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Inductors in Parallel

Inductors are said to be connected in parallel when two terminals of an inductor respectively connected to each terminal of other inductors or inductor. Similar to the parallel connection of resistors, the total inductance in parallel connection of inductors is somewhat lesser than smallest inductance of an inductor in that connection.

When the inductors are connected in parallel, the current flow through each inductor is not exactly equal to the total current, but the sum of each individual current through parallel inductors gives the total current (as it divides among parallel inductors).

If the current flow through the each inductor is less than the total current, the magnetic field generated by each inductor is also less than that of field generated by total current through it.

In case of resistors in parallel, most of the current flows through the smallest resistor as it offer the least opposition to the current flow than larger resistor.

Likewise, if the inductors are connected in parallel, current chooses least opposition path of inductor when current in that circuit is decreased or increased while each inductor individually opposes that change (increase or decrease of current).

In the parallel connection, the voltage across each inductor is equal and also if the total current is changed, the voltage drop across each individual inductor will be less as compared with series connection. For a given rate of change of current, less will be the inductance in less voltage.

These are the basic points to be considered in parallel connection of inductors. Now we will discuss the parallel connection of inductors with and without considering mutual coupling between these inductors.

Inductors Connected in Parallel (Without Magnetic Coupling)

As we discussed above, one end of inductors is connected to a node and other ends of inductors are collectively connected to another node in parallel connection. The parallel connection of n inductors is shown in below figure.

Consider that there is no magnetic coupling between inductors and hence the total inductance equal to the sum of reciprocals of the individual inductances. Let us discuss how this statement can be obtained.

We know that, in a parallel network the voltage remains constant and the current divides at each parallel inductor. If IL1, IL2, IL3 and so on ILn are the individual currents flowing in the parallel connected inductors L1, L2 and so on Ln, respectively, then the total current in the parallel inductors is given by

ITotal = IL1 + IL2 + IL3 . . . . + In

If the individual voltage drops in the parallel connection are VL1, VL2, VL3 and so on VLn, then the total voltage drop between the two terminals VT is

VTotal = VL1 = VL2 = VL3 . . . . = Vn

The voltage drop in terms of self inductance can be expressed as V = L di/ dt. This implies total voltage drop,

VT = LT di/dt

⇒ LT  d/dt (IL1 + IL2 + IL3 . . . . + In)

⇒ LT ( (di1)/dt + (di2)/dt + (di3)/dt . . . .)

Substituting V / L in place of di/dt, the above equation becomes

VT = LT (V/L1+ V/L2 + V/L3 . . . .)

As the voltage drop is constant across the circuit, then v = VT. So we can write

1/LT = 1/L1 + 1/L2 + 1/L3 . . . . .

This means that the reciprocal of total inductance of the parallel connection is the sum of reciprocals of individual inductances of all inductors. The above equation is true when there is no mutual inductance is affect between the parallel connected coils.

For avoiding complexity in dealing with fractions, we can use product over sum method to calculate the total inductance. If two inductors are connected in parallel, and if there is no mutual inductance between them, then the total inductance is given as

LT = (L1× L2)/(L1+ L2)

Example of Inductors Connected in Parallel

If a circuit has 2 inductors of 20 Henry and 30 Henry connected in parallel, what will be the total inductance of the parallel arrangement?

Sol: We know that the formula for total inductance of series, 1/LT = 1/L1+ 1/L2

Given that L1 = 20 Henry

L2 = 30 Henry

LT = (L1* L2)/(L11+ L2 ) = ((20*30))/((20+30) ) = 600 / 50 = 12

The total inductance is LTotal = 12 Henry.

Mutually Coupled Inductors in Parallel

When there exist magnetic coupling between the inductors, the above derived formula for total inductance must be modified because total inductance can be more or less depending on the magnetic field directions from each inductor. The magnetic flux produced by the parallel connected inductors will link with each other.

When the fluxes produced are in the same direction of magnetic flux, the mutual inductance will increase; these coils are called “Aiding” coils. If the flux is in opposite direction to the magnetic flux, the mutual inductance will decrease; these coils are called “Opposing” coils. This mutual inductance will depend upon the placed distance of two coils.

Consider that two inductors are connected in parallel with self inductances L1 and L2, and which are mutually coupled with mutual inductance M as shown in below figure.

Parallel Aiding Inductors

Consider the figure (a), in which inductors L1 and L2 are connected in parallel with their magnetic fields aiding. The total current through the circuit is given as

i = i1 + i2

di/dt = (di1)/dt + (di2)/dt …………. (1)

The voltage across the inductor or parallel branch is given as

V = L1 (di1)/dt + M (di2)/dt or L2 (di2)/dt + M (di1)/dt

L1 (di1)/dt + M (di2)/dt = L2 (di2)/dt + M (di1)/dt

(di1)/dt (L1– M) = (di2)/dt (L2– M)

(di1)/dt = (di2)/dt ((L2 – M))/((L1 – M)) …………. (2)

Substituting equation 2 in equation 1, we get

di/dt = (di2)/dt ((L2– M))/((L1– M)) + (di2)/dt

di/dt = (di2)/dt { (L2– M))/((L1– M)) + 1} …………. (3)

If LT is the total inductance of the parallel inductor circuit , then voltage is given by

V = LT di/dt

LT di/dt = L1 (di1)/dt + M (di2)/dt

di/dt = 1/ LT { L1 (di1)/dt + M (di2)/dt }

Substituting equation 2 in above equation, we get

di/dt = 1/ LT { L1 (di2)/dt (L2– M))/((L1– M)) + M (di2)/dt }

di/dt = 1/ LT { L1 (L2– M))/((L1– M)) + M }(di2)/dt …………. (4)

Equating equations 3 and 4 we get

(L2– M))/((L1– M)) + 1 = 1/ LT { L1 (L2– M))/((L1– M)) + M }

Simplifying the above equation, results

LT = (L1 L2– M2)/(L1+ L1 )-2M)

Here 2M represents the magnetic flux of L1 on L2 or L2 on L1. If the magnitude of two inductances is equal with a perfect magnetic coupling between them, the equivalent inductance of two inductors is L because LT = L1 = L2 = M. In such case, if the mutual inductance is zero, the total inductance would be L ÷ 2.

Parallel Aiding Inductors Example

If the two inductors 25mH and 45mH are connected in parallel aiding, calculate the total inductance of the parallel combination. The mutual inductance is given as 20mH.

SolGiven that, L1 = 25 mH

L2 = 45 mH

M = 20 mH

Applying the formula for total inductance of aiding inductors, LT = (L1 L2– M2)/(L1+ L1 )-2M)

LT = (25*45- 202)/(25+45-2*20)

= (1125-400)/(70-40)

= 725/30

= 24.166mH

Therefore the total inductance is 24.166 milli Henry.

Parallel Opposing Inductors

Similarly, if we consider the figure (b), in which inductors L1 & L2 are connected in parallel with their magnetic fields opposing, the total inductance is given as

LT = (L1 L2– M2)/(L1+ L2)+ 2M)

In opposing parallel inductors, if the magnitudes of two inductances are equal with the perfect magnetic coupling, the equivalent inductance of two inductors will be zero, as they cancel each other. If two inductors effectively allows the current to flow through them, the total inductance is given as (L ± M) ÷ 2.

Parallel opposing Inductors Example

Ex: If the two inductors 25mH and 45mH are connected in parallel opposing, calculate the total inductance of the parallel combination. The mutual inductance is given as 20mH.

Sol: Given that, L1 = 25 mH

L2 = 45 mH

M = 20 mH

Applying the formula for total inductance of aiding inductors, LT = (L1 L2 M2)/(L1+ L2 )+2M)

LT = (25*45-202)/(25+45+2*20)

= (1125-400)/(70+40)

= 725/110

= 6.59mH

Therefore the total inductance is 6.59 milli Henry.

Summary

• Connecting the two terminals of inductor respectively to other inductor or inductors terminals, then that connection is referred as “parallel connection of inductors”.
• When the fluxes produced by individual inductors are in the same direction, the mutual inductance will be increased; then these coils are called “Aiding” coils. Total inductance for aiding coils is LT = (L1 L2– M2)/(L1+ L2 )-2M).When the fluxes produced by individual inductors are in the opposite direction of magnetic flux, the mutual inductance will be decreased; then these coils are called “opposing” coils. Total inductance for aiding coils is LT = (L1 L2– M2)/(L1+ L2 )+2M)

USB MP3 Player

Many of us have the hobby of listening to music in leisure time.Mp3 players which are USB compatible are very useful to carry any where and listen to music at anytime.

This small sized, portable and low cost USB MP3 player, can play audio files on USB. For implementing this portable device, you just need very few basic electronics components and hence you require less assembling time.

Now we are going to design and build a compact size USB MP3 player module which can read and play the audio files stored in MP3 format on USB. This circuit uses a digital IC to convert the digital voice data into analog audio that can be heard through headphones or audio speakers

Circuit Components

• R1, R5- R7- 1.5K
• R2- R4 – 22K
• R8 – 1M
• C1, C2 – 33uF
• C3- C8 – 100nF
• Crystal Oscillator – 12MHz
• PCM2704CDB IC
• Push Buttons
• Headphones – 32ohms
• USB interface (Female)

Circuit Diagram of USB MP3 Player

USB MP3 Player Circuit Design

1. This design uses PCM 2704CDB IC chip as a heart of this circuit. This 28 pin IC is a stereo audio digital to analog converter with USB 2.0 interface compatible. This IC can operate in an adaptive asynchronous mode for playback functions. It is to be noted that it is a plug and play device which requires no software code.
2. Connect the pins of USB female connector to the respective pins of PCM 2704CDB IC (D+, D- and VBUS) making sure that D+ and D- pins are connected through 22 ohm resistor.
3. Connect each push button to Human Interface Device (HID) pins (HID0, HID1 and HID2) through 1.5K ohm resistors by making a common connection to suspend flag pin. And also connect the 12 MHz crystal oscillator and all remaining capacitors to the pins as shown in the figure.
4. Connect a headphone to headphone jack by taking inputs from Vout R (15) and Vout L (14) of the IC.
5. Connect the power supply 5V (VBUS) to the respective pins shown in the figure.

USB MP3 Player Circuit Operation

Turn ON the power supply to the circuit by giving two respective DC voltages to the pins. Once the circuit is powered, it gets audio data and control data from USB through D- and D+ pins. Due to high speed transceivers, all this data transfer is carried out at full speed.

This IC decodes the received data and converts to the analog audio which will be available at the VoutR and VoutL pins of the IC. It is also possible to increase/decrease the volume and also mute by pressing corresponding push buttons connected to HID pins.

We can also connect audio speaker in place of headphones using appropriate power amplifier circuit between Vout pins and audio speaker. We can extend this circuit for remote control operation by adding an IR sensor with the corresponding receiver circuit.

Advantages of USB MP3 Player

Portable device as it consist less number of components
Much easier construction
Facilitates headphones as well as audio speaker output
Most feasible and low cost alternative to commercial MP3 players

Note:

• +5V DC and +3.3V DC power supplies have to be derived from respective low dropout (LDO) regulators.
• Loudspeakers or Low impedance headphones must need an amplifier circuit at Vout terminals to make audio signals audible.
• It is recommended to go through the data sheet of PCM 2704CDB IC before going to build this circuit. You can download this datasheet by clicking here.

Digital Alarm Clock

Alarm Clocks are very useful devices in today’s busy life. They are designed to make a signal / alarm at a specific time. The use of digital alarm clocks has increased over years with development in electronics.

The advantage of digital alarm clocks over analogue alarm clocks is that they require less power, the time can be set or reset easily and displays the time in digits. Design of a simple digital alarm clock is explained here.

Using this digital alarm clock, time can be displayed in 24 HR format using an LED display and alarm can be set to a specific time. Alternatively, the digital alarm clock circuit can also be used to turn ON/OFF and electrical appliance after a specific time.

Digital Alarm Circuit Diagram

Digital Alarm Circuit Components

• IC1 – LM 8560
• IC2 – LM 386
• Duplex LED Display
• T1 – 0-12V / 500mA Transformer
• D1, D2, D3 – IN4001
• R1 – 100K
• R2, R3 – 120
• R4 – 91K
• R5 – 10
• P1 – 10K Potentiometer
• C1 – 470µF
• C2, C3 – 10nF
• C4 – 10µF
• C5, C6 – 100nF
• C7 – 100µF
• B1 – Buzzer
• S1, S2, S3, S4 – Switches

Digital Alarm Circuit Design

LM8560

The heart of the digital alarm clock is the IC LM8560. It is an alarm equipped digital clock IC. The drivers required to drive a duplex LED display to show the time are in-built to the IC. The functions of the IC LM8560 are display of current time, alarm support, snooze function and supports 12 HR and 24 HR time format.

LM386

It is a low voltage audio power amplifier. It is used in the circuit to drive the alarm buzzer. The alarm output of the LM8560 is given to power amplifier through a potentiometer.

Duplex LED Display

In the digital alarm clock, to display the information like current time, alarm notification etc. we use a duplex LED display. There is no need for any extra drivers to drive the display unit as all the required drivers are built in to the digital clock IC.

Push Button Switches (S1 – S4)

These four switches are used to set time, set alarm and turn off the alarm. When setting time and alarm in the digital alarm clock, both Hours and Minutes can be set separately using these switches.

Diodes (D1, D2, and D3)

Diode D1 in combination with the capacitor C1 will produce a DC voltage that is supplied to the IC1 (LM8560). Diodes D2 and D3 act as switch signal generator to the cathode of the duplex display. They work in relation to the input of the LM8560.

Buzzer (B1)

The main function of the digital alarm clock is to signal an alarm at specified time. A buzzer is used for this purpose which is driven by an audio power amplifier.

Transformer (T1)

A transformer is used to step down the 220-240 V AC supply to 12 V AC supply.

Digital Alarm Circuit Working

A digital alarm clock is a very useful device as it can display the current time as well as function as an alarming device. The circuit shown above can act as a simple digital alarm clock without requirement of any programming.

The components required and the complexity of the circuit are less. The working of the above shown digital alarm clock is as follows.

As explained in the component description, the heart of the digital alarm clock circuit is the digital clock IC LM8560. The input from the mains is given to the transformer which converts the input voltage to 12V AC.

A diode (D1) and a capacitor (C1) are used to convert this AC voltage to DC and supplied to the Vdd (Pin 20) of the IC1. IC1 (LM8560) works at a frequency of 50/60 Hz. This signal can be supplied from the mains supply using the resistor R1 to the Pin 25 of the IC1.

In relation to the input of the IC1, corresponding inputs to the cathode of the display are given through the diodes (D2 and D3). After switching on the power, we need to set the time. This is the current time to be displayed. In order to set the current time, we use the switches S1 and S2.

Switch S1 is used to set Hours and S2 is used to set Minutes. LM8560 automatically triggers the display connected to it and the time will be displayed on the duplex LED display.

The duplex display is a 4-bit seven segment LED display, where two bits are for displaying the Hours and two bits are for displaying Minutes and the Hour and Minute bits are separated by a colon.

Once the time is set, we can set the alarm using the switches S1, S2 and S3. In order to differentiate between the setting of current time and alarm time in the digital alarm clock, we use the switch S3 which internally triggers the alarm display of the LM8560.

In order to set the alarm, we need to hold the switch S3 and set the Hour and Minutes of the alarm time by using switches S1 and S2 respectively. Once the alarm is set successfully, the digital clock IC waits for the alarm time and once the alarm time is reached, an alarm output at the Pin 3 is generated.

This can be connected to a small buzzer directly for alarm output. But in order to get special alarm tones from the digital alarm clock, we can use special alarm signals. But to drive these special buzzers, we need an audio power amplifier. Hence, the IC LM386, which is a power audio amplifier, is used in the digital alarm clock circuit.

The alarm output of the IC1 (LM8560) is given to the input (Pin 3) of the IC2 (LM386) through a potentiometer P1. The potentiometer is used to control the volume and pressure of the alarm signal from the buzzer.

Once, the digital alarm clock starts the alarm signal, we need to stop this alarm. For this, we use the switch S4. The switch S4 can also be used to cancel the alarm any time before triggering.

This digital alarm clock can also be provided with snooze and repeat alarm facilities. Additional Pins of the IC1 (LM8560) have to be used in the digital alarm clock circuit in order to use these functionalities.

Guidelines & Alternative Components

• The digital alarm clock can be used to view the current time as well as act as an alarming device.
• Even though the supply given here is from mains, a battery powered device can also be designed. In order to design the digital alarm clock as a battery powered device, we need to use an oscillator to generate a 50Hz Sine wave for the digital clock IC (LM8560) to work.
• Some of the alternatives to the digital clock IC’s are TMS 3450, LM8361 and MM5387.
• The duplex display described in this digital alarm clock circuit is Longtek 6052X-S. Other alternatives can also be found.
• To use the snooze and repeat alarm functionalities of the digital alarm clock, we need to add two more switches to the digital clock IC (LM8560).
• The digital alarm clock circuit can also be used as a timer for electric appliances where they can be turned ON/OFF after a specific time. For this, we need to use an additional pin (Pin 23) on the digital clock IC (LM8560) with a switch. This pin has a sleep functionality.