Inverted LED driver circuit

Question

I am looking for an LED driver circuit that would illuminate when a voltage is at zero, and which would gradually shut off as a voltage rises to 3.3v, with a (somewhat) direct relationship between brightness and voltage level.

I have already designed this circuit using an inverting op amp to invert and offset the control voltage signal with a negative reference, and then a second op amp to drive the LED using the feedback path.

I am hoping that someone might know of a simpler design, that would be lower cost and fewer parts. As it stands, this will cost me about $0.30 and uses 5 parts (not counting the LED itself). I need 16 of these in a single circuit, and I was hoping to be able to reduce the cost to below $2 total, ideally below $1

And here is the signal analysis, where the Green line is the 0-3v3 control signal, and the Blue line is the current used by the LED itself.

UPDATE 1:

One additional constraint I forgot to mention is the control input must be high impedance, as that signal will be needed for other purposes as well

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

The 3.3V comes from a regulator for the LEDs (one required). If you reduce the positive rail on the op-amps to +8V or so you could use quad LM324 op-amps and avoid the diodes, so total 8 parts (4 LM324 + 4 resistor networks) + 2 regulators if you use x4 resistor networks. Plus the LEDs.

That's about $0.03 USD per driver and 0.5 parts per driver (not counting regulators or LEDs) in 250 unit quantity- Digikey list price.

Note that inputs are high impedance and can withstand input voltage from a bit below the negative rail up to 32V.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Voltage controlled LED current.

How it works:

• R3 and R4 with OA2 provide a common 3.3 V reference for each channel.
• With Control at 3.3 V the op-amp will be stabilise with the inverting input at 3.3 V, no current through R1 and, therefore, zero current through the LED.
• With Control at 0 V the op-amp will again stabilise with the inverting input at 3.3 V. This time there will be 4.8 mA through R1 and, therefore, through D1.

I have not considered what will happen at power-up and whether any protection is required to prevent high-current spikes or through the LEDs. Reverse voltage on the LEDs is limited by grounding the op-amp negative supply rather than using -12 V as in your design.

The minimum output voltage required of the op-amps is 3.3 V so rail to rail op-amps are not required.

Answer 3

A very crude circuit would be a PNP darlington with collector to ground, Vc to the base, and the LED and it's resistor between the emitter and a reference voltage. This would be very nonlinear at maximum Vc but this could be improved with a bias resistor from the emitter to V+. Two or three cheap parts.

Answer 4

This works for Red Yellow LEDs that are typically Vf=2.0 to 2.1V with a very dim or off threshold at 1.7V such that 3.3+1.7=5.0V

It doesn't get much simpler than this.

Rev A