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I am attempting to add a signal activity LED to my dummy load box. I have constructed the following circuit and so far it does what it is supposed to (lights an LED with signal applied and does not interfere with overall load impedance). I am now trying to tune the circuit to turn on at appropriate level and not over-current the LED at higher signals. I am using a 3mm red LED with max current of 30mA, which is what I get when the load box sees a 100V peak to peak signal, about the largest signal the load box will see. The LED currently does not turn on until load box sees ~12.6V peak to peak signal, I would like the LED to turn on at ~5V peak to peak. Is there a way to accomplish this while still allowing 100V peak to peak signals without toasting the LED?

This is my current circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

If possible the LED would need to come on at ~5V peak to peak and then increase in light intensity up until 100V peak to peak. My main issue as noted is that I'm already nearing the LEDs maximum current (30mA) at 100V peak to peak and the LED does not turn on until ~12.6V peak to peak.

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  • \$\begingroup\$ Do you want the LED to glow more brightly the higher the input voltage? Or is it acceptable to switch the LED on at ~5V peak to peak, and maintain a fairly constant brightness from there upwards? \$\endgroup\$
    – gbulmer
    Sep 25, 2014 at 14:28
  • \$\begingroup\$ Would like to LED to glow more brightly as input increases. \$\endgroup\$
    – disorder
    Sep 25, 2014 at 14:44
  • \$\begingroup\$ en.wikipedia.org/wiki/Transconductance \$\endgroup\$ Sep 25, 2014 at 14:48
  • \$\begingroup\$ Okay. Please keep adding requirements to the question as they are discovered. It makes it easier for folks to understand the question, and comments may get trimmed off in the future. \$\endgroup\$
    – gbulmer
    Sep 25, 2014 at 14:48
  • \$\begingroup\$ Seems like you need an inverse log amplification biggest increase at beginning, less drastic as input rises, Instead of linear amplification. \$\endgroup\$
    – Passerby
    Sep 25, 2014 at 16:38

1 Answer 1

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You could return the LED to a fixed negative voltage of (say) -1V to control where it turns on. Or (better) put the LED in a feedback loop so that you get mA out for volts in.

The latter will eliminate the LED 'on' threshold entirely- if that's acceptable (it might glow very slightly with zero input).

Or add an additional resistor to get a threshold, as below:-

schematic

simulate this circuit – Schematic created using CircuitLab

The maximum current is set (mostly) by R1 so if the maximum input voltage is 13V, you'd have a maximum current of about 25mA minus about 0.8mA or about 24mA

Threshold current is set (roughly) by R2 so if the LED starts to come on at around 1.5V, and you want it to light up at about 650mV in you could set R2 to about 2.2K

Edit: D2 is just to prevent the op-amp from imposing a high negative voltage across the LED if it's presented with a negative voltage at the input. If R2 is not present, even a tiny voltage (or the offset of the op-amp) would be enough to rail the op-amp which would exceed the reverse voltage rating on the LED.

The op-amp drives a constant current proportional to the input voltage through the parallel combination of D1||D2||R2. I = Vin/R1. For positive input voltage we can ignore D2 since it will be reverse biased.

D1 needs about 1.5V before it starts conduct much current (and thus to light). Without R2, the voltage across D1 will rise to whatever is required to get even a tiny current because of the feedback mechanism. R2 provides an alternative path when the current is low so, say at 0.1mA, the voltage across R2 will be only 220mV, so the LED will not light at all. So a minimum current of Vf(min)/R2 is required to get the LED to begin to light significantly.

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  • \$\begingroup\$ This is a good idea but as noted I am already at the LEDs max current of 30mA when at my largest signal of 100V peak to peak. I can increase opamp gain to get LED to turn on at 5V but at 100V current would be too high then. \$\endgroup\$
    – disorder
    Sep 25, 2014 at 14:58
  • \$\begingroup\$ Sorry, I changed the half-finished answer after dealing with a meeting etc.. Please see the new answer, which is quite different. \$\endgroup\$ Sep 25, 2014 at 16:01
  • \$\begingroup\$ This is interesting. Could you help me understand how R2 and D2 operate? I understand that placing the LED in the feedback loop eliminates turn on threshold, similar to how a precision rectifier (super diode) works. \$\endgroup\$
    – disorder
    Sep 25, 2014 at 17:25
  • \$\begingroup\$ Yes, that's exactly right. Please see edit. \$\endgroup\$ Sep 25, 2014 at 17:55

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