The unit of density in the International System of Units is tesla.

Is given by:

where B is the magnetic flux density generated by a load moving with velocity v at a distance r from the load, and ur is the unit vector connecting the load to the point where B is measured (point g).

or :

where B is the magnetic flux density generated by a conductor through which flows a current I, at a distance r.

The formula of this definition is called the Biot-Savart, and magnetism is equivalent to Coulomb’s law of electrostatics, as used to calculate the forces acting on moving charges.

The field induction, B, or magnetic flux density (the three names are equivalent) is essential in the field H electromagnetism, since it is responsible for the forces on the moving charges and is, therefore, the physical equivalent to E.

Current divider formula for two resistors

When have two resistors in parallel, which are subject to the same voltage, it is possible to calculate the current path one of the two resistors.

To calculate one of these two currents ( in our case I1 or I2 ), however, you must know the total current ( here is the current I ) flowing through these resistors and know the value of resistance ( ie, we know the value of R1 and R2 ).

Since these two resistors are subject to the same voltage, then:

Similarly, we can know the voltage U as a function of I1 and R1, which gives:

Using the two formulas above views, we can deduce this formula:

The formula of dividing bridge current is:

Voltage divider circuit

• Unloaded Voltage Divider (two resistors in series)
• oaded Voltage Divider

Unloaded voltage divider

An unloaded voltage divider comprises two series-connected resistors R 1 and R 2.
The current and voltage distribution in the unloaded voltage dividers is identical to the series circuit.Here, the same formulas and rules.
Unloaded voltage divider calculatation
The following formulas are used to calculate the partial voltages (voltage divider rules). They apply only if both resistors by the same current flows, so there is an unloaded voltage divider. In this case, calculate the voltage across the three-part set.

Loaded voltage divider

A loaded voltage divider consists of a series circuit of resistors R 1 and R 2. In addition, one of the two resistors by a consumer, in this case the resistance R L load (load resistance). The circuit is a series connection to a mixed circuit of parallel connection (R 2 | | R L) and series (R 1+ ( R 2| | R L)).

Loaded voltage divider calculation
The following formula is used to calculate the parallel resistance in the loaded voltage divider.

If the voltage divider loaded with a resistance, shall be held in the circuit following changes:

* The total resistance of the circuit is small.
* As a result, increases the total current I tot .
* The voltage drop U 1 at the resistor R 1 increases.
* The supply voltage U 2 across resistor R 2 becomes smaller.

Variable loads can be powered by low voltage divider with a fairly stable voltage. However, a voltage divider should not be burdened by a very small resistance. This leads to changes in the current and voltage distribution within the circuit. Thus, the voltage divider is useless.

DC to DC converter circuit is an electronic circuit which can be used either to transform one voltage to different level/value; or to provide an isolation barrier for a voltage bus. DC to DC converters are typically being used in power distribution system to provide a local voltage conversion, point of load voltage regulation or power bus isolation.

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This is a basic theory document for power supply design by Teuvo Suntio. You’ll find reference about basic electrical math there…

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A power supply or UPS circuit can be a device that supplies electrical power to one or far more electrical loads. The term is most generally applied to devices that convert one form of electrical energy to an additional, though it could also refer to devices that convert another form of energy (e.g., mechanical, chemical, solar) to electrical energy. A regulated power source is one that controls the output voltage or present to a specific value; the controlled worth is held almost constant regardless of variations in either load electric current or the voltage supplied by the power supply’s power source.

A frequent power supply circuit typically includes in the following modul circuit:

Transformer
Transformers convert AC electricity from one voltage to another with little loss of energy. Transformers function only with AC and this is one of the factors why mains electrical energy is AC.

Step-up transformers boost voltage, step-down transformers decrease voltage. Most energy supplies use a step-down transformer to reduce the dangerously high mains voltage (230V in UK) to a safer reduced voltage.

The input coil is called the primary along with the output coil is referred to as the secondary. There’s no electrical connection between the two coils, instead they’re linked by an alternating magnetic field designed in the soft-iron core of the transformer. The two lines in the middle of the circuit symbol represent the core.

Transformers waste very small power so the power out is (almost) equal to the power in. Note that as voltage is stepped down electric current is stepped up.

The ratio of the number of turns on every coil, called the turns ratio, determines the ratio of the voltages. A step-down transformer features a big quantity of turns on its main (input) coil that’s connected towards the high voltage mains source, plus a small variety of turns on its secondary (output) coil to give a reduced output voltage.

Rectifier
You’ll find a number of ways of connecting diodes to make a rectifier to convert AC to DC. The bridge rectifier will be the most critical and it generates full-wave varying DC. A full-wave rectifier may also be produced from just two diodes if a centre-tap transformer is implemented, but this method is hardly ever used now that diodes are more affordable. A single diode might be used as a rectifier but it only uses the positive (+) parts in the AC wave to generate half-wave varying DC.

Smoothing
Smoothing is performed by a large worth electrolytic capacitor connected across the DC source to act as a reservoir, supplying current for the output when the varying DC voltage from the rectifier is falling. The diagram exhibits the unsmoothed varying DC (dotted line) and also the smoothed DC (solid line). The capacitor charges quickly close to the peak with the varying DC, and then discharges as it supplies electric current for the output.

Voltage regulator ICs are accessible with fixed (normally 5, 12 and 15V) or variable output voltages. They’re also rated by the maximum current they can pass. Negative voltage regulators are available, mainly for use in dual supplies. Most regulators consist of some automatic protection from excessive current (‘overload protection’) and overheating (‘thermal protection’).

A switched-mode power supply (switching-mode power supply, SMPS, or just switcher) is an electronic power supply that incorporates a switching regulator in order to be highly efficient in the conversion of electrical power. Like other kinds of power supplies, an SMPS transfers power from a source like the electrical power grid to a load (e.g., a individual computer) whilst converting voltage and current characteristics. An SMPS is usually employed to efficiently supply a regulated output voltage, generally at a level diverse from the input voltage. Unlike a linear power supply, the pass transistor of a switching mode supply switches very quickly (typically between 50 kHz and 1 MHz) between full-on and full-off states, which minimizes wasted energy. Voltage regulation is supplied by varying the ratio of on to off time. In contrast, a linear power supply should dissipate the excess voltage to regulate the output. This higher efficiency will be the chief benefit of a switched-mode power supply.

Switching regulators are utilized as replacements for the linear regulators when greater efficiency, smaller size or lighter weight are needed. They are, nevertheless, far more complex, their switching currents can trigger electrical noise problems if not cautiously suppressed, and simple designs might have a poor power factor.

Explanation

A linear regulator provides the desired output voltage by dissipating excess power in ohmic losses (e.g., in a resistor or inside the collector-emitter region of a pass transistor in its active mode). A linear regulator regulates either output voltage or current by dissipating the excess electric power inside the form of heat, and hence its maximum power efficiency is voltage-out/voltage-in given that the volt difference is wasted. In contrast, a switched-mode power supply regulates either output voltage or existing by switching ideal storage elements, like inductors and capacitors, into and out of distinct electrical configurations. Ideal switching elements (e.g., transistors operated outside of their active mode) have no resistance when “closed” and carry no present when “open”, and so the converters can theoretically operate with 100% efficiency (i.e., all input power is delivered to the load; no power is wasted as dissipated heat).

For instance, if a DC source, an inductor, a switch, and also the corresponding electrical ground are placed in series and the switch is driven by a square wave, the peak-to-peak voltage of the waveform measured across the switch can exceed the input voltage from the DC source. This is since the inductor responds to adjustments in current by inducing its own voltage to counter the alter in current, and this voltage adds to the source voltage even though the switch is open. If a diode-and-capacitor combination is placed in parallel to the switch, the peak voltage might be stored in the capacitor, and also the capacitor can be used as a DC source with an output voltage greater than the DC voltage driving the circuit. This increase converter acts like a step-up transformer for DC signals. A buck-boost converter works in a similar manner, but yields an output voltage which is opposite in polarity to the input voltage. Other buck circuits exist to boost the average output existing having a reduction of voltage.

In an SMPS, the output present flow depends on the input power signal, the storage elements and circuit topologies utilised, and also on the pattern used (e.g., pulse-width modulation with an adjustable duty cycle) to drive the switching elements. Normally, the spectral density of these switching waveforms has energy concentrated at fairly high frequencies. As such, switching transients, like ripple, introduced onto the output waveforms could be filtered with tiny LC filters.

Advantages and disadvantages

The major advantage of this approach is higher efficiency simply because the switching transistor dissipates little power when it is outside of its active region (i.e., when the transistor acts like a switch and either has a negligible voltage drop across it or a negligible existing by way of it). Other positive aspects include smaller size and lighter weight (from the elimination of low frequency transformers which have a high weight) and lower heat generation because of greater efficiency. Disadvantages consist of higher complexity, the generation of high-amplitude, high-frequency energy that the low-pass filter must block to prevent electromagnetic interference (EMI), along with a ripple voltage at the switching frequency and also the harmonic frequencies thereof.

Extremely low cost SMPSs may couple electrical switching noise back onto the mains power line, causing interference with A/V equipment connected to the same phase. Non-power-factor-corrected SMPSs also trigger harmonic distortion.

The sample of switching power supply: 5V DC / 10A Offline Switching Power Supply

An H bridge is an electronic circuit that enables a voltage to be applied across a load in either direction. These circuits are usually applied in robotics and various applications to enable DC motors to move forwards and reverse. H bridges are ready as integrated circuits (IC chip – for example: L298N), or can be built from discrete components (transistors).

General
The term H bridge is derived from the typical graphical representation of such a circuit. An H bridge is built with four switches (solid-state or mechanical). When the switches S1 and S4 (based on the first figure) are closed (and S2 and S3 are open) a positive voltage will probably be applied across the motor. By opening S1 and S4 switches and closing S2 and S3 switches, this voltage is reversed, allowing reverse operation of the motor.

Employing the nomenclature above, the switches S1 and S2 need to by no means be closed at the very same time, as this would trigger a short circuit on the input voltage source. The identical applies to the switches S3 and S4. This condition is known as shoot-through.

Operation
The H-bridge arrangement is normally employed to reverse the polarity of the motor, but can also be utilized to ‘brake’ the motor, where the motor comes to a sudden quit, as the motor’s terminals are shorted, or to let the motor ‘free run’ to a quit, as the motor is successfully disconnected from the circuit. The following table summarises operation, with S1-S4 corresponding to the diagram above.

Construction
A solid-state H bridge is typically constructed using opposite polarity devices, including PNP BJTs or P-channel MOSFETs connected to the high voltage bus and NPN BJTs or N-channel MOSFETs connected to the low voltage bus.

The most efficient MOSFET designs use N-channel MOSFETs on both the high side and low side due to the fact they usually have a third of the ON resistance of P-channel MOSFETs. This needs a more complex style given that the gates of the high side MOSFETs need to be driven positive with respect to the DC supply rail. Nonetheless, many integrated circuit MOSFET drivers incorporate a charge pump within the device to obtain this.

Alternatively, a switch-mode DC-DC converter can be used to offer isolated (‘floating’) supplies to the gate drive circuitry. A multiple-output flyback converter is well-suited to this application.

An additional method for driving MOSFET-bridges will be the use of a specialised transformer referred to as a GDT (Gate Drive Transformer), which gives the isolated outputs for driving the upper FETs gates. The transformer core is normally a ferrite toroid, with 1:1 or 4:9 winding ratio. Nevertheless, this technique can only be utilised with high frequency signals. The style of the transformer is also extremely essential, as the leakage inductance ought to be minimized, or cross conduction may happen. The outputs of the transformer also have to be generally clamped by zener diodes, because high voltage spikes could destroy the MOSFET gates.

A widespread variation of this circuit uses just the two transistors on one side of the load, comparable to a class AB amplifier. Such a configuration is called a “half bridge”. The half bridge is utilised in some switched-mode power supplies that use synchronous rectifiers and in switching amplifiers. The half-H bridge kind is frequently abbreviated to “Half-H” to distinguish it from full (“Full-H”) H bridges. One more frequent variation, adding a third ‘leg’ to the bridge, creates a 3-phase inverter. The 3-phase inverter will be the core of any AC motor drive.

A further variation is the half-controlled bridge, where one of the high- and low-side switching devices (on opposite sides of the bridge) are replaced with diodes. This eliminates the shoot-through failure mode, and is frequently utilized to drive variable/switched reluctance machines and actuators where bi-directional present flow just isn’t needed.

A “double pole double throw” relay can usually accomplish the exact same electrical functionality as an H bridge (taking into consideration the usual function of the device). An H bridge could be preferable to the relay where a smaller physical size, high speed switching, or low driving voltage is needed, or where the wearing out of mechanical parts is undesirable.

H-Bridge Circuit sample

Amplifies a voltage (multiplies by a constant greater than 1)

• Input impedance
  • The input impedance is at least the impedance between non-inverting ( + ) and inverting ( − ) inputs, which is typically 1 MΩ to 10 TΩ, plus the impedance of the path from the inverting ( − ) input to ground (i.e., R₁ in parallel with R₂).
  • Because negative feedback ensures that the non-inverting and inverting inputs match, the input impedance is actually much higher.

• Although this circuit has a large input impedance, it suffers from error of input bias current.
  • The non-inverting ( + ) and inverting ( − ) inputs draw small leakage currents into the operational amplifier.
  • These input currents generate voltages that act like unmodeled input offsets. These unmodeled effects can lead to noise on the output (e.g., offsets or drift).
  • Assuming that the two leaking currents are matched, their effect can be mitigated by ensuring the DC impedance looking out of each input is the same.
    • The voltage produced by each bias current is equal to the product of the bias current with the equivalent DC impedance looking out of each input. Making those impedances equal makes the offset voltage at each input equal, and so the non-zero bias currents will have no impact on the difference between the two inputs.
    • A resistor of value

      

    which is the equivalent resistance of R₁ in parallel with R₂, between the V_in source and the non-inverting ( + ) input will ensure the impedances looking out of each input will be matched.

    • The matched bias currents will then generate matched offset voltages, and their effect will be hidden to the operational amplifier (which acts on the difference between its inputs) so long as the CMRR is good.
  • Very often, the input currents are not matched.
    • Most operational amplifiers provide some method of balancing the two input currents (e.g., by way of an external potentiometer).
    • Alternatively, an external offset can be added to the operational amplifier input to nullify the effect.
    • Another solution is to insert a variable resistor between the V_in source and the non-inverting ( + ) input. The resistance can be tuned until the offset voltages at each input are matched.
    • Operational amplifiers with MOSFET-based input stages have input currents that are so small that they often can be neglected.

An inverting amplifier makes use of negative feedback to invert and amplify a voltage. The R_in,R_f resistor network permits a few of the output signal to be returned towards the input. Because the output is 180° out of phase, this value is properly subtracted from the input, thereby lowering the input into the operational amplifier. This cuts down the overall gain with the amplifier and is dubbed negative feedback.

• Z_in = R_in (because V ₋ is a virtual ground)
• A third resistor, of value , added between the non-inverting input and ground, while not necessary, minimizes errors due to input bias currents.

The gain of the amplifier is determined by the ratio of R_f to R_in. That is:

The presence with the negative sign is really a convention indicating that the output is inverted. For instance, if R_f is 10 kΩ and R_in is 1 kΩ, then the gain would be -10 kΩ/1 kΩ, which is -10.

Theory of operation: An Perfect Operational Amplifier has two characteristics that imply the operation of the inverting amplifier: Infinite input impedance, and infinite differential gain. Infinite input impedance implies there isn’t any current in either with the input pins simply because current cannot flow through an infinite impedance. Infinite differential gain implies that both the (+) and (-) input pins are in the same voltage because the output is equal to infinity occasions (V+ – V-). Because the output approaches any arbitrary finite voltage, then the term (V+ – V-) approaches 0, therefore the two input pins are at the exact same voltage for any finite output.

To start analysis, first it’s noted that with the (+) pin grounded, the (-) should also be at 0 volts possible due to implication two. using the (-) at 0 volts, the present via Rin (from left to correct) is given by I = Vin/Rin by Ohm’s law. Second, because no present is flowing into the op amp via the (-) pin due to implication 1, all of the current via Rin should also be flowing via Rf (see Kirchoff’s Present Law). Consequently, with V- = 0 volts and I(Rf) = Vin/Rin the output voltage given by Ohm’s law is -Vin*Rf/Rin.

Actual op amps have each finite input impedance and differential gain, nevertheless each are high enough as to induce error that’s considered negligible in most applications.