Voltage Regulator Circuit Calculations

In summary, you would need to measure the current draw of the motors when supplied with 6 volts to find out how much power they will require.
  • #1
Alexander1
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Hey Guys, so I'm just in need of some help in relation to finding the output current after the voltage has been dropped to 6V. The circuit has an input voltage of 24V and I've attached a photo of my circuit diagram. Hopefully I haven't made any errors in that, but I've tested the circuit and it does work effectively. The purpose of the circuit is to be able to flick between motors and also change their direction of rotation. The reason I need to know the output current is so I can find the power that each motor will require but I'm just a bit confused. Thanks.
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  • #2
You would have to measure the currents.

Unless the motor comes with information about how much current it draws when supplied with 6 volts and with different loads, the only way you can find out is to measure it.

You could put an ammeter in series with each motor, but this might cause the current to be reduced due to the resistance of the meter.

Or, you could put the meter directly after the 24 volt supply and measure the current there.

You could measure without the motors to work out how much is being used by the resistors and the regulator. Then subtract this from the figure you get with the motors running.
 
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  • #3
Thanks very much. The motors are micro metal gear motors so I'll have a look and see if they have that information.
 
  • #4
Alexander1 said:
Hey Guys, so I'm just in need of some help in relation to finding the output current after the voltage has been dropped to 6V.

As VK6KRO has said, just measure it

Be aware that the LM317 like most other linear regulators have inbuilt current limiting --- check the specifics in the datasheet
it's somewhere around 1.5A
 
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  • #5
How much current is drawn depends on motor characteristics as VK6KRO and davenn have said, but also upon how much mechanical load is present at the motor shaft. The motors will have a full load current rating, and (possibly) a locked rotor rating.

Keep in mind that when flipping the direction switch the motor will pull LRA (Locked Rotor Amps) as it spins down before reversing itself. This can be quite a bit more than full load current, and may be enough to put the LM317 into overload limiting. If operating anywhere near the 1.5 amp rating bolt a fairly hefty heat sink to the regulator, or chances are good that it'll fry.
 
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1. How do I calculate the voltage output of a regulator circuit?

The voltage output of a regulator circuit can be calculated by using the formula Vout = Vin - (Iload * R), where Vin is the input voltage, Iload is the load current, and R is the regulator resistance. Make sure to use the correct units for each variable.

2. What is the purpose of a voltage regulator circuit?

A voltage regulator circuit is used to maintain a constant voltage output regardless of changes in the input voltage or load current. This is important for protecting electronic components from damage and ensuring proper functioning of the circuit.

3. How do I choose the appropriate regulator for my circuit?

The appropriate regulator for your circuit will depend on the input voltage, desired output voltage, and load current. You can use a datasheet for the regulator to find its specifications and choose one that meets your requirements.

4. Can I use multiple regulators in a circuit?

Yes, you can use multiple regulators in a circuit. This may be necessary if you have different voltage requirements for different components. However, be careful to properly calculate the voltage outputs and ensure that the regulators can handle the total load current.

5. How do I account for voltage drops in my regulator circuit calculations?

To account for voltage drops in your regulator circuit calculations, you can use the formula Vout = Vin - (Iload * R) - Vdrop, where Vdrop is the voltage drop across any components in the circuit, such as diodes or transistors. Make sure to include all relevant voltage drops in your calculations for an accurate result.

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