Parallel Resistors | Formula Of Parallel Resistors

Resistors

Parallel Resistors | Formula Of Parallel Resistors

apai    August 04, 2022   

I have discussed series connected resistor , it is time to learn more about parallel connected resistor. First I would like to mention the definition of parallel circuits.

Two components are connected in such a way that they have two common connections. A parallel circuit is a circuit in which all components are connected in parallel.

Similarly, two resistors are connected in parallel if they have two common terminals. It is often confusing to identify parallel circuits. Beginners will have to redesign the circuits.

Properties of parallel circuits:

• The voltage remains the same across each resistor connected in parallel.
• The current is divided proportionally between all the resistors. The higher the value of the resistor, the less current will flow through the resistor.

Formula for parallel forces:

illustration 1

Consider the circuit in Figure 1. A voltage source is connected in parallel with three resistors. As I explained earlier, the voltage of each resistor connected in parallel remains the same. The current is divided between the resistors. Therefore, using Ohm's law, we can easily calculate the current flowing through each resistor.

Current through R1:

i1 = v / R1

Current through R2:

i2 = v / R2

Current through R3:

i3 = v / R3

The total current in the circuit:

i = i1 + i2 + i3

i = v/R1 + v/R2 + v/R3

i = v{1/R1 + 1/R2 + 1/R3}

i/v = 1/R1 + 1/R2 + 1/R3

According to Ohm's law, i/v = 1/Req. Substitute into the above equation

1 / Required = 1 / R1 + 1 / R2 + 1 / R3

Request = R1 || R2 || R3

The easiest way to solve the above equation is the reciprocal method. First solve the individual fractions and then add them.

Worked examples:

Example 1:

Determine the total or equivalent resistance and the current per resistor using Ohm's law.

Equivalent resistance: It is easy to solve with the previous formula.

1 / Required = 1 / R1 + 1 / R2 + 1 / R3

1 / Required = 1 / 1k + 1 / 2k + 1 / 3k

Required = 545.5 Ω

Operation:

leave it

i1 = current through R1

i2 = current through R2

i3 = current through R3

The voltage remains across the three resistors because they are connected in parallel.

.

i1 = v1 / R1

i1 = 10/1000

i1 = 10mA

i2 = v1 / R2

i2 = 10/2000

i2 = 5mA

i3 = v1 / R3

i3 = 10/3000

i3 = 3.33 mA

Conclusion:

• The total/equivalent resistance of the circuit is less than the minimum resistance in the parallel circuit
• If we continuously add resistance in parallel, the total resistance of the circuit will decrease
• The higher the resistance, the lower the current. Or you could say that current always follows the path of least resistance.
• There is another simpler method to find the current through resistors in parallel. This method is known as the current divisor rule .

apai    August 04, 2022   

Binary translator. introduction with examples

Translation tools are still very popular and handy for enabling different people to understand native language concepts. We have experience with such useful tools to reduce the level of difficulty and increase the understanding of the concept.

In this article we will talk about a translator that improves the security of our data, as well as some examples of how to change it and a brief explanation of how to change this resource.

What is a binary system?

The binary system is a base 2 number system. Only two numbers are used: 0 and 1. This system is used in physics to say that a system is decoupled or disconnected. We have already transformed a digital system into another digital system.

We have a basic understanding of binary to decimal and decimal to binary conversion.

Two out of ten.

If we have a decimal number in the base 10 system, we can convert it to a binary system. We need to divide the given decimal number by 2 and write the mnemonic.

Example (20) 10 = (10100) 2

What is ASCII?

ASCII is the abbreviation of American Standard Code for Information Interchange. It is a character encoding standard in electrical communications. Most character encodings are based on ASCII. ASCII is a language of codes, which are special forms in the native language. We use the ASCII system to write the binary language in English.

Binary in English.

We use binary code and translate it to English using ASCII. We have a binary number to convert to English. We have a binary code that we need to translate into English.

Example 1.

binary code

011101111 01100101 00100000 01100001 01110010 01100101 00100000 01110100 0110000 01100101 00100000 01100111111111110010010011111111100100 01111111111111111001100

Translation

We are 7th graders.

Example 2.

binary code

0110110100 01100100 01100100 011110 0110110 0110110 001100110 001100 00100 00100 00100

000 01101000 01101111 01110101 01110010 01110011 00101110 00001010

Translation:

We play for 10 hours.

We can use the ASCII table to convert the computer binary system into English, which takes a long time to translate. Use a binary translator to save time and decode the binary code provided in English .

We write the binary code on the left and press the text button

Example 3.

English letter	ASCII Decimal	A binary string
D:	100	01100100
Oh	111:	01101111
VS :	103:	01100111

01100100 01101111 01100111 = dog

Why is ASCII important?

Since ASCII is now the same on all computers, it acts as an interface between our computer screen and the hard drive.

ASCII used;

Computer text is converted to human text using ASCII.

Every computer uses binary, which is a series of 0s and 1s. However, computers have their own language versions, just like English and Spanish may use the same alphabet but have completely different words for the same objects. .

Applications of binary numbers.

Since then, many different applications have used the binary number system. These include various software for digital signal processing, recording high-quality HD music and movies, storing millions of pieces of data, and image processing. A binary converter is a tool that guarantees the success of these applications.

Summary:

In this article, we learned about the ASCII binary system and translated the binary system into English. Now after reading the above article, you can convert any binary code to English and improve your data security.

DC Analysis BJT part 3

BJT BJT Numericals

DC Analysis BJT part 3

apai    August 04, 2022   

This is another tutorial on DC BJT analysis. This tutorial is important by design. How to set circuit parameters to operate in active mode, saturation mode or shutdown mode. For example, changing the base and collector resistances (RC and RB ) will change the base and collector currents . This change will directly affect the operation of the transistor. Design problems are a bit more complicated. Let's begin.

Example 1: Calculation of R B , RC and RE .

\ [I_ {CQ} = \frac {1} {2} I_ {C (sat)} \]

\ [I_ {C (sat)} = 8mA = 4mA \]

\ [V_C = 18 V \]

\[\beta = 110 \]

Apply KVL to the input circuit.

\ [- V_B + I_BR_B + V_ {BE} + I_ER_E = 0 \text {Equation 1} \]

[I_C = \beta I_B\]

\ [I_B = 36 \mu A \]

\ [I_E = I_C + I_B \]

\ [I_E = 4.036 mA \]

Calculate the RC .

\ [R_C = \frac {V_ {CC} -V_C} {I_ {CQ}} \]

\[R_C = 2.5k \omega\]

and RE determines the value of the Q point and the slope of the load line .

\ [I_ {C (sat)} = \frac {V_ {CC}} {R_C + R_E} \]

\ [R_C + R_E = \frac {28} {8m} = 3.5k \Omega\]

\ [R_C = 3.5k-2.5k = 1k \ Omega \]

Insert RE into equation 1 and calculate RB .

\ [- 28 + 36 \ mu R_B + 0.7 + 4.036 m * 1k = 0 \]

\[R_B = 646k\omega\]

example n. 2: Calculate V CC , RC , R B

From load lines,

\ [V_ {CC} = 20 V \]

\ [I_ {C (low)} = 8 mA \]

\ [I_ {BQ} = 40 \ mu LA \]

\ [I_ {C (sat)} = \frac {V_ {CC}} {R_C} \]

\[R_C = 2.5k \omega\]

Apply KVL to the input circuit for RB .

\ [- V_ {CC} + I_B R_B + V_ {BE} = 0 \]

\ [- 20 + 40 \ mu R_B + 0.7 = 0 \]

\[R_B = 482k\omega\]

example n. 3: Design of simple base NPN transistor with following parameters.

\ [I_E = 1.5 mA \]

\ [V_ {CB} = 7.5 V \]

\[V_{CC}=15 V\]

\ [V_ {CE} = - 5V \]

Apply KVL to the input circuit.

\ [- V_ {EE} + V_ {BE} + I_ER_E = 0 \]

\ [R_E = \frac {V_ {EE} -V_ {BE}} {I_E} \]

\ [R_E = \frac {5-0.7} {1.5 m} = 4.5 k \Omega\]

KVL at the output of the loop.

Note that the common base configuration has unique current gain and therefore I C = I E .

\ [- V_ {CC} + V_ {CB} + I_CR_C = 0 \]

\ [R_C = \frac {V_ {CC} -V_ {CB}} {I_C} \]

\[R_C = \frac {15-7.5} {1.5 m} = 5k \Omega\]

Example #4: Calculation of RC and RE .

\ [I_E = 2 mA \]

\[V_{CB}=9V\]

KVL in the input loop.

\ [- V_ {EE} + V_ {BE} + I_ER_E = 0 \]

\ [R_E = \frac {V_ {EE} -V_ {BE}} {I_E} \]

\ [R_E = \frac {10-0.7} {2m} = 4.7k \Omega \]

KVL at the output of the loop.

\ [- V_ {CC} + V_ {CB} + I_CR_C = 0 \]

\ [R_C = \frac {V_ {CC} -V_ {CB}} {I_C} \]

\ [R_C = \frac {20-9} {2m} = 5.5k \Omega\]

Example #5: Calculation of R B and RC .

Offset point value.

\[V_{CE}=6V\]

\ [I_C = 2 mA \]

\ [I_B = \frac {I_C} {\beta} = 20 \mu A \]

Find the possible values ​​of the polarization point for β = 50 - 150.

Evaluate KVL and RB in the input loop.

\ [- V_ {CC} + I_BR_B + 0V_ {BE} = 0 \]

\ [R_B = \frac {V_ {CC} -V_ {BE}} {I_B} \]

\ [R_B = \frac {12-0.7} {2 \mu} = 565k \Omega \text {equation 1} \]

Estimate KVL and RC in the output circuit .

\ [- V_ {CC} + I_CR_C + V_ {CE} = 0 \ text { equation 2} \]

\ [R_C = \frac {V_ {CC} -V_ {CE}} {I_C} \]

\ [R_C = \frac {12-6} {2m} = 3k \Omega \]

For β = 50 and β = 150: calculate the minimum and maximum I C and V CE .

From equation 1:

\ [I_B = \frac {12-0.7} {565k} = 20 \mu LA \]

\[I_{C(min)} = \beta_{min}I_B\]

\ [I_ {C (min)} = 50 * 20 \ mu = 1 mA \]

\[I_{C(max)} = \beta_{max}I_B\]

\ [I_ {C (max)} = 150 * 20 \ mu = 3 mA \]

From equation 2:

\ [V_ {CE (min)} = V_ {CC} -I_ {C (max.)} R_C \]

\ [V_ {CE (min)} = 12-3 m * 3k = 3V \]

\ [V_ {CE (max)} = V_ {CC} -I_ {C (min)} R_C \]

\ [V_ {CE (max)} = 12-1 m * 3k = 9V \]

I C can vary between 1 mA and 3 mA.

V CE can vary between 3V and 9V.

Example 6: Find R B and R C .

\[V_{CE}=5V\]

\ [I_C = 1 mA \]

\[V_{CC}=15 V\]

\[\beta = 100\]

\ [I_B = 10 \ mu LA \]

Find the possible values ​​of the polarization point for β = 30 - 150.

Evaluate KVL and RB in the input circuit.

\ [- V_ {CC} + I_BR_B + 0V_ {BE} = 0 \]

\ [R_B = \frac {V_ {BB} -V_ {BE}} {I_B} \]

\ [R_B = \frac {15-0.7} {10 \mu} = 1.4M \Omega \text {equation 1} \]

Estimate KVL and RC in the output circuit.

\ [- V_ {CC} + I_CR_C + V_ {CE} = 0 \ text { equation 2} \]

\ [R_C = \frac {V_ {CC} -V_ {CE}} {I_C} \]

\[R_C = \frac {15-5} {1m} = 10k \Omega\]

For β = 30 and β = 150: Calculate the minimum and maximum IC and VCE.

From equation 1:

\ [I_B = \frac {15-0.7} {1.4M} = 10 \mu LA \]

\[I_{C(min)} = \beta_{min}I_B\]

\ [I_ {C (min)} = 30 * 10 \ mu = 0.3 mA \]

\[I_{C(max)} = \beta_{max}I_B\]

\ [I_ {C (max)} = 150 * 10 \ mu = 1.5 mA \]

From equation 2:

\ [V_ {CE (min)} = V_ {CC} -I_ {C (max)} R_C \]

\ [V_ {CE (min)} = 15-1.5 m * 10k = 0V \]

\ [V_ {CE (max)} = V_ {CC} -I_ {C (min)} R_C \]

\ [V_ {CE (max)} = 15-0.3 m * 10k = 12V \]

I C can vary between 0.3 mA and 1.5 mA.

V after Christ

apai    September 23, 2021   

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