Kirchhoffs Circuit Law

Kirchhoffs First Law – The Current Law, (KCL)

Kirchhoffs Current Law or KCL, states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero, I(exiting) + I(entering) = 0. This idea by Kirchhoff is commonly known as the Conservation of Charge.

Kirchhoffs Current Law

kirchhoffs current law

Here, the three currents entering the node, I1, I2, I3 are all positive in value and the two currents leaving the node, I4 and I5 are negative in value. Then this means we can also rewrite the equation as;

I1 + I2 + I3 – I4 – I5 = 0

The term Node in an electrical circuit generally refers to a connection or junction of two or more current carrying paths or elements such as cables and components. Also for current to flow either in or out of a node a closed circuit path must exist. We can use Kirchhoff’s current law when analysing parallel circuits.

Kirchhoffs Second Law – The Voltage Law, (KVL)

Kirchhoffs Voltage Law or KVL, states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchhoff is known as the Conservation of Energy.

Kirchhoffs Voltage Law

kirchhoffs voltage law

Starting at any point in the loop continue in the same direction noting the direction of all the voltage drops, either positive or negative, and returning back to the same starting point. It is important to maintain the same direction either clockwise or anti-clockwise or the final voltage sum will not be equal to zero. We can use Kirchhoff’s voltage law when analysing series circuits.

When analysing either DC circuits or AC circuits using Kirchhoffs Circuit Laws a number of definitions and terminologies are used to describe the parts of the circuit being analysed such as: node, paths, branches, loops and meshes. These terms are used frequently in circuit analysis so it is important to understand them.

Common DC Circuit Theory Terms:

  • • Circuit – a circuit is a closed loop conducting path in which an electrical current flows.
  • • Path – a single line of connecting elements or sources.
  • • Node – a node is a junction, connection or terminal within a circuit were two or more circuit elements are connected or joined together giving a connection point between two or more branches. A node is indicated by a dot.
  • • Branch – a branch is a single or group of components such as resistors or a source which are connected between two nodes.
  • • Loop – a loop is a simple closed path in a circuit in which no circuit element or node is encountered more than once.
  • • Mesh – a mesh is a single open loop that does not have a closed path. There are no components inside a mesh.

Note that:

    Components are said to be connected together in Series if the same current value flows through all the components.

    Components are said to be connected together in Parallel if they have the same voltage applied across them.

A Typical DC Circuit

kirchhoffs circuit law

Kirchhoffs Circuit Law Example No1

Find the current flowing in the 40Ω Resistor, R3

kirchhoffs law example

The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.

Using Kirchhoffs Current LawKCL the equations are given as:

At node A :    I1 + I2 = I3

At node B :    I3 = I1 + I2

Using Kirchhoffs Voltage LawKVL the equations are given as:

Loop 1 is given as :    10 = R1 I1 + R3 I3 = 10I1 + 40I3

Loop 2 is given as :    20 = R2 I2 + R3 I3 = 20I2 + 40I3

Loop 3 is given as :    10 – 20 = 10I1 – 20I2

As I3 is the sum of I1 + I2 we can rewrite the equations as;

Eq. No 1 :    10 = 10I1 + 40(I1 + I2)  =  50I1 + 40I2

Eq. No 2 :    20 = 20I2 + 40(I1 + I2)  =  40I1 + 60I2

We now have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2 

Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps

Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps

As :    I3 = I1 + I2

The current flowing in resistor R3 is given as :    -0.143 + 0.429 = 0.286 Amps

and the voltage across the resistor R3 is given as :    0.286 x 40 = 11.44 volts

The negative sign for I1 means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the 20v battery is charging the 10v battery.

Application of Kirchhoffs Circuit Laws

These two laws enable the Currents and Voltages in a circuit to be found, ie, the circuit is said to be “Analysed”, and the basic procedure for using Kirchhoff’s Circuit Laws is as follows:

  • 1. Assume all voltages and resistances are given. ( If not label them V1, V2,… R1, R2, etc. )
  • 2. Label each branch with a branch current. ( I1, I2, I3 etc. )
  • 3. Find Kirchhoff’s first law equations for each node.
  • 4. Find Kirchhoff’s second law equations for each of the independent loops of the circuit.
  • 5. Use Linear simultaneous equations as required to find the unknown currents.

As well as using Kirchhoffs Circuit Law to calculate the various voltages and currents circulating around a linear circuit, we can also use loop analysis to calculate the currents in each independent loop which helps to reduce the amount of mathematics required by using just Kirchhoff’s laws. In the next tutorial about DC circuits, we will look at Mesh Current Analysis to do just that.

Op-amp Circuits


Amplifiers or any operational amplifier for that matter and these are.

  • No Current Flows into the Input Terminals
  • The Differential Input Voltage is Zero as V1 = V2 = 0 (Virtual Earth)

Then by using these two rules we can derive the equation for calculating the closed-loop gain of an inverting amplifier, using first principles.

Current ( i ) flows through the resistor network as shown.

resistor feedback circuit
inverting op-amp gain formula

Then, the Closed-Loop Voltage Gain of an Inverting Amplifier is given as.

inverting operational amplifier gain equation

and this can be transposed to give Vout as:

inverting operational amplifier gain


non-inverting operational amplifier

In the previous Inverting Amplifier tutorial, we said that for an ideal op-amp “No current flows into the input terminal” of the amplifier and that “V1 always equals V2”. This was because the junction of the input and feedback signal ( V1 ) are at the same potential.

In other words the junction is a “virtual earth” summing point. Because of this virtual earth node the resistors, Rƒ and R2 form a simple potential divider network across the non-inverting amplifier with the voltage gain of the circuit being determined by the ratios of R2 and Rƒ as shown below.

Equivalent Potential Divider Network

non-inverting amplifier potential divider

Then using the formula to calculate the output voltage of a potential divider network, we can calculate the closed-loop voltage gain ( AV ) of the Non-inverting Amplifier as follows:

non-inverting op-amp gain

Then the closed loop voltage gain of a Non-inverting Operational Amplifier will be given as:

non-inverting operational amplifier gain


summing amplifier

In this simple summing amplifier circuit, the output voltage, ( Vout ) now becomes proportional to the sum of the input voltages, V1, V2, V3, etc. Then we can modify the original equation for the inverting amplifier to take account of these new inputs thus:

summing amplifier formula

However, if all the input impedances, ( RIN ) are equal in value, we can simplify the above equation to give an output voltage of:

Summing Amplifier Equation

summing amplifier equation

Differential Amplifier

differential amplifier circuit

By connecting each input in turn to 0v ground we can use superposition to solve for the output voltage Vout. Then the transfer function for a Differential Amplifier circuit is given as:

differential amplifier transfer function

When resistors, R1 = R2 and R3 = R4 the above transfer function for the differential amplifier can be simplified to the following expression:

Differential Amplifier Equation

differential amplifier equation

Op-amp Integrator Circuit

op-amp integrator amplifier circuit
op-amp integrator equation

To simplify the math’s a little, this can also be re-written as:

simplified integrator equation

if Vin is constant voltage so Rin voltage drop always equal Vin so it is feeding C with constant current source for example if we have capacitor with 1H and constant current source 1A every seconds the capacitor it will increase 1 volt constantly Vin/(Rin.C)

Op-amp Differentiator Circuit

op-amp differential amplifier
op-amp differentiator equation

Operational Amplifier Basics

Operational amplifiers are linear devices that have all the properties required for nearly ideal DC amplification and are therefore used extensively in signal conditioning, filtering or to perform mathematical operations such as add, subtract, integration and differentiation.

An Operational Amplifier, or op-amp for short, is fundamentally a voltage amplifying device designed to be used with external feedback components such as resistors and capacitors between its output and input terminals. These feedback components determine the resulting function or “operation” of the amplifier and by virtue of the different feedback configurations whether resistive, capacitive or both, the amplifier can perform a variety of different operations, giving rise to its name of “Operational Amplifier”.

An Operational Amplifier is basically a three-terminal device which consists of two high impedance inputs. One of the inputs is called the Inverting Input, marked with a negative or “minus” sign, ( – ). The other input is called the Non-inverting Input, marked with a positive or “plus” sign ( + ).

A third terminal represents the operational amplifiers output port which can both sink and source either a voltage or a current. In a linear operational amplifier, the output signal is the amplification factor, known as the amplifiers gain ( A ) multiplied by the value of the input signal and depending on the nature of these input and output signals, there can be four different classifications of operational amplifier gain.

  • Voltage  – Voltage “in” and Voltage “out”
  • Current  – Current “in” and Current “out”
  • Transconductance  – Voltage “in” and Current “out”
  • Transresistance  – Current “in” and Voltage “out”

Since most of the circuits dealing with operational amplifiers are voltage amplifiers, we will limit the tutorials in this section to voltage amplifiers only, (Vin and Vout).

The output voltage signal from an Operational Amplifier is the difference between the signals being applied to its two individual inputs. In other words, an op-amps output signal is the difference between the two input signals as the input stage of an Operational Amplifier is in fact a differential amplifier as shown below.

Differential Amplifier

The circuit below shows a generalized form of a differential amplifier with two inputs marked V1 and V2. The two identical transistors TR1 and TR2 are both biased at the same operating point with their emitters connected together and returned to the common rail, -Vee by way of resistor Re.operational amplifier basics the differential input

Differential Amplifier

The circuit operates from a dual supply +Vcc and -Vee which ensures a constant supply. The voltage that appears at the output, Vout of the amplifier is the difference between the two input signals as the two base inputs are in anti-phase with each other.

So as the forward bias of transistor, TR1 is increased, the forward bias of transistor TR2 is reduced and vice versa. Then if the two transistors are perfectly matched, the current flowing through the common emitter resistor, Re will remain constant.

Like the input signal, the output signal is also balanced and since the collector voltages either swing in opposite directions (anti-phase) or in the same direction (in-phase) the output voltage signal, taken from between the two collectors is, assuming a perfectly balanced circuit the zero difference between the two collector voltages.

This is known as the Common Mode of Operation with the common mode gain of the amplifier being the output gain when the input is zero.

Operational Amplifiers also have one output (although there are ones with an additional differential output) of low impedance that is referenced to a common ground terminal and it should ignore any common mode signals that is, if an identical signal is applied to both the inverting and non-inverting inputs there should no change to the output.

However, in real amplifiers there is always some variation and the ratio of the change to the output voltage with regards to the change in the common mode input voltage is called the Common Mode Rejection Ratio or CMRR for short.

Operational Amplifiers on their own have a very high open loop DC gain and by applying some form of Negative Feedback we can produce an operational amplifier circuit that has a very precise gain characteristic that is dependant only on the feedback used. Note that the term “open loop” means that there are no feedback components used around the amplifier so the feedback path or loop is open.

An operational amplifier only responds to the difference between the voltages on its two input terminals, known commonly as the “Differential Input Voltage” and not to their common potential. Then if the same voltage potential is applied to both terminals the resultant output will be zero. An Operational Amplifiers gain is commonly known as the Open Loop Differential Gain, and is given the symbol (Ao).

Equivalent Circuit of an Ideal Operational Amplifier

ideal operational amplifier

Op-amp Parameter and Idealised Characteristic

  • Open Loop Gain, (Avo)
    • Infinite – The main function of an operational amplifier is to amplify the input signal and the more open loop gain it has the better. Open-loop gain is the gain of the op-amp without positive or negative feedback and for such an amplifier the gain will be infinite but typical real values range from about 20,000 to 200,000.
  • Input impedance, (ZIN)
    • Infinite – Input impedance is the ratio of input voltage to input current and is assumed to be infinite to prevent any current flowing from the source supply into the amplifiers input circuitry ( IIN = 0 ). Real op-amps have input leakage currents from a few pico-amps to a few milli-amps.
  • Output impedance, (ZOUT)
    • Zero – The output impedance of the ideal operational amplifier is assumed to be zero acting as a perfect internal voltage source with no internal resistance so that it can supply as much current as necessary to the load. This internal resistance is effectively in series with the load thereby reducing the output voltage available to the load. Real op-amps have output impedances in the 100-20kΩ range.
  • Bandwidth, (BW)
    • Infinite – An ideal operational amplifier has an infinite frequency response and can amplify any frequency signal from DC to the highest AC frequencies so it is therefore assumed to have an infinite bandwidth. With real op-amps, the bandwidth is limited by the Gain-Bandwidth product (GB), which is equal to the frequency where the amplifiers gain becomes unity.
  • Offset Voltage, (VIO)
    • Zero – The amplifiers output will be zero when the voltage difference between the inverting and the non-inverting inputs is zero, the same or when both inputs are grounded. Real op-amps have some amount of output offset voltage.

From these “idealized” characteristics above, we can see that the input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”). It is important to remember these two properties as they will help us understand the workings of the Operational Amplifier with regards to the analysis and design of op-amp circuits.

However, real Operational Amplifiers such as the commonly available uA741, for example do not have infinite gain or bandwidth but have a typical “Open Loop Gain” which is defined as the amplifiers output amplification without any external feedback signals connected to it and for a typical operational amplifier is about 100dB at DC (zero Hz). This output gain decreases linearly with frequency down to “Unity Gain” or 1, at about 1MHz and this is shown in the following open loop gain response curve.

Open-loop Frequency Response Curve

operational amplifier frequency response

From this frequency response curve we can see that the product of the gain against frequency is constant at any point along the curve. Also that the unity gain (0dB) frequency also determines the gain of the amplifier at any point along the curve. This constant is generally known as the Gain Bandwidth Product or GBP. Therefore:

GBP = Gain x Bandwidth = A x BW

For example, from the graph above the gain of the amplifier at 100kHz is given as 20dB or 10, then the gain bandwidth product is calculated as:

GBP = A x BW = 10 x 100,000Hz = 1,000,000.

Similarly, the operational amplifiers gain at 1kHz = 60dB or 1000, therefore the GBP is given as:

GBP = A x BW = 1,000 x 1,000Hz = 1,000,000. The same!.

The Voltage Gain (AV) of the operational amplifier can be found using the following formula:

op-amp voltage gain

and in Decibels or (dB) is given as:

op-amp gain in decibels, dB

An Operational Amplifiers Bandwidth

The operational amplifiers bandwidth is the frequency range over which the voltage gain of the amplifier is above 70.7% or -3dB (where 0dB is the maximum) of its maximum output value as shown below.

op-amp frequency response curve

Here we have used the 40dB line as an example. The -3dB or 70.7% of Vmax down point from the frequency response curve is given as 37dB. Taking a line across until it intersects with the main GBP curve gives us a frequency point just above the 10kHz line at about 12 to 15kHz. We can now calculate this more accurately as we already know the GBP of the amplifier, in this particular case 1MHz.

Operational Amplifier Example No1.

Using the formula 20 log (A), we can calculate the bandwidth of the amplifier as:

37 = 20 log (A)   therefore, A = anti-log (37 ÷ 20) = 70.8

GBP ÷ A = Bandwidth,  therefore, 1,000,000 ÷ 70.8 = 14,124Hz, or 14kHz

Then the bandwidth of the amplifier at a gain of 40dB is given as 14kHz as previously predicted from the graph.

Operational Amplifier Example No2.

If the gain of the operational amplifier was reduced by half to say 20dB in the above frequency response curve, the -3dB point would now be at 17dB. This would then give the operational amplifier an overall gain of 7.08, therefore A = 7.08.

If we use the same formula as above, this new gain would give us a bandwidth of approximately 141.2kHz, ten times more than the frequency given at the 40dB point. It can therefore be seen that by reducing the overall “open loop gain” of an operational amplifier its bandwidth is increased and visa versa.

In other words, an operational amplifiers bandwidth is inversely proportional to its gain, ( A 1/∞ BW ). Also, this -3dB corner frequency point is generally known as the “half power point”, as the output power of the amplifier is at half its maximum value as shown:

half power at corner frequency


  • The concept of circuit loading
  • The emitter-follower circuit
  • Calculation of the effective (load) input impedance Zin in the emitter-follower circuit
  • Calculation of the effective (source) output impedance Zout in the emitter-follower circuit
  • An alternative derivation of the output and input impedances.
  • Making stiffer sources: Emitter follower
  • Biasing the emitter follower
  • The input impedance of the emitter-follower circuit
  • The output impedance of the emitter-follower circuit
  • Matching impedance: measuring the 50 ohm output impedance.

pdf file

input impedance

output impedance

Norton’s Theorem

What is Norton’s Theorem?

Norton’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin’s Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Simplifying Linear Circuits

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

simplifying linear circuits

. . . after Norton conversion . . .

norton equivalent circuit diagram

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

Thevenin’s Theorem vs. Norton’s Theorem

As with Thevenin’s Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin’s Theorem are the steps used in Norton’s Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton).

Identify The Load Resistance

As before, the first step is to identify the load resistance and remove it from the original circuit:

identify the load resistance

Find The Norton Current

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin’s Theorem, where we replaced the load resistor with a break (open circuit):

find the norton current

With zero voltage dropped between the load resistor connection points, the current through R1 is strictly a function of B1‘s voltage and R1‘s resistance: 7 amps (I=E/R). Likewise, the current through R3 is now strictly a function of B2‘s voltage and R3‘s resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (INorton) in our equivalent circuit:

norton equivalent circuit diagram

Find Norton Resistance

Remember, the arrow notation for current source points in the direction of conventional current flowTo calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

find the norton resistance

Now our Norton equivalent circuit looks like this:

norton equivalent circuit diagram

Determine The Voltage Across The Load Resistor

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

voltage across the load resistor

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R2; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin’s Theorem apply to Norton’s as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit, again and again, applying nothing more complex than simple parallel circuit analysis to determine what’s happening with each trial load.



  • Norton’s Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.
  • Steps to follow for Norton’s Theorem:
    • Find the Norton source current by removing the load resistor from the original circuit and calculating the current through a short (wire) jumping across the open connection points where the load resistor used to be.
    • Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
    • Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
    • Analyze voltage and current for the load resistor following the rules for parallel circuits.