Skip to main content

How Does Noise Affect Circuits? - Understanding Noise Part II

In the first part of the series we made an introduction about the noise and how we can categorize them into device noise, emitted noise, and conducted noise. We first discussed about the most common noise, the Johnson noise coming from a resistor. This noise is also a white noise, due to the fact that it is present at all frequencies. This noise is produced whether or not a resistor is connected in a circuit.

When a resistor is connected to a circuit, say to an amplifier, this resistor will be a source of noise to the circuit. Let's take an inverting operational amplifier as an example that has two resistors connected to it, one at the input and the other one as feedback. The noise due to the resistors can be modeled as noise sources in series with the resistors.

Figure 1 Resistor Noise

In order to appreciate its effect in the circuit, we must be able to understand how these noise affect the behavior and output of the amplifier. Modeling the noise sources as in the figure above will help us easily understand the concept. The ensuing analysis should be straightforward using KVL and good mastery of the operational amplifier. Let's assume at the moment that the op amp is ideal and doesn't contribute noise to the system.  The noise due the resistor R1, eN1, will appear as signal at the input and will be gained up by the factor (RF/R1), because it appears as if an input to the inverting amplifier. But because noise doesn't have polarity, there is no negative sign to it.  Notice that all inputs at are ground when we do the noise analysis, using the superposition technique. If the noise is given as spectral density, Vrms/root Hz, we then proceed to calculating the spectral noise density at the output due to the individual noise sources. The noise due to the feedback resistor, RF, will then just be as it is without any amplification, because the negative input of the op-amp is a virtual short to ground.

The total noise at the output of the amplifiers due to the resistor noise, in  will then be:



If we know the bandwidth  with which we are going to operate our circuit, we can calculate the noise at the output in terms of RMS.

The noise in RMS can be obtained by multiplying the spectral noise density by the square root of the bandwidth. This is a straightforward task because we are dealing with white noise where noise power level is the same at all frequencies. But there is a correction factor that needs to be used because the bandwidth, and the filters as such, aren’t ideal brick wall. The following correction factors:

1.57fC for single pole filter
1.2 fC for the double pole

This means multiplying your bandwidth by 1.57 if the circuit is a single-pole low pass filter in order to obtain the equivalent noise bandwidth. This also means that a single-pole filter allows more noise that would a brick wall filter allows at the same cut-off frequency.

The discussion on noise density, rms noise, and peak-to-peak, are in order in the Part III of this series.


Reference:
MT-048 Tutorial, Op Amp Noise Relationships: 1/f Noise, RMS Noise, and Equivalent Noise Bandwidth 

Labels:

Comments

Post a Comment

Popular posts from this blog

Arduino Transistor DC Motor Control

Arduino boards have PWM (Pulse Width Modulation) outputs that can be used to control like the speed of a DC motor. On a Pro Micro, those outputs are encircled white on the board. You can program it such that it outputs on a scale of 1-255, 1 being the slowest and 255 being the fastest. PWM gives out pulses whose width is varied (modulated) while the period is constant. The longer it is high, the higher power it delivers to the circuit. PWM can be used to drive a transistor (switch) which in turn drives the motor ON and OFF. The longer the switch is ON, the higher the power delivered to the motor, and thus faster. The circuit consists of a transistor Q1 (TIP31) NPN transistor, driven by the PWM from Arduino. The resistor is to limit the current to the base, but enough to operate the transistor in saturation when the input is high. The diode is to protect the motor from any back emf that might come from the motor when the current is cut off. The circuit is supplied by a 9V battery.
3 comments
Read more

What Can You Do With Two Transistors (BJT)? Part 2 - Transistor Regulator

In the first installment of this series, we talked about what exciting things we can do with just two transistors, in particular BJT transistors. Some of us might probably think that we don't talk a lot about BJT transistors nowadays, not with the thrill, excitement, and trend brought about by high advancements in the technology in the Internet of Things, Machine Learning, Machine Intelligence, Advanced Algorithm, etc.. But we argue that these little circuits and components are constantly in the background of the advancements we see around us and we should not ignore them, as much we will not ignore what we currently enjoy and what the future holds for us. Let's continue to look at the Two-Transistor Series Regulator we talked about in Part 1 . The circuit below is a different version, now using the transistor to provide feedback from the output. This makes the output voltage higher. The output is still derived from the zener voltage. The output voltage is now a s
6 comments
Read more

How Does The Howland Current Source Work?

Current source programs an output current as defined by the user, regardless of the load resistance. We know that when load resistance is larger, the voltage across it also increases. An ideal source would maintain a constant amount of current even if the load resistance changes, or in other words even if the output voltage changes. Such behavior describes a very good current source that has high output impedance. Remember, the output voltage changes, but very little or no change in output current. That means high output resistance or impedance. We use the term impedance when dealing with changes. Howland circuit in the figure is a classic current source circuit.   When R 1 is made equal to R 2 , R F equal to R 3,  the output current is given by V IN /R 1 where V IN is V 1 minus V 2 . A simple implementation is grounding V 2 and taking V 1 as the V IN . Figure 1 Basic Howland Current Source This circuit is such a clever manipulation of the op amp, such
Post a Comment
Read more