[Circuits] Designing a simple voltage-divider

In summary, a voltage divider can be designed to provide a voltage of 5V (±10%) across a variable load with a current demand of 0 to 5mA using a circuit with R1 = 2R2 and a maximum load current of 5mA corresponding to a minimum voltage of 4.5V. The circuit should also have a ground, 15V, and 5V connection.
  • #1
ainster31
158
1

Homework Statement



Design a voltage divider to provide a voltage of about 5V (±10%) across the variable load. The load-current demand varies in the range of 0 to 5 mA and the available dc-supply voltage is 15V.

Homework Equations





The Attempt at a Solution



Here is the circuit given (I've added ground, 15V, and 5V):

h_1390930293_9756129_f0b4069220.png


First, I assumed that there was no load. This gave me the following relationship: ##R_1=2R_2## but I'm not sure how to proceed.
 
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  • #2
The maximal current for the load will correspond to the lowest acceptable voltage (can you calculate that?). This allows to determine R1 and R2 together with the equation you found.
 
  • #3
mfb said:
The maximal current for the load will correspond to the lowest acceptable voltage (can you calculate that?). This allows to determine R1 and R2 together with the equation you found.

The lowest acceptable voltage would be 4.5V, right?

How do you know that maximum current would correspond to lowest voltage? Does that mean lowest current would correspond to highest voltage? Should I have used 5.5V instead of 5V in my first calculation then?
 
  • #4
For no load set Vout at 5.5V, Now with 5mA load set Vout at 4.5V.
 
  • #5
Jony130 said:
For no load set Vout at 5.5V, Now with 5mA load set Vout at 4.5V.
That is even better.

ainster31 said:
The lowest acceptable voltage would be 4.5V, right?
Right.

How do you know that maximum current would correspond to lowest voltage? Does that mean lowest current would correspond to highest voltage? Should I have used 5.5V instead of 5V in my first calculation then?
Smaller load resistances lower the voltage and increase the current, so the smallest current corresponds to the largest voltage.
 

FAQ: [Circuits] Designing a simple voltage-divider

1. What is a voltage-divider?

A voltage-divider is a simple circuit that is used to divide a voltage into smaller parts. It consists of two resistors in series, with the output voltage taken from the connection between the two resistors.

2. How does a voltage-divider work?

A voltage-divider works by using the concept of voltage division, where the total voltage across a series circuit is distributed among the different components based on their resistance values. In a voltage-divider, the output voltage is proportional to the ratio of the two resistors used.

3. What are the applications of a voltage-divider?

A voltage-divider is commonly used in electronic circuits to provide a specific voltage level for different components, such as sensors, transistors, and microcontrollers. It is also used in power supplies to regulate the output voltage and in audio circuits to adjust the volume level.

4. How do I design a simple voltage-divider?

To design a voltage-divider, you will need to know the desired output voltage and the input voltage. Using Ohm's law, you can calculate the resistance values for the two resistors needed to achieve the desired output voltage. The formula for a voltage-divider is Vout = Vin * (R2 / (R1 + R2)).

5. What are some important considerations when designing a voltage-divider?

When designing a voltage-divider, it is important to consider the power rating of the resistors used, as well as their tolerance and temperature coefficients. It is also essential to make sure the output voltage does not exceed the maximum rating of any components connected to the circuit. Additionally, the values of the resistors should be chosen to minimize the effect of any input or output impedance on the circuit's performance.

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