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Basic Electrical Concepts

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bluffmaster

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Hi,

I am trying to learn basic electronic concepts. Voltage divider produces the output voltage which is lower than the input voltage. If am making a circuit where I use 9V battery but the load uses only 5 volts, whats the way to reduce voltage. Voltage divider or Voltage Regulator like LM78M05? Which one should be the way to go and why?

Also...
If I am trying to light a led. I am using 9V battery. I would use respective resistor to limit the current. But I would also need to reduce the voltage using voltage divider or voltage regulator, right?

Resistor limits the current. Whereas current divider reduces the current. Which one to use in what case?

Thanks!
 
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MikeMl

Well-Known Member
Most Helpful Member
Hi,

I am trying to learn basic electronic concepts. Voltage divider produces the output voltage which is lower than the input voltage. If am making a circuit where I use 9V battery but the load uses only 5 volts, whats the way to reduce voltage. Voltage divider or Voltage Regulator like LM78M05? Which one should be the way to go and why?
If you need to draw significant current at the reduced voltage, then an active-circuit voltage regulator is the way to go because the reduced voltage is held constant even with a varying load.

If very little current is required, then a voltage divider might be ok. The voltage divider has an effective output resistance determined from the two resistors which make it up. If you connect a high current load, the additional voltage drop across the effective output resistance will give you a much lower voltage than you expected. If the load varies, the output voltage will vary, too.

Also...
If I am trying to light a led. I am using 9V battery. I would use respective resistor to limit the current. But I would also need to reduce the voltage using voltage divider or voltage regulator, right?
No. just use a simple series circuit and make the current limiting resistor bigger. Say you are starting from 9V, and you want to supply 10mA to a red LED. Look up the spec sheet for the LED which says the forward voltage drop across the LED at 10mA is 1.65V. If you put a resistor between the 9V source and the LED, there will be 9-1.65 = 7.35V across the resistor. Since you want 10mA through it, by Ohms law, R= E/I = 7.35/0.01 = 735Ω. In a series circuit, the current through the resistor is the same as the current through the LED.

Resistor limits the current. Whereas current divider reduces the current. Which one to use in what case?!
By transposing Ohms law, we see that I = E/R, which says the larger R is, the smaller the current, for a fixed V. So to reduce the current, we just make R bigger. Say that in the example above we want to reduce the current from 10mA to 5mA. Simply doubling the value of R to 1470Ω would do that.

Resistors in parallel divide current. But in the case of your LED, why bypass current to ground which is wasted? Just make the single resistor bigger.
 
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