Voltage is definitely more than the motor likes because it says on the body that it's a 13.2v motor whereas this motor is tied directly to system rails. If on AC, I can see 18+ volts. Running a few large diodes in series can help drop the voltage down a bit to the motor and make for a longer lasting motor.baddboybill said:Have you checked voltage to deck? Maybe your getting to much
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Dave,
I would be happy to take a look at it for you, but there are the following issues:
- I am in the USA so shipping back/forth from UK would cost a few bucks & shipping will add some time
- As I would be testing it on the bench, I'm not sure it would e possible for me to determine the actual motor speed
- Consequently, the scope of what I can do would be to confirm that all the components are within spec and confirm that theres nothing amiss in the circuitry.
However...
Before we go drastic here. May I suggest that you take a step back, maybe don't think about this for a few days or a week before taking another fresh look? I do that a lot when I need a break and next time, I see from a different perspective. I know that it's frustrating but there really isn't all that many components here and whatever is the problem probably is easily within your abilities to correct. For example, it looks like replacing the pot from 3k to 50k was a setback since the symptoms got worse. But..... But, you missed the silver lining. Whereas before, nothing you did made any appreciable difference in the motor speed, this swap here DID make a difference. What does this tell you? See if you look at the schematic, you'll see that the power circuit runs from (+) to motor, and then to ICp3 which must find ground through ICp2. Although Pitch(-) is also connected at motor(-), it is merely a resistor network (8-12k) and not connected directly to ground. IF there somehow was a short to ground through that network, then the motor would find an alternate ground and bypass the IC thereby rendering it non-effective in throttling the motor speed. However, you said that you've already tested the network and there is no continuity to ground. Additionally, you've tested the static resistance of motor (-) to ground and found 396 ohms. Were it shorted, then it could explain why the motor would run fast with no ability to vary the speed. However with 396 ohms and because the mere addition of 47 ohms to the circuit (when you replaced inductor with resistor) rendered the motor virtually unusable, it seems like the static unpowered resistance from ICp3 to ICp2 does change when powered up and therefore, probably working.
So, I think we can surmise that by changing the feedback to the IC, we were able to illicit a variation in the motor speed, even if in the opposite direction. Also, the more I look at the two circuits (yours & the sample circuit), the more I see that they are basically the same. What ?! you say.
Look at it this way. In the sample diagram, the control is basically pin 4 through a voltage divider network, set by the voltage between rail (+) and the voltage after the motor (with associated voltage drop). Varying the potentiometer wiper position from one end to the other varies the signal to pin 4 somewhere between those 2 voltages.
Now, let's move on to YOUR circuit, based on Eric's drawing. We see the exact same thing except that instead of the end of the internal pot going directly to pin 3, as in the sample circuit, it first goes to (+) pitch, through a resistor network (8k-12k variable), and then to (-) pitch, which, yep, goes to pin 3 of IC. So basically, it's the exact same thing except that it diverts a portion of the resistance network externally.
Somehow, the values in this complex resistance voltage divider network (ending at pin 4 of the control IC) is off, and therefore the reason why you are losing the ability to control the speed. In fact, the front mounted pitch pot is a critical part of the network and you would lose the ability to control the speed if it was disconnected. In the sample circuit, if the wiper is set at the midpoint of a 50k pot, then you should expect about 25k from either voltage sources to pin 4. A different pot can be used since again, it is merely a voltage divider and it is the "balance" between the resistance from both ends of the wiper from the voltage sources that is important except that varying the pot value could alter the current to pin 4 and it's unclear if pin 4 is monitoring voltage or current.
I recommend at this point for you to re-check and verify the resistances, they probably need to be spot on with little deviation. Now that I understand the purposes of those resistors as voltage dividers, then it's clear that the values are critical and even small deviations will have an effect.
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Now, not to complicate things, if this motor was in my hands right now, what I would do is the following:
- Disconnect the wires from pitch (-) and pitch (+)
- Tack a ~10k resistor across pitch(-) and pitch(+)
- Check/record voltage from ground to pin3
- Check/record voltage from ground to the union of the 3-resistors
- Check voltage at pin 4, and adjust wiper to see if it's possible to vary the voltage from a balanced range between those 2 voltages.
If the voltage readings at pin 4 consistently stays at or near one end of those 2 ranges, then that's probably why you aren't getting any adjustment out of your pot. Also, the voltage range between those 2 readings is probably very low so minute changes makes a difference. Lastly, I'm not sure if it's possible for you to take readings while running the motor since we need "live" readings (it doesn't have to be on deck for this test). If accessibility to the board is limited when assembled for testing, you can tack some pigtails that extend out for the testing.
Eric, can you comment on what I wrote above and see if you concur that the control pot appears to be a voltage divider for pin 4 between rail and the voltage dropped source after the motor? Refer to the sample circuit which is more basic but with careful study, can see that the actual circuit is essentially the same except that pin 1 in this case gets power through a diode (and it's associated voltage drop).