## How to Make 10V DC Power Supply (PCB included)

Hi friends, in this short tutorial we are going to learn how to make 10V DC power supply using LM350 IC. You can use this circuit as your first or second year engineering electronics mini project. This circuit is capable of producing 10V DC output at a current rating of maximum 3A. It uses LM350, a three terminal positive voltage regulator. It can produce an output voltage ranging from 1.2V to 33V for the input voltage of 14V to 24 V obtained from the AC mains through a step-down transformer. The capacitor C2 ensures the stability of the output from the regulator. To get different levels of the output voltage, the values of R2 and C3 can be varied accordingly.

### Components required

• C1 = 0.1uF
• C2 = 1uF
• C3 = 10uF
• D1,D2,D3,D4 = IN5001
• F1 = 3A
• J1,J2 are screw terminals
• R1 = 240Ω
• R2 = 1.65K
• U1 = LM350

### Features of 10V DC Power Supply

Input(V): 14VAC to 24VAC
Output(V): 10 VDC
PCB:68mmX35mm

Note: Use a ccl with 2oz Cu thickness for PCB fabrication.

## Shunt Voltage Regulator – Working Principle

A zener diode forms an integral part of any voltage regulator. Before we go ahead, we know, that in all cases, the voltage across a zener diode will remain constant. i.e. ∆VZ = 0. In all cases, we indicate load resistance by RL.

### Regulator using zener diode only

• Across RL we have: V = VZ = ILRL                                                                (Equation 1)
• From current law: I = IZ + IL                                                              (Equation 2)
• From KVL along indicated path: VS = I*R + VZ                             (Equation 3)

Equation 1 tells that output voltage VO will always be constant = VZ.

Assume two cases:

• Assume supply current I change by dI
From Equation 2: ∆I = ∆IZ + ∆IL
From Equation 1: ∆VZ = ∆ILRL ; or, ∆IL = 0 (since ∆VZ = 0)
Thus ∆I = ∆IZ. This shows that excess current is bifurcated through the zener diode.
• Assume load RL changes by ∆RL (with VS constant)
Output voltage VO will remain constant, but change in IL will be compensated by change in IZ
From Equation 3: ∆VS = ∆I*R+ ∆VZ ;      or, 0 = ∆I*R + 0 ;               or, ∆I = 0
From Equation 2: ∆I = ∆IZ + ∆IL ;            or, 0 = ∆IZ + ∆IL  ;             or, ∆IL = – ∆IZ

Thus if IL increases, IZ decreases and vice versa.

### Regulator using transistor and zener diode

Few points:

Correlating VO and indicated path from point X to GND: VO = VX = VZ + VBE          (Equation 1)

I = IB + IC + IL ;   or, I = IC + IL(since IB is very small)                                                     (Equation 2)

The increase in VBE causes more collector current IC to flow.

• Assume current I increase by dI keeping VS constant (opposite analysis will take place of I decreases)∆I is positive. VS – I*R = VX ;         or, 0 – ∆I*R = ∆VX ; (since VS is constant)
i.e. VX = VO  decreases on increase in I.                                                                  (Effect 1: VO tends to decrease)

Next, from Equation 1: ∆VO = ∆VZ + ∆VBE ;            or, ∆VO = 0 + ∆VBE ;
i.e VBE also decreases on decrease in VO
As VBE decreases, IC decreases.

From Equation 2: ∆I = ∆IC + ∆IL
If ∆I = positive (assumed);           ∆IC = negative (as VBE decrease);           so IL must increase.
The voltage across load VL = IL*Rincreases.                                                       (Effect 2: VO tends to increase)

The Effect 1 and Effect 2 neutralize and VO is constant.

• Assume supply voltage VS is increased keeping current I constantThe analysis will take place just as done previously.
VS – I*R = VX ;    or, ∆VS – 0 = ∆VX ; (since I is constant)
i.e. VX = VO increases on increase in VS.                                                                 (Effect 1: VO tends to increase)

Next, from Equation 1: ∆VO = ∆VZ + ∆VBE ;            or, ∆VO = 0 + ∆VBE ;
i.e VBE also increases on increase in VAs VBE decreases, IC increases.

From Equation 2: ∆I = ∆IC + ∆IL
If ∆I = 0 (assumed);         ∆IC = positive (as VBE increase);  so IL must decrease.
The voltage across load VL = IL*RL decreases.                                                     (Effect 2: VO tends to decrease)
The Effect 1 and Effect 2 neutralize and VO is constant.

## Linear Voltage Regulator – Series and Shunt type

Hi friends, in this article, we will take a basic overview of a linear voltage regulator and its types. This includes the block diagram and working principles.

### What is voltage regulator?

A voltage regulator prevents the varying of the voltage across a load in spite of variation in the supply. It is also used to regulate or vary the output voltage of the circuit.

Two terms:

• Regulation: compensates for variation in the mains (line voltage)
• Stabilization: compensates for variation in load current

However, in practice, both the terms loosely used for the same meaning of voltage regulation.

### Types of voltage regulator

There are mainly two types:

1. Series voltage regulator
2. Shunt voltage regulator

### Series voltage regulator:

A simple block diagram is as follows

The series voltage regulator controls variation in voltage (DVS) across the load by providing a voltage in series with the load.

A further more detailed block diagram is shown. A series regulator has its regulating unit in series with the load.

There is always a voltage drop in the regulating unit (VR). The output voltage VO (or VL) is:

VL = VS – VR

Series voltage regulator usually has a negative feedback system. If load voltage (VL) tends to fall, smaller feedback decreases controlling unit resistance thereby allowing more current to flow in the load (VR decreases) and increasing VL. Opposite happens when VL increases.

### Shunt voltage regulator

Block diagram is as follows

Shunt voltage regulator controls the voltage across the load by varying the current flowing through the load (IL) and through the regulating unit (IR).

A further detailed block diagram is shown below. A shunt regulator has its regulating unit in parallel to the load.

I = IR + IL

The stability in the voltage across the load RL is brought about by ensuring a steady current flow through it.

When the current across RL tends to increase, regulating unit prevents it by allowing the excess current to flow through it. Since current I is constant, IL decreases.

Same happens when current IL tends to decrease. Regulating unit prevents it by decreasing current flow (IR) through it, thereby increasing IL.

Tags: LM317, adjustable voltage regulator, zener diode voltage regulator, voltage regulator 7805, vrm, avr, ldo.

## Voltage to Current Converter using Op Amp

Hi friends, in this post we will see how to convert voltage into current using simple circuitry. In most of the cases we get the output of measuring devices in the form of voltage. It is not good to transmit this output voltage to the destination directly. Due to addition of noise and wire impedance the output voltage may get corrupted. So in such cases we have convert that voltage into current form. So let us see voltage to current converter.

#### Voltage to Current Converter using Op Amp

Following circuit shows the voltage to current converter using operational amplifier. It consist of simple resistance connected to the inverting and non inverting terminals of op amp.

In this circuit the load is grounded and the current through the load can be calculated as follows.

$I_{1}=\left (\frac{V_{in}-V_{2}}{R} \right )$

$I_{2}=\left (\frac{V_{o}-V_{2}}{R} \right )$

The current through the load is given by,

$I_{L}=I_{1}+I_{2}=\left (\frac{V_{in}-V_{2}}{R} \right )+\left (\frac{V_{o}-V_{2}}{R} \right ) =\frac{V_{in}+V_{0}-2V_{2}}{R}$

The gain of the amplifier is

$\left(1+\frac{R}{R}\right)=2$

So

$V_{0}= 2V_{2}$

Substituting this value in above equation we get,

$I_{L}=\left (\frac{V_{in}}{R} \right )$

Thus the current is directly proportional to the applied voltage and the resistance used in the circuit. it should be noted that all the resistances used in the circuit are equal to R.

See alos: Current to Voltage Converter

I hope you are satisfied with the above given explanation of voltage to current converter. If you liked this article please share it with your friends and like our facebook page for such articles.

## Superposition Theorem with Solved Example

Hi friends, here are going to learn a superposition theorem. We will also solve on simple example using superposition theorem.

### Statement:

Superposition theorem states that, the responce in any element of LTI linear, bilateral network containing more than one sources is the sum of the responses produced by the sources each acting independently.

The superposition theorem is not applicable for the power, as power is directly proportional to the square of the current which is not a linear function.

### Steps:

1)      Select any one source and short all other voltage sources  and open all current sources if internal impedance is not known. If known replace them by their impedance.

2)      Find out the current or voltage across the required element, due to the source under consideration.

3)      Repeat the above steps for all other sources.

4)      Add all the individual effects produced by individual sources to obtain the total current in or across the voltage element.

Example: Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in figure below.

Solution:

Step 1: Remove the 8V power supply from the original circuit, such that the new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 2K are in parallel, therefore resultant resistance will be 1.245K.

Using voltage divider rule voltage across 1.245K will be

V1= [1.245/(1.245+4.7)]*5 = 1.047V

Step 2: Remove the 5V power supply from the original circuit such that the new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be 1.938K.

Using voltage divider rule voltage across 1.938K will be

V2= [1.938/(1.938+2)]*8 = 3.9377V

Therefore voltage drop across 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847

### You may also like:

Tags: superposition theorem solved problems, superposition theorem examples, superposition principle.

## RC and LC Filters:

Hi friends, today we are going to learn some basic filter circuits like RC filter and LC filter.

### RC filter:

In the above figure two sections of RC filter are shown. These are connected between the input capacitor and the load resistor. The value of R should be at least 10 times greater than the capacitive reactance Xc. Therefore the ripple is dropped across the series resistors instead of across the load resistor. Each section reduces the ripple by a factor of 10. Therefore ac components are removed and at the output we get a steady dc voltage.

The main disadvantage of RC filter is the loss of dc voltage across each R. Therefore RC filter is suitable only for light loads. i.e. small load current.

### LC filters:

In this type inductor L is in series and capacitor C is in shunt with load. The choke (L) allows the dc component to pass through easily because its dc resistance R is very small. The capacitive reactance Xc is very high for dc and it acts as open circuit. All dc current passes through across which dc output voltage is obtained.

The inductive reactance XL = 2πfL is high for ac components. Therefore the ripples are reduced. Even if any ac current passes through L, it flows through the capacitor because of its low capacitive reactance.

1) In choke input filter, current flows continuously. Therefore the transformer is used more efficiently.

2) Ripple content at the output is low.

3) It is less dependent on the load current.

4) DC voltage drop across L is much smaller because its de resistance R is very small.

1) Large size and weight of inductors,

2) More cost,

3) External hold is produced by inductor.

For providing smoothest output voltage π type (capacitor input) filter can be used.

### You may also like:

Tags: RC filter working principle, LC filter working principle.

## Series inductor filter

This post provides an information about series inductor filter. Before going to study series inductor filter it is necessary to understand what is meant by filter circuit?

Filter Circuit – Most of the electronic circuits require a dc voltage that is constant, similar to the voltage from a battery but the rectifiers cannot provide ripple free dc voltage. They provide a pulsating dc. The circuit used for filtering or smoothing out the ac variations from the rectified voltage is called as Filter circuit.

Series Inductor Filter:

The circuit diagram and waveforms of series inductor filter is shown below:

Series Inductor Filter – An inductor opposes any change in the current flowing through it. Whenever the current through an inductor tends to change, a back emf is induced in it. This prevents the change in current.

Inductive reactance XL = 2π* f*L. For dc, f = 0, therefore direct current easily passes through inductor to the load. Only opposition to dc is due to internal resistance of choke. The reactance increases with frequency. Therefore ac component is opposed. The output waveform shows a large dc component and a small ac component.

The operation of a series inductor filter depends upon the current through it. The higher the current flowing through it, the better is its filtering action. An increase in load current reduces the ripples.

## Capacitor input filter:

This post provides an information about capacitor input filter, its working principle, and applications.

What do you mean by filter circuit?

Filter Circuit – Most of the electronic circuits require a dc voltage that is constant, similar to the voltage from a battery but the rectifiers cannot provide ripple free dc voltage. They provide a pulsating dc. The circuit used for filtering or smoothing out the ac variations from the rectified voltage is called as Filter circuit’.

Types of Filters –

• 1)
• 2)      Inductor filter
• 3)      RC filter,

### Capacitor Input Filter:

In this type a capacitor C is connected in shunt with load resistor R. Capacitor acts as open circuit for dc. Therefore direct current passes through the load. Capacitor provides low-reactance path to the ac components of current. AC gets bypassed to the ground. Only a small part of the ac component passes through the load producing a small ripple voltage.

When the rectifier output is increasing, the capacitor charges to the peak voltage Vp. When the rectifier voltage becomes slightly less than Vp, the capacitor starts to discharge through the load (point B). This prevents the load voltage from falling to zero. The capacitor discharges until the rectifier voltage becomes more than the capacitor voltage (point C). The capacitor again charges to the peak value Vp. The current is maintained through the load all the time.

For getting steadier output voltage, the time constant C*R should be large. Then load current is small and capacitor does not discharge very much and the average load voltage Vdc, is slightly less than the peak value Vp.

### You may also like:

Tags: capacitor input filter rectifier.