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Old 30th April 2009, 12:00 AM   #1
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Default Driving UV LEDs with a constant current source

Okay, not *exactly* an audio question, but for those circuit enthusiasts out there, I figure this would be a good question.

We are trying to drive ultraviolet LEDs with a voltage-to-current converter (see schematic). We want a constant current of about 15mA through the LED. The transconductance gain (current out for a given voltage in) of the attached circuit schematic is set by adjusting the gain of the current sense amp (the current sense amp has an external pot for gain setting that I didn't draw in). The input voltage is from a voltage reference. The output current is the current measured through the LED using a current meter in series with the LED.

This circuit IS functional. When we install one LED, we adjust the current sense amplifier gain so that 15mA flows through the LED. After making this adjustment, when we try another UV LED in the same circuit, the output current is different. We have measured currents from as low as 12mA to as high as 18mA without making any changes or adjustments to the circuit.

Why is this? I know that LEDs are nonlinear resistive elements, but if we are using voltage-to-current converter, shouldn't the same current flow through any LED that we try (assuming we have enough voltage headroom, which we do)?
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Old 30th April 2009, 12:29 AM   #2
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That's because you don't have a V-I converter there.

You need the diode in the FB loop. Swap your resistor and diode.

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Old 30th April 2009, 12:35 AM   #3
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Is the CMRR of the sense amp high enough? If the voltage across the sense resistor is very small, then the common mode error might be significant.

Why put things in this order anyway? Wouldn't it be better to have the LED connected to the supply voltage, then the MOSFET, then the sense resistor referenced to ground?
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Old 30th April 2009, 12:42 AM   #4
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Quote:
Why put things in this order anyway? Wouldn't it be better to have the LED connected to the supply voltage, then the MOSFET, then the sense resistor referenced to ground?
--> Between the cathode of the LED and ground, there is actually an N-FET switch that I didn't draw in the schematic (to keep the schematic simple). This N-FET switch is very low resistance (about 20 milliohms, according to the datasheet), but it is low-side referenced, meaning the LED itself can't be ground referenced because this Nfet switch is "in the way".

(I know, it sounds like a complicated way of doing things, but it was the circuit we were given, and we have to work with it at the moment)
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Old 30th April 2009, 12:54 AM   #5
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One word of caution on UV sources -- even leds -- don't fry your eyeballs.

Quite frankly, I use mercury vapor lamps for my radiators, and they'd give you a suntan in milliseconds ....
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Old 30th April 2009, 01:01 AM   #6
Mr Evil is offline Mr Evil  United Kingdom
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Quote:
Originally posted by rtarbell


--> Between the cathode of the LED and ground, there is actually an N-FET switch that I didn't draw in the schematic (to keep the schematic simple). This N-FET switch is very low resistance (about 20 milliohms, according to the datasheet), but it is low-side referenced, meaning the LED itself can't be ground referenced because this Nfet switch is "in the way".

(I know, it sounds like a complicated way of doing things, but it was the circuit we were given, and we have to work with it at the moment)
Are you not allowed to rearrange anything? You could at least do what Geek said and swap the resistor and diode, then the common-mode voltage will be reduced and be independent of the number of diodes. If you are feeling clever, you could even choose a FET for the switch with an on-resistance a bit higher and use it as a sense resistor instead of an actual resistor.
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Old 30th April 2009, 01:18 AM   #7
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Are you not allowed to rearrange anything?
Correct, because we have to demo the circuit next week. For future versions, we do intend on changing this circuit (this circuit has some obvious flaws), but we can't change anything now due to time constraints.

The V-to-I gain of this circuit is given by this equation:

G = finite gain of current sense amp (adjustable by external gain pot, not shown)

I out = (Vin/G) / Rsense
OR
Iout = Vin / (G * Rsense) [simplifying above equation]

The current sense amp measures the voltage across the sense resistor to get the current through the sense resistor, which is the same current through the LED. In an ideal circuit, this current is independent of the load, but in practice, our output current IS showing some dependency on the load.

At this point, we can not swap the diode and the sense resistor because the PC board for this round has already been made.
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Old 30th April 2009, 03:05 AM   #8
Mr Evil is offline Mr Evil  United Kingdom
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Well, check to see if it is common-mode error. If it is, then you can either change your current sense amp to one with better CMRR, or use a larger sense resistor.
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