equivalent resistor R7 (200Ω) with R4:
Finally, adding up R2, R8 (66.66Ω) and R6:
R9 is calculated to be 266,66Ω. Finally, these are in parallel. This gives one simple resistor of 72.72Ω.
The current from the 0.1A source through a 72.7Ω resistor, using Ohm's Law gives a voltage of 7.27V.
Now, we need to go back to our original circuit.
If we find the current though R1 by Ohm's law, then we can subtract it from the 0.1A source using KCL and find the current through R2.
Now we can use Ohm's law and KVL to solve for Vx and Vz. For Vx we take Vw and subtract the drop on R2:
Next, for Vz we know that I2 rejoins and flows though R6 to ground, so we can use Ohm's law again:
Finally, we can solve for I3 using Ohm's law again. Here is a simplified diagram.
We simply take the voltage over BOTH resistors R3 and R5 divided by the resistance value of BOTH V3 and V5. This is easier than using I3=V3/R3 since we do not know V3. By KVL, we know V3 + V5 is equal to the the potential difference over both resistors.
Putting this together:
Finally, the power is:
