Solenoid valve control circuit

Thread Starter


Joined Jun 13, 2020
Good morning, I have found this circuit in a medical ventilator service manual. BL1 is the control signal from the microprocessor.
I did not understand the purpose of the zener diode in the circuit and why two MOSFET are used.
Thank you.


Joined Sep 17, 2013
The zener protects the Q10 gate from excessive voltage.
Surely the solenoid valve pin 2 should connect to the Q10/Q11 drains, not to the micro control signal /BL1 ?


Joined Sep 17, 2013
What is the source of the /BL1 voltage? If the solenoid pin 2 is connected directly to an MCU output pin then the MCU is likely to be destroyed by the kickback spike when the solenoid turns off, as there is no protection circuitry evident.
An MCU can usually source or sink only two or three tens of mA. How much current does the solenoid valve draw?
In the circuit shown, the only purpose served by Q10 and Q11 seems to be to charge and discharge C1. Is that correct?


Joined Jun 4, 2014
Surely the solenoid valve pin 2 should connect to the Q10/Q11 drains,
I think it should connect to the 0V of the 24V supply, presumably this is the earth connection shown on the diagram.
Then if the uC supplies a square wave, the mosfets with C1 and D1 can boost the solenoid voltage.


Joined Aug 1, 2013
Yes-really, and yes-kinda.

When the coil is off, C1 charges up through D1 and Q11. When /BL1 goes low, Q11 is turned off through 10 K resistor R12, while Q10 is turned on more slowly through 100K resistor R10. This prevents cross-conduction. Q10 turns on and connects the cathode of C1 to the 24 V rail. This charge-pumps V1, the solenoid voltage, up to over 40 V (guess), and C1 starts discharging through the solenoid coil.. When the charge in C1 runs out, the coil holding current is maintained through D1 by the 24 V rail.

The coil current return path is through the TI driver. When it turns off, its internal suppression components kick in to absorb / reroute the solenoid inductive kick energy. That's why there is no explicit suppression diode in the circuit schematic. Functionally, your D6 is inside the driver.

This is different from the normal diode across the load to its source, as in a ULN2003-4-x. With suppression across the driver rather than across the load, there is no direct connection necessary between the driver chip and the load's voltage source. In the TI part, the output driver transistor is used to shunt some of the inductor energy to GND. Cute. This doesn't happen until the voltage across the output rises to over 50 V, which lets the solenoid turn off (release) more quickly.

Someone had their thinking cap on.

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