I would like to modify a blood pressure pump to light an LED while the blood pressure test is in progress.
The pump has a motor, two smaller solenoids acting as control valves, and a larger solenoid for the dump valve. When the test starts, the BP pump closes the valves and runs the inflation motor until the max pressure is achieved in the cuff. The control valves slowly release pressure as the user's pulse and pressure range are measured. At the end of the test, the dump valve is opened to fully deflate the cuff.
I would like to add an LED that will turn on when the test starts, and I want this LED to stay on until the dump valve is opened. The dump valve closes when it receives 5V from the Atmel micro, and it opens when that voltage is not present. Basically, I want the LED to be on when the 5V to the dump valve is present.
I plan to solder a wire into the dump valve's power line and connect it to the base of a BJT to switch the LED. The LED (C503B-RCN-CW0Z0AA1) will be powered by the 12V line to the BP pump, so I will need a current limiting resistor. I think I'll use a 2N3904 for switching, so I'll also need a base resistor. I've read that the base emitter current should ideally be 1/15 the load current. Is this correct for this application?
LED resistor
Input voltage = 12V
LED Vf = 2.1
2N3904 collector emitter saturation voltage = 0.2V
Target LED current = 20mA (or 0.02A)
(12V – 2.1V) - 0.2V = 9.7V to drop across resistor
9.7V =0 .02A*R
R = 485Ω (closest value 487Ω)
(9.7V^2) / 487Ω = 0.19W power dissipation of LED resistor
Base resistor
Input voltage = 5V
Base collector saturation = 0.65V
Target base emitter current 0.02A(1/15) = 1.3mA (or 0.0013A)
(5V – 0.65V) / 1.3A = 3.26kΩ (closest value 3.24kΩ)
(4.35V^2) / 3,240Ω = 0.006W power dissipation of base resistor
I need this to be reliable, so I hope someone will tell me if I have something wrong. I've attached an image of the schematic as it stands now.
The pump has a motor, two smaller solenoids acting as control valves, and a larger solenoid for the dump valve. When the test starts, the BP pump closes the valves and runs the inflation motor until the max pressure is achieved in the cuff. The control valves slowly release pressure as the user's pulse and pressure range are measured. At the end of the test, the dump valve is opened to fully deflate the cuff.
I would like to add an LED that will turn on when the test starts, and I want this LED to stay on until the dump valve is opened. The dump valve closes when it receives 5V from the Atmel micro, and it opens when that voltage is not present. Basically, I want the LED to be on when the 5V to the dump valve is present.
I plan to solder a wire into the dump valve's power line and connect it to the base of a BJT to switch the LED. The LED (C503B-RCN-CW0Z0AA1) will be powered by the 12V line to the BP pump, so I will need a current limiting resistor. I think I'll use a 2N3904 for switching, so I'll also need a base resistor. I've read that the base emitter current should ideally be 1/15 the load current. Is this correct for this application?
LED resistor
Input voltage = 12V
LED Vf = 2.1
2N3904 collector emitter saturation voltage = 0.2V
Target LED current = 20mA (or 0.02A)
(12V – 2.1V) - 0.2V = 9.7V to drop across resistor
9.7V =0 .02A*R
R = 485Ω (closest value 487Ω)
(9.7V^2) / 487Ω = 0.19W power dissipation of LED resistor
Base resistor
Input voltage = 5V
Base collector saturation = 0.65V
Target base emitter current 0.02A(1/15) = 1.3mA (or 0.0013A)
(5V – 0.65V) / 1.3A = 3.26kΩ (closest value 3.24kΩ)
(4.35V^2) / 3,240Ω = 0.006W power dissipation of base resistor
I need this to be reliable, so I hope someone will tell me if I have something wrong. I've attached an image of the schematic as it stands now.
Comment