I said I was done, so I don't know why I'm being drawn into it.
Probably wasting my time, but hey-ho.
The following only requires an understanding of ohms law and basic potential dividers. Potential dividers used to be taught very early on in secondary school. For anybody that can't do potential dividers or ohms law, maybe stop here.
This is a basic representation of the valve and its drive.
VRV by
shcm, on Flickr
R1 represents the total resistance of the cable harness path from the battery, through a relay (refer to the Toyota document I posted) to the one pin of the VRV. This resistance should be low. Fractions of an ohm.
R2 represents the internal resistance of the VRV. The VRV is basically a coil of wire (inductor) that will produce force (current through an inductor produces force) and that force, via mechy bits, will provide a valve type action. Ideally the VRV would have no resistance, but all that wire making up the inductor, means it has some. The equivalent circuit is shown for the VRV. i.e. an ideal inductor (no resistance) in series with a resistance. Typically the VRV resistance (R2) will be around 12 Ohms. Here's an example:
zatonevkredit.ru/repair_manuals/raw_content/RM000003THT002X.xml#RM000003THT002XR3 represents the cable (harness) resistance and some connector resistance from the VRV to the ECU pin. This again should be low. Fractions of an ohm.
Q1 is the drive in the ECU. Simplified, it can be thought of as a fast switching switch (SW1). SW1 (or Q1) has some resistance when in the "switch closed" state, but it is a very low "on resistance". Again, normally, assume SW1 "on resistance" is fractions of an ohm.
There will be, also in series, a ground return path from the ECU to chassis/battery, which I will ignore for now.
When SW1 is open, no current flows. The voltage at A,B,C and D has to be 12V. (Current through a resistance, causes a voltage drop, i.e. the voltage drop is ohms law V=IR. Since there is no current, there is no voltage drop, so B,C and D are all at the battery voltage (A)).
When SW1 is closed, current flows. The voltage at D is now very close to ground, but not quite ground because the switch has some small resistance. V=IR, so V in this case is usually small (the current here isn't massive and the resistance isn't large), so point D has to be a little bit (maybe say 0.2V) above ground.
These two voltage levels at point D are normal and are shown in the waveform in Toyota's document, that I attached previously.
It's similar for point C. R3 is small, so C is just a little bit higher again than D, voltage-wise. By the time we get to B, we've gone beyond the most dominant resistance here, i.e. R2 - the VRV resistance. Point B will therefore be almost 12V, but not quite, because of the small voltage drop across R1. (Resistance R1 is very much smaller than the dominant VRV resistance R2).
What happens if say SW1 "on resistance" becomes say a faulty 12 Ohms? Well when SW1 is off point D is still 12V. When SW1 is on, point D rather than being just a bit above ground is now probably around 6V. SW1's faulty on resistance is now roughly the same resistance as R2, while the other resistances R3 and R1 are very low. So you've essentially got a potential divider with roughly 2 equal resistances, so point D is going to be half the supply (half of 12V) voltage. The waveform at point D will therefore have a "high level" at 12V (SW1 off) and a low level of 6V (SW1 on) rather than a normal low level of close to 0V (GND). So, any significant rise in the low level waveform voltage @ D indicates a problem with SW1 (or actually an ECU ground return problem too). Refer again to the waveform plot in Toyota's document.
By the same reasoning, similar applies @ point C, should R3 have higher resistance that normal. The "low part" of the switching waveform @ C will be higher than expected.
The above explains:
With oscilloscope at pin 2 of the VRV, if the low part of the PWM (pulse width modulated) waveform there doesn't pull close to ground and it does pull to ground with the oscilloscope at the ECU pin, that would also indicate a harness/connector problem. It may indicate insufficient drive current for the VRV (not due to the ECU, but due to an ageing harness or connector or corroded/dirty connector)." Should R1's resistance become significant, then instead of being close to 12V, when SW1 closes, there will be a voltage drop across R1, so you will see a waveform @ B.
The above covers:
"Anything lower than approx "battery voltage" at pin 1 of the VRV at any time (probably needs an oscilloscope, although a DMM may display an average) would indicate a harness/connector problem. You can get some indication of what's going on with a multimeter, but it just averages the voltage reading. You need to be sure that SW1 is switching (which produces an averaged force variation for the valve control) rather than it just sitting there acting as a resistor. A multimeter will show both the switching case and the fault case (i.e. the switch just sitting closed at some resistance), as the same thing.
Now, you could just take the cables off and measure the resistances. I have suggested it. However, the tolerance on the VRV is only +/- 0.7 ohm. Any resistance variation more than that in this circuit, is likely to become significant, meaning the VRV maybe doesn't get quite the current it needs and therefore maybe doesn't produce the force/control required. With just a simple multimeter, measuring to that kind of resistance accuracy is maybe just possible, but becoming borderline. Looking at voltage drops and waveforms is more informative. Try measuring resistances with starter motor cables for example, whereas a relative voltage drop measurement will help pin point the problem. It's similar but on a slightly different scale here.
You could also measure the average current, but again the "faulty fixed resistance" versus switching problem, as above, applies. You'd need a "current clamp" into an oscilloscope.
A cable resistance change here of one or two ohms could make all the difference and a partial internal cable break or iffy connector is more than capable of producing that.
(By coincidence, we have a PC that acts a a TV and media server for the whole house. This morning it starts behaving erratically. The problem? A power connector not quite seated properly. Must have been like it for a few years, since purchase.)
The ECU may well be monitoring VRV current and the waveform @ D for fault detection. It is done in many cases.
Maybe measure the VRV resistance too.
Now, I don't think the problem is probably any of the above either, but if everything else has been changed and is assumed not to be faulty, by a process of elimination, what's left? - You have to at least try to look elsewhere. Quickly measuring a harness cable resistance to see if it is more than a few ohms is not exactly a "whoosh moment".
As an example, if Charlie were to offer advice about a roof it would not be questioned. If Don were to offer advice about brakes it would not be questioned. I suggest maybe just quickly measuring a harness resistance, or describe what waveforms to expect at certain points...........hmmmm. Maybe the time has come not to bother at all.
Like I said, 30 years involved with this stuff, clearly isn't enough.
Well, all of the above is some of my life probably spent pointlessly, which I won't get back.
Enjoy!