Thanks MikeML, dl324 and ak.

But you forgot that the when I increased Rb to 200Ω, I simultaneously added 2V to its supply, so that I(D3) is approximately the same as I(D2), so the LED brightness would be the same.

Thanks but what I meant was if in the circuit shown in #36 you increased Rb to 400 Ohms, I(D3) would be dimmer, but the proportion at which the current increases as the temperature rises would be slightly less than if Rb was still 200 Ohms?

Regards,
Robert

 

If you have the components in the following schematic, you can do some experiments with operating LEDs at constant and variable current (variable voltage with fixed resistor that we've been discussing) to solidify your understanding:

If you operate the circuit at a few volts (around 3V) to 12V, you'll see that the brightness of LED1 doesn't change noticeably, but LED2 will. This is the difference between operating at constant current (LED1) and with a current limiting resistor (LED2) with a varying voltage. Unless the supply voltage is varying over a wide range, a current source (current sink in this case) isn't necessary. In fact, few LED indicator circuits will use a current source because they'll likely be operating from a well regulated power supply. If the supply voltage drops too low for the current sink to maintain the desired current, LED1 will dim.

The resistor values are not very critical. You want R2 small enough so the current in LED2 isn't affected much by the base current for Q1. R1 sets the current in LED1; 100 ohms gives about 10mA.

As an additional experiment, you can place another LED or 2 in series with LED1 and observe that the brightness stays the same; assuming the supply provides enough voltage. In the circuit we've been discussing, replacing the LED with a different one or increasing/decreasing temperature would change the current in the LED; possibly enough to cause a noticeable change in brightness (luminous intensity). With a current sink, changes due to voltage and/or temperature would be reduced.

As a further experiment, you can measure the voltage on the collector of the transistor and observe that it will change as needed to maintain a constant current in the LEDs. This is how constant current sources/sinks work.

Even within the same "lot", LEDs will exhibit a range of forward voltages. However, luminous intensity is specified at a specific current (usually 20mA). A current source makes those small, or large, variations in forward voltage a don't care. With a current limiting resistor, series LEDs could operate at different voltages resulting in different brightnesses; though probably not noticeable to the human eye.

 

Hello All,

If I had a simple circuit with a 12V battery and an LED that drops 2V and needs 20mA, therefore I'd need a resistor to drop 10V and set the current at 20mA, so I'd do R = 10/0.020 = 500 Ohms.

1. Is it possible to 'use up' the voltage before it reaches the LED. For example if you placed the resistor before the LED, and wanted it to drop the whole 12V, then if I used a 600 Ohm resistor, would it drop 12V and then leave nothing for the LED?

2. If I wanted to 40mA to flow through the LED, then I could use R = 10/0.040 = 250 Ohm resistor. Looking at the voltage-current curve for an LED, to increase the current through it I'd also have to increase the voltage (albeit slightly), but I'd still only have 2V after the resistor. How can the LED have a current of 40mA, when it only has 2V available after the resistor, it goes against the voltage-current curve?

If anyone could help clear things up a bit it would be appreciated.

Thanks.

The LED will drop about 2 V across it at just about any current. It will be a bit less at 1 mA and a bit more at 40 mA. The rest of the 12 V is dropped across the resistor and that determines the current in the circuit.
Play with it and see what happens.

 

The LED will drop about 2 V across it at just about any current. It will be a bit less at 1 mA and a bit more at 40 mA. The rest of the 12 V is dropped across the resistor and that determines the current in the circuit.
Play with it and see what happens.

The OP hasn't posted to this thread in almost 4 months. Either he got all the information he wanted or decided to move on. Either way, this thread should be allowed to go dormant.

 

The OP hasn't posted to this thread in almost 4 months. Either he got all the information he wanted or decided to move on. Either way, this thread should be allowed to go dormant.

That makes sense unless you consider that others may come along with an interest in the same subject. I kind of assumed the posts were retained for history since we encourage others to search for similar posts.

 

That makes sense unless you consider that others may come along with an interest in the same subject. I kind of assumed the posts were retained for history since we encourage others to search for similar posts.

If you're correcting or adding useful information I could see it. But, if you're just adding another option; as I said, the OP has vacated the thread.

 

If you're correcting or adding useful information I could see it. But, if you're just adding another option; as I said, the OP has vacated the thread.

For future reference should I consider the OP to be the only reader and future readers don't matter?

 

For future reference should I consider the OP to be the only reader and future readers don't matter?

That isn't what I said. If you're correcting something posted or adding something of value, feel free to post.

If you're just stating the obvious, it's better to let the thread die if it's older than a month or so. I participated in several of the threads you bumped today and was compelled to read them to see if the OP had some additional questions.

 

That isn't what I said. If you're correcting something posted or adding something of value, feel free to post.

If you're just stating the obvious, it's better to let the thread die if it's older than a month or so. I participated in several of the threads you bumped today and was compelled to read them to see if the OP had some additional questions.

Okay, thanks for the wisdom.