Background Application: 1995 Ford Ranger Manual Control Blend Door Actuator

This 1 page pdf guide shows the block level function of a manual blend door actuator (my component is from a Ford Ranger.)

When the user changes the control potentiometer position the blend door actuator moves a baffle (blend door) to blend air from hot and cold sources in the heating and cooling system.

In my blend door actuator pin 8 ground also connects inside to pin 4. Pin 7 at 12 volts is input to Q1 shown below. Pin 6 is what I marked as Vreg output, however, based on feedback, I think this output is not regulated, but rather, Q1 provides either current limiting or crude thermal shutdown function.

Problem: Q1 NPN Darlington Malfunction and Inability to Understand Design

Reverse engineering effort produces this circuit (note 18 ohm "dummy load" is used to eliminate complexity of the actual op amp servo motor control circuit):


simulate this circuit – Schematic created using CircuitLab

The "Darlington" component Q1 is marked ST Mal KSD1692. I am unable to find an ST part by this indication. The Fairchild KSD1692 datasheet shows the pinout that I am using.

Based on diode drop tests, the Darlington pass transistor Q1, as shown in the circuit, failed inside the blend door actuator. Zener diode D1 is rated 9.1 volts.

Because the pass transistor is bad, and I don't have a replacement blend door yet, I do not know the design value for regulator output Vreg. Does this bias circuit makes sense? If so, then how does the bias circuit work to set the value for output voltage Vreg? Edit: The new question is now how does the Zener and BJT provide an effective or not so effective thermal shutdown function?

PS - The parts in the circuit seem to be 12-14 volt tolerant, so I removed the parts D1, Q1, and Q2 from the malfunctioning blend door and applied 12 volts to the node indicated Vreg. This makes the blend door actuator work as expected in a bench test.

New Material - Edit

The "dummy load" should be 120 ohms for measured motor resistance instead of the 18 ohm load shown above. A stalled motor would give the armature resistance at zero rotations per minute and the current limit, if provided, should protect the motor during stall conditions.

This is a single sided board with components on the face and traces on the bottom. There are 12 resistors on the circuit board all large values 3.4k or more. I don't see a low value current sense resistor that would otherwise connect to the base of Q2.

In the additional diagram below (which relates to a new question) there is a possible sense path indicated at the bottom right. The short trace shown to ground below is actually 2-3 inches of black trace on the face of the board. This trace goes under the pin header so I can't follow it back to the base of Q2. Maybe it provides some sort of control function keeping Q2 off or in the active region until some fault condition occurs. I can't make sense of it yet.

enter image description here

In the new diagram the user manual control, the "Hot/Cold" knob is shown at the top right and the top and bottom wipers, etched on the face of the circuit board in black trace material, are shown at the bottom right. A conductive "comb" device attaches to the final output gear which moves the blend door. The comb fingers make contact with both wipers so the comb voltage picked up on the top wiper transfers to the bottom wiper.

The input voltage Vs = Vreg (above) is the output from the circuit above, but note there must be some feedback path that keeps the base of Q2 either at zero volts or biased in the active region during normal operations otherwise Vs would not be in the expected operational range.

If one disconnects the "Hot/Cold" knob the L2722 control chip forces node A to half the source voltage shown as node A = 0.5Vs. The L2722 forces the motor to move in the direction that moves the comb to the center of the two wipers. So with no sense input from the "Hot/Cold" knob the L2722 design drives the motor until the comb moves to the center and node A settles at 0.5Vs.

If the operational "Hot/Cold" knob is set exactly to the middle position then there is 0.5Vs on both sides of the 10k and 93k resistors and the wipers are in the middle position too.

If the user sets the "Hot/Cold" knob to some other pick-up voltage, but not in the center of the pot, then the L2722 drives the motor, which drives the comb position, until the voltage at node A is restored to 0.5Vs, and then the motor stops. Node A only connects to a high impedance op amp input so current flowing through the 10k resistor should also flow through the 93k resistor.

I think the Top and Bottom wiper can be replaced in SPICE model as follows. The Top wiper would be a voltage controlled voltage source. The Bottom wiper would be a variable resistor. But so far I am stumped by how this servo feedback circuit works.

  • \$\begingroup\$ What you have here looks more like something intended to connect R4 to the 12 V supply only when the 12 V supply is below about 10.4 V -- a low battery, perhaps? It is otherwise activating Q2 as a switch, which pulls the Darlington down and therefore OFF, so far as R4 is concerned. It's certainly not a regulator. \$\endgroup\$
    – jonk
    Aug 21, 2021 at 18:53
  • \$\begingroup\$ If R3 connected to the output (Vreg) rather than 12V you might have a regulator. But stability could be an issue without some compensation. \$\endgroup\$
    – user4574
    Aug 23, 2021 at 18:03
  • \$\begingroup\$ R3 is definitely connected to the automotive battery input side of the darlington. When the alternator is charging the lead acid battery a typical high voltage is 14.4 volts on this input. Nothing in this design requires a local regulated voltage because all components are voltage and power tolerant well beyond 14.4 volts. However if there is a stall torque fault (stuck blend door) then the motor would draw maximum current and melt varnish on the rotor coils due to high temperature. There is third connection to base of Q2, a via with solder joint, but I can't trace it under the input jack. \$\endgroup\$ Aug 23, 2021 at 18:58
  • \$\begingroup\$ What I am trying to say is that a current limit circuit that holds the base of Q2 low or keeps Q2 in the linear region to control current flowing through Q1 makes the most sense. However so far I cannot find a clear control path to the base of Q2 by tracing the circuit board. \$\endgroup\$ Aug 23, 2021 at 19:04

3 Answers 3


I have done my best to trace the circuit layout

The feedback operation would be more obvious if the top of R3 was connected to the output of the circuit, i.e. the emitter of the pass transistor, Q1. Are you sure you have identified the collector and emitter connections of Q1 correctly?

With R3 moved to the output I'd expect the regulated voltage to be about 10.3 V.

  • \$\begingroup\$ The "darlington" component is marked ST Mal KSD1692. I am unable to find an ST part by this indication. However Fairchild KSD1692 datasheet mouser.com/datasheet/2/149/KSD1692-69376.pdf shows the pinout that I am using. The middle pin (collector) connects to 12 volts input and the top of R2 clearly on a trace on the reverse of the board. Then D1 connects to the other side of R2 and cathode of D1 to base of BJT. However I agree there is something strange with this circuit and will keep checking my reverse engineering effort. \$\endgroup\$ Aug 21, 2021 at 19:01
  • \$\begingroup\$ @jonk's suggestion that the circuit only turns on the output when the battery voltage falls below 10.4 V is another valid option. The battery voltage will fall this low when the starter motor is cranking. I'm not familiar with a blend door. Is it something that needs to be activated during starting? As an A/C component it's not obvious that it needs to do anything while the engine is being started. \$\endgroup\$
    – Graham Nye
    Aug 21, 2021 at 19:36

That is a nice capture!
That circuitry works as a current regulator/current limit, but not a voltage regulator.

You must missed the door switch connected to the ... likely on R3 & D1 node. Please check it and tell me I am correct. :-)

When the switch is pressed, Q2 turns off.
R2 supplies current (some amount of) to Q1 base.
Ie = Ib x Beta1 x Beta2.

BTW, how do you flip the direction? You must missed another wiring, too.

  • \$\begingroup\$ I think you may be right this would make sense as a current limit circuit. There is no switch only a master potentiometer attached to the hot-cold control knob which puts a voltage back to the actuator on voltage sense pin 3. This sense input goes to the L2722 op amp. The L2722 provides motor voltage and reversing voltage to drive the servo until the wiper on the slave "pot" matches the voltage at the sense pin and then stops driving the motor. I think a current limit circuit makes sense to protect the motor from a mechanical stall condition on the actuator output shaft. \$\endgroup\$ Aug 21, 2021 at 19:36
  • \$\begingroup\$ @SystemTheory , Thanks for the update. I thought, sorry, you were dealing with your truck door opener. :-) Yes, a voltage regulator will burn stuck failing motor, fuse, or the circuit. \$\endgroup\$
    – jay
    Aug 21, 2021 at 19:45

The circuit you have shown does not regulate anything except shutoff power when it should be on above 10.5V.

The load ought to be low side switched for an NPN darlington.

In order to make it a DC OK current limited power switch, I made some modifications.

enter image description here

I have no idea what you have, but should only disable power for low voltage.

However you know it works with 12V out,so using the Emitter follower , the Darlington gets hot and has no flyback protection shown here to Gnd.

It drops 2.8V from Vbat and the threshold is 10 V.

none if this makes sense for efficiency.

enter image description here


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