Accurate Power Control of the MAX3740A Laser Driver
The best circuit for a particular application will depend on the designer's requirements. Things that need to be considered include the photodiode current to achieve the desired laser power level, adjustment range, and the required resolution. With the information in this article the designer will be able to determine the best option for a particular application.
ChallengesThe MAX3740A data sheet specifies the use of a resistor between the reference pin (REF) and the power monitor photodiode (MD) to set the photodiode current. The power control loop drives the laser diode to an intensity that yields this photodiode current, and therefore the average operating power level. The problem is the nominal voltage at MD is 1.6V and the nominal voltage at REF is 1.8V. This only gives 0.2V across the resistor to set the photodiode current. Digital potentiometers like those in the DS1859 can have minimum resistances up to 1K. This would give a maximum current of only 200µA. You can increase this current level with a fixed resistor but you can't change the limited adjustment range, nonlinearity and poor resolution at high currents.
Another challenge with using the DS1859 is the wide variation of voltages at MD and REF on the MAX3740A. The delta between these two points is very stable at 0.2V but the common mode can change up and down by ±0.5V. This large variation of MD relative to ground means the current defined by a resistor between this node and ground will have the same large variation.
Solution #1In this solution the DS1859 variable resistor and the 806Ω fixed resistor are placed between REF and MD, please refer to Figure 1. The fixed resistor caries 0.2V/806 = 248µA. The variable resistor caries 0.2V/50K (DS1859 max value) = 4µA to 0.2V/1K (DS1859 min value) = 200µA. The sum of the currents though the fixed and variable resistors determine total current through the photodiode. This technique gives a nonlinear response with good resolution at low currents. It is not affected by voltage variations at MD.
Figure 1. Solution #1 schematic.
Figure 2. Solution #1 simulation results (X axis DS1859 resistance in Ω).
Figure 3. Solution #1 summary.
Solution #2In this solution the DS1859 variable resistor is put between MD and ground, and a 243Ω fixed resistor is put between REF and MD, please refer to Figure 4. The current in the fixed resistor is 0.2V/243 = 823µA. The current in the variable resistor is 1.6V (MD nominal)/50K (DS1859 max) = 32µA to 1.6V (MD nominal)/1K (DS1859 min) = 1600µA. The total current through the photodiode is the current through the fixed resistor minus the current through the variable resistor. This technique gives good resolution at high currents. The current values are strongly dependent on voltage variations at MD. Note that the photodiode current can go negative for low values of resistance. Figure 5 shows the photodiode variation with MD and REF varied through a ±5V range.
Figure 4. Solution #2 schematic.
Figure 5. Solution #2 simulation results with MD and REF varied ±0.5V (X axis DS1859 resistance in Ω).
Figure 6. Solution #2 summary.
Solution #3In this solution a small form factor (SC70 package) op amp, the MAX4245, is added to the circuit between REF and MD, please refer to Figure 7. Utilizing the same power supply as the DS1859 and MAX3740A, the op amp generates a voltage, VO, proportional to the value of MD, REF-MD, and the DS1859 resistance value. This generates a current through R2 proportional to the voltage difference between VO and MD. The effects of the voltage change at the MD pin cancel out so the current through R2 is only dependent on REF – MD, a stable 0.2V, and the resistance of the DS1859. The current through the photodiode is equal to the current thru R1 (803µA) + the current thru R2. The photodiode current is a linear function of potentiometer value. This circuit can work with any value potentiometer and provide current over any range. Its only limitations are the current drive capability of the MAX4245 op amp.
Voltage at the op amp output (VO) is derived and calculated as follows. Make sure the voltage at VO does not exceed the op amps maximum output swing.
VO = REF × - (DS1859/R3) + MD × (1 + DS1859/R3)The current through R2 is derived and calculated as follows:
VO = -REF × DS1859/R3 + MD + MD × DS1859/R3
VO = MD + (MD - REF) × DS1859/R3
VO = MD - 0.2V × DS1859/R3
VO = 1.0V (min) - 0.2V × 1K (DS1859 min)/10K = 0.98V
VO = 1.0V (min) - 0.2V × 50K (DS1859 max)/10K = 0V
VO = 2.0V (max) - 0.2V × 1K (DS1859 min)/10K = 1.98V
VO = 2.0V (max) - 0.2V × 50K (DS1859 max)/10K = 1.0V
I(R2) = (VO - MD)/R2
I(R2) = VO/R2 - MD/R2
I(R2) = (MD - 0.2V × DS1859/R3)/R2 - MD/R2
I(R2) = MD/R2 - 0.2V × DS1859/R2 × R3 - MD/R2
I(R2) = -0.2V × DS1859/R2 × R3
I(R2) = -DS1859/62,000,000
I(R2) = -1K(DS1859 min)/62,000,000 = -16µA
I(R2) = -50K(DS1859 max)/62,000,000 = -806µA
Figure 7. Solution #3 schematic.
Figure 8. Solution #3 simulation results (X axis DS1859 resistance in Ω).
Figure 9. Solution #3 summary.