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Can a transistor array supply current to a single LED using multiple outputs pins? In the following 5 volt circuit, an LED is connected to two outputs of a M54561P current supply transistor array. The configuration is intended to render two different brightness levels from the LED (1 triggered, or 1 & 2 triggered). Will this circuit function, or will current flow back into the array (through the output clamping diodes maybe) if only input 1 is triggered?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Looks like it will work. \$\endgroup\$ – mkeith Apr 12 '16 at 6:01
  • \$\begingroup\$ By the way, thank you for including a schematic with reference designators and part numbers. \$\endgroup\$ – mkeith Apr 12 '16 at 6:02
  • \$\begingroup\$ I suggest using different resistor values. Power dissipation will be different on each resistor, but then you have way more brightness levels. \$\endgroup\$ – Wesley Lee Apr 12 '16 at 6:04
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Yes. You have a proper understanding of how it will work. When both outputs are enabled, the two 500 ohm resistors will be in parallel, working like a single 250 ohm resistor, and the current will increase appropriately.

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  • \$\begingroup\$ No, the U-I curve is not linear, so the current is not proportional to the voltage. \$\endgroup\$ – Simon Richter Apr 12 '16 at 6:07
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    \$\begingroup\$ @SimonRichter: This is nitpicking. Given the example in your own answer the voltage over any resistor will only change by 1.5% to 3% depending whether you half or double the nominal current of 20 mA. So the current is in a close linear relationship to the effective conductance of the used resistors. \$\endgroup\$ – Ariser Apr 12 '16 at 10:57
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May I suggest the following?

schematic

simulate this circuit – Schematic created using CircuitLab

Then instead of 2 levels: 250R and 500R

You get 3: ~260R, 430R, 670R

Of course you'd have to adjust the resistors for the respective power ratings of the components and the brightnesses you want but.. This way you get a bit more versatility for "free".

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    \$\begingroup\$ Although my example was valid, I agree that your design describes a more effective system. \$\endgroup\$ – Hoytman Apr 12 '16 at 16:12
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You'd essentially only create a switch between two different resistor values.

If full brightness is 20mA at 2V, this means 500 Ohms is correct for a supply voltage of 12V, and R1 drops 10V. If you'd switch R1 and R2 in parallel, you'd end up with 250 Ohms, which is way too small, and the magic smoke will escape.

If you increase the resistor values, e.g to 1kOhm, full brightness would work, but with only one of the outputs active, the LED would be rather dim, because the U-I curve is not exactly linear.

You can use two different resistor values to try to reach a sensible setting, but that doesn't address manufacture tolerances, and values that work for one LED may not even lead to discernible differences in brightness for another.

Ideally, you'd try to pulse the LED to dim it. If the control is set by a microcontroller, it may have PWM logic that can generate pulses on its own already.

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  • \$\begingroup\$ You are right, U-I curve is far from being linear. But this does not mean, you cannot control the output power approximately linear. In fact, if you reduce the current through the LED by 50%, output radiation will drop by nearly 50%, deviations depending on droop and TDP of the die. \$\endgroup\$ – Ariser Apr 12 '16 at 8:18
  • \$\begingroup\$ Right, so the resistor values need to be chosen so that one of them can be used for 50% current, and both together for 100%. \$\endgroup\$ – Simon Richter Apr 12 '16 at 10:35
  • \$\begingroup\$ This will be achieved with a close approximation with 1 kOhms for both. \$\endgroup\$ – Ariser Apr 12 '16 at 10:58
  • \$\begingroup\$ I edited the question to focus on a 5 volt system \$\endgroup\$ – Hoytman Apr 12 '16 at 17:13
  • \$\begingroup\$ In a 5V system, you need smaller resistors. \$\endgroup\$ – Simon Richter Apr 12 '16 at 19:04

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