The noninverting voltage amplifier is based on SP negative feedback. An example is given in Figure 4.2.1. Note the similarity to the generic SP circuits of Chapter Three. Recalling the basic action of SP negative feedback, we expect a very high Zin, a very low Zout, and a reduction in voltage gain. With Ohm’s law we calculate the resistance value of the ballast resistor: R = V −V LED1 −V LED2 I = 12−2−4.5 0.03 = 183.3Ω R = V − V L E D 1 − V L E D 2 I = 12 − 2 − 4.5 0.03 = 183.3 Ω. The resistor must have a value of at least 183.3 Ω. Note that the voltage drop across the resistor is 5.5 V. EDIT: Since you did measure that the charger bridged the two 3.6 V packs in series, forming a 6S1P configuration during charging, the 9 V and 0.2 A makes perfect sense. It’s actually so low “overvoltage” (9-6*1,45=0.25 V) I’m suspecting the charger is delta V sensing or “smart charger”. Each light takes 4 C-cell batteries in line. That is they are stacked one on top of the other with positive terminal touching negative. I think that means I'd need to add the voltages together meaning 4 x 1.5V = 6V required. I have no idea how much power each light draws (specs don't say anywhere) and the math I did doesn't seem to make sense. VCC < VBAT 17 50 VCC < VUV 17 40 VBAT Termination output voltage V CC = 4.3V to 6.5V, RPROG =10KΩ 4.158 4.2 4.242 V IBAT BAT pin current Current mode RPROG=10KΩ 90 100 110 mA Current mode RPROG=2KΩ 465 500 535 mA Standby mode V BAT=4.2V 0 -2.5 -6 µA Shutdown mode (RPROG not connected), T J=25°C ±1 ±2µA Sleep mode, VCC=0V, TJ=25°C ±1 A 1/4 W resistor will last forever. or the input impedance for the 4-20 mA input is 200 ohms. Almost certainly 250 Ω (not 200) because you would be jumpering in the resistor across the 0 - 5 V input so that 20 mA gives 5 V. R = V I = 5 20m = 250 Ω R = V I = 5 20 m = 250 Ω. The 4 - 20 mA will be read as 1 - 5 V by the analog input. Share A silicon diode has roughly a 0.65V drop. Putting it in series with the 4.95V from USB will get you close to 4.5V: simulate this circuit – Schematic created using CircuitLab. An LDO will also work, and will have better regulation (voltage will vary less with current) and could be closer to 4.5V. To estimate power supply needs, multiply the number of pixels by 20, then divide the result by 1,000 for the “rule of thumb” power supply rating in Amps. Or use 60 (instead of 20) if you want to guarantee an absolute margin of safety for all situations. For example: 60 NeoPixels × 20 mA ÷ 1,000 = 1.2 Amps minimum. With a 5v constant supply the series "ballast" resistor will drop (absorb) 2.0V instead of 1.5v. A 5V input results in a permanent increase in the current through the Leds, in the ratio of a 2 volts drop compared to the original 1.5 volts drop across the ballast resistor giving 1.33 times the battery current.. It works for oil, fuel, water or air pressure and can be used in used in oil tank, gas tank, etc. Features: Brand new pressure transducer. High quality material made. Suitable for oil, fuel, water or air pressure. Specifications: Supply voltage:5 VDC. Output voltage :0.5 ~ 4.5 V. Working pressure :0 ~ 3.5 MPA. Performance Guarantee pressure I6N1cI.