Correcting Voltage Drop by Adding Resistor L Pad

[Idea04]

I have a source of 800 Kilo Ohms and I want to power an LED with that source. Which I would presume the LED would have a resistance of 9.5 to 10 Ohms if I'm correct. When I connect the LED to the source the voltage drops from 12 volts DC to around 1 volts DC. If my thinking is correct the reason this is happening is because of the impedance mismatching in which a high impedance source is driving a low impedance load and the voltage drops to a much lower value.

To correct this I want to put a resistor L pad between the source and load to hopefully minimize the voltage drop. The value of resistor I calculated being would be 800 Kilo Ohms resistor in series with the source and a 10 ohms resistor in parallel with the load.

Is this correct? And if not where am I wrong.

 

[sophiecentaur]

You can't say that a diode has fixed "resistance" because it doesn't follow Ohm's Law. The voltage/current relationship is very non-linear. With an LED, you would normally assume that it has a more or less constant voltage across it and you would pass a current through it which would be defined, usually, by a supply voltage in series with a resistance.

Your "800kΩ" source needs to have an emf (voltage) specified before you can say what current will pass through the diode.
Tell us the specific context of this with a circuit diagram, if possible, and we can probably help you.
As it stands, your question can't be answered except to say that a 12V source with 800kΩ in series will only pass 12/800e3 Amps, or 15microamps. That will not light an LED!

 

[vk6kro]

This is how you normally set up a LED to work from a battery:

[PLAIN]http://dl.dropbox.com/u/4222062/LED%20circuit.PNG

You can see that the value of R is not given, but you can calculate it.

It has 10 volts across it and it has 20 mA flowing in it.
So, its resistance must be 10 volts / 0.020 Amps or 500 ohms.

If you really have to light a LED from a 12 Volt source with 800 K series resistance, you can use a circuit like this:

[PLAIN]http://dl.dropbox.com/u/4222062/LED%20circuit%20with%20transistor.PNG

By adding a high gain transistor and a power source for the LED, you can use the transistor to control the LED.
You might just get enough current to light a high efficiency LED.

There will be about 15 µA of base current in the transistor. If it had a gain of 300 the collector current would be 4.5 mA. This would be visible on some LEDs.

 

[Jiggy-Ninja]

Your "800kΩ" source needs to have an emf (voltage) specified before you can say what current will pass through the diode.
Tell us the specific context of this with a circuit diagram, if possible, and we can probably help you.
As it stands, your question can't be answered except to say that a 12V source with 800kΩ in series will only pass 12/800e3 Amps, or 15microamps. That will not light an LED!

Depending on what he measured it with, his meter might not have enough input resistance to accurately measure a voltage source with 800kΩ of Thevinin resistance.