DC Steady State and Transient Analysis

Circuit Analysis is a big topic to be sure. However, understanding the basics real well enables one to solve most problems without resorting to an overwhelming set of calculations. Important characteristics of the key components: resistors, inductors, and capacitors; are presented. Examples of DC and Transient Analysis for these components are given.

DC Steady State Analysis

Resistors:
Resistors play the major role in DC Analysis. Capacitors are an open circuit at DC and Inductors are a short circuit and as such this is the only role they play in the analysis. Note in the diagrams below that the capacitor plays no role because it acts as an open circuit; but, the inductor plays a significant role because it shorts the output of the resistor to ground. The role may change depending on the orientation and placement within in a circuit.
L and C at DC steady state

Focusing only on the resistor then, this is the most used equation in electronics – Ohms Law:

EQN 1) $\displaystyle { V = {I}{R} }$
(upper case letters signify DC steady state analysis)

EQN 1 means the voltage across a resistor is directly and instantaneously proportional to the current passing through it. Georg Ohm, a German physicist, determined this equation through experimentation and published his result in 1827. This characteristic is used to create voltage references and limit the current in other devices.

Voltage Divider
Consider the the simple voltage divider and determine the value of Vout.
Resistor Voltage Divider
Begin by recognizing that the voltage across R2 is the same as Vout. Also note that the current is the same through both resistors. Use EQN 1 to determine the current through both resistors. Then use that value of the current to find the voltage across R2.
$\displaystyle V = IR \Rightarrow I = \frac{V}{R} \Rightarrow I = \frac{V_{in}}{R_1+R_2}$
Using this value for I provides the solution.
$\; \displaystyle V = I R \Rightarrow V_{R_2} = (\frac{V_{in}}{R_1+R_2}) R_2 \; = V_{out}$

Given: Vin = 5V, R1 = 10K, R2 = 5K then
$\; \displaystyle V_{out} = 5V \frac{5K}{15K} = 1.67 V$

Current Limiter
Consider the resistor as a current limiting device needed to bias an LED.
LED with Current Limiting Resistor
A tyical LED will have a forward voltage of roughly 1.5V. This means that the voltage across the Resistor will be 3.5V (5V-1.5V) regardless of the value of the resistor. Using a 1KΩ resistor, EQN 1 tells us that the current through R, and also the LED, will be 3.5mA (3.5V/1000Ω). According to the data sheet the LED is rated for full luminosity at 20mA. So with 3.5mA this LED will be dimly lit. For full luminosity R should be 175Ω. The maximum rating for the LED is 30mA. So any value for R less than 117Ω would exceed the LED’s limits.

EQN 2) $\displaystyle { P = {I^2}{R} }$

EQN 2 tells us that a resistor dissipates power in the form of heat. The more current that passes through a resistor the hotter it gets. Using a 100Ω resistor to achieve 35mA in the LED circuit would not only exceed the maximum rating of the LED but it might exceed the power rating of the resistor as well. The power dissipated by that resistor will be 0.123 Watts. ( .035A * .035A * 100Ω ). A typical resistor has a power rating of 0.1 Watts. So 0.123 would exceed that rating. A typical axial lead resistor is rated at 0.25 Watts. Though acceptable for the resistor it’s still not acceptable for the LED. It’s important to calculate the amount of power being dissipated by each resistor and to check each components operating conditions to make sure that their limits are not being exceeded.

In summary, resistors resist the flow of electrons, drop the voltage in the circuit and radiate heat in the process. Other devices also radiate heat when powered, but it’s really the intrinsic resistance within the device that’s doing this. Small amounts of parasitic resistance are part of every component in electronics.

Transient Analysis

Inductors and capacitors don’t consume power like a resistor. They store energy in one moment of time only to return all the energy to the circuit at a later time. In order to do this, these components must operate in a transient environment. As noted above these components are either open or short in a DC Steady State circuit. Transients within electronic circuits occur as bits toggle from low to high and high to low. All micro-controllers are clock driven and every clock cycle causes a large number of bit transitions to occur. Many of the problems we are forced to deal with are related to the undesired effect of parasitic values of these 3 components. The tracks and cables that connect one part of a circuit to another are largely parasitic inductors. The semiconductor gates used in all logic circuits have a significant parasitic capacitor on the input. And small amounts of parasitic resistance are everywhere. There are also some very useful applications for these components that have great value in embedded systems. We will look at both in the examples that follow.

Inductors
EQN 3) $\displaystyle \; v_L = L\frac{ di }{ dt }$
(lower case symbols signify transient analysis)
EQN 3 tells us that voltage is only present across an inductor when the current is changing. Notice how current and time are represented in this equation. This is a classical differential equation as seen in calculus. The “d” means a small difference or a small change. So to use this equation you must know how much the current has changed and over what period of time. Another thing to take from this equation is that “it takes time” for the current to change through an inductor – it cannot change instantly. It’s part of the process of storing energy in the device in the form of a magnetic field. The voltage, however, can change instantly and reverse direction instantly. The magnetic field opposes any change in current and it does this by “inducing” a voltage across the inductor whose polarity opposes the change. It’s a property that get’s exploited in certain applications. And, of course, the scaling factor in all this is the value of the inductance L. The bigger the L the bigger the voltage for the same change in current.
Parasitic Inductance
Inductive Transient
The micro-controller is being powered by a 5 volt adapter that has a cable 3 feet long. The self-inductance of this cable is given as 1.5uH. The rise time for the clock is 7.5ns. When the controller toggles its state on the leading edge of the clock, how much change in current would cause a 3 volt change across the cable?
Solution
$\displaystyle v = L\frac{di}{dt} \Rightarrow di = v\frac{dt}{L} \Rightarrow di = 3\frac{7.5E-9}{1.5E-6} \Rightarrow di = 15mA$
To put this in perspective, 15mA is the current needed to light an LED. If the micro-controller turns one LED ON, while keeping everything else the same, the increased current demand on the power supply causes a transient across the cable that drops the voltage from 5V to 2V effectively powering the micro-controller off. When the transient subsides the micro-controller would be in reset. Similarly, if the micro-controller turned one LED OFF, this would reduce the current demand on the power supply causing a transient across the cable that would spike the voltage from 5V to 8V, exceeding the maximum rating of the device and likely destroying it.

The harmful nature of parasitic inductance associated with power supplies brought about the need for the decoupling capacitor. It is necessary to “decouple” a device from the power supply. The power supply must only provide an average current and never supply a transient current. Transient currents are to be supplied by the decoupling capacitor.

Capacitors
EQN 4) $\displaystyle \; i_C = C\frac{ dv }{ dt }$
EQN 4 tells us that there is no current flow to and from a capacitor unless the voltage across the capacitor is changing. To use this equation we must know how much the voltage has changed and over what period of time. When current flows into a capacitor the voltage increases. When current flows out of a capacitor the voltage decreases. The voltage across a capacitor can not change instantly – it takes time to change. The current, however, can change amplitude and direction instantly.

Decoupling Capacitor

The micro-controller needs to power 10 LEDs on and off as a blinking indicator. Each LED requires 20mA of current. The rise time for the clock is 7.5ns. A .001uF capacitor is used for power supply decoupling. When the controller toggles its state on the leading edge of the clock, how much will the voltage of the decoupling capacitor drop?
Solution

$\displaystyle i_C = \frac{ C dv }{ dt } \Rightarrow dv = i_C\frac{C}{dt} \Rightarrow dv = 0.2\frac{.001E-6}{7.5E-9} \Rightarrow dv = 27mV$

The decoupling cap did its job by supplying a 200mA transient within a 7.5ns window and only dropped the voltage to the micro-controller by 27mV in the process. The cap will need a recharge, but the time frame available to do that will be much longer – many, many clock cycles. Effectively, the power supply no longer has to deal with transients.

The Boost Voltage Convertor
Voltage Pump
This circuit finds application where a higher voltage is needed. The transistor switch can create a current surge through the inductor while the diode restricts the direction that currents are allowed to flow.
The desired voltage out is 12V. The output voltage is monitored such that when the voltage drops to 11V or less the circuit is activated to pump the voltage back to 12V. The pumping algorithm turns Q1 on for 10us and then off for 10us. When Q1 is on, L1 is shorted to GND and 5V is impressed across L1. This causes the current through L1 to ramp up according to the equation:
$\displaystyle \; v_L = L \frac{ di }{ dt } \; \Rightarrow \; di = v_L \frac{dt}{L} \; \Rightarrow \; di = 5 \frac{10E-6}{680E-6} = 73.5mA$

It will ramp from 0 to 73.5mA in 10us. The only limit to the current’s value is the amount of time it’s allowed to ramp and the maximum limits of Q1. After 10us, Q1 is turned off and the magnetic field established by the current begins to collapse. The collapsing magnetic field will attempt to maintain the current as it was. The current, which is now ramping down, can only go through D1 and charge up C2. The voltage across L1 is now:
$\displaystyle \; V_L = V_{out} - V_{D1} -V_{in} \; =\; 11 - .5 - 5 \; = \; 5.5V$

The pumping algorithm only kicks on at 11V or less and the Forward Voltage of D1 at 73mA is about 0.5V. The slope of the current’s down ramp is also dictated by the voltage across L1. This affects the amount of time it takes for the current to ramp down to zero.
$\displaystyle \; v_L = L \frac{ di }{ dt } \; \Rightarrow \; dt = L \frac{ di }{v_L } \Rightarrow \; dt = 680E-6 \frac{ 73.5E-3 }{ 5.5 } = 9.1us$

Once the current stops the Voltage across L1 drops to 0 and D1 becomes reverse-biased preventing the charge stored in C2 from leaking out. The average current during the charge transfer was 73.5mA / 2 during the 9.1us. How much does the voltage increase in C2 during a single pumping action?
Solution
$\displaystyle \; i_C = C \frac{dv}{dt} \; \Rightarrow \; dv = i_C \frac{dt}{C} \Rightarrow \; dv = 36.75E-3 \frac{9.1E-6}{47E-6} \; = \; 7mV$
Neglecting the leakage through R2 and R3, It will take 143 cycles of this pumping action to return C2 to 12V. How much is the leakage through R2 and R3? How many more pumping cycles are needed to compensate?