perpetualmotion

complacentsee, your circuit is similar to the standard variable voltage divider implemented using a simple potentiometer. The advantage of the potentiometer is that you can easily vary the voltage across the load simply by turning the ****.

I messed up what I wrote, though. When I wrote:

"When the sliding contact is fully at one end, the full resistance of the potentiometer is in parallel with the load. The series resistance then determines the voltage drop across both the potentiometer and the ECU circuit, and since that series resistance will be somewhat less than the resistance of the ECU by itself, the voltage drop will be less."

I should have said "parallel resistance ... parallel resistance".

A voltage divider done this way is so standard that it even has a name: "L-pad". The total resistance of the potentiometer is split between two parts. One part is in parallel with the load, the other part is in series with that parallel part. The ratio of the voltage across the load to the total voltage is the same as the ratio of the parallel resistance (that includes the load) to the total series resistance involving that part.

You want the resistance of the potentiometer to be several times greater than the load by itself, to insure that when the potentiometer is fully to the end that results in all of its resistance being in parallel with the load, the parallel resistance will not be appreciably less than that of the load itself. You can of course address that concern by putting another fixed resistor in series with the potentiometer. Suppose the resistance of the potentiometer is ten times greater than that of the load (the input resistance of the ECU circuit). The parallel resistance will then be (1/ (1 + .1)) times that of the load, or 91% as great as the load by itself. The voltage across the load will drop by an amount that depends to an extent on what fraction of the total series resistance was made up by the load. When the sliding contact on that same potentiometer is moved to the other end, there will then be a resistor ten times as great as the load resistance, in series with the load, and the voltage across the load will drop to approximately 10% of what it was previously.

If I were doing this, I would estimate the resistance across the load by putting an ammeter in series with it and measuring the current and then dividing the measured voltage across the load by the measured current, and then I would find a log-taper potentiometer with ten times that much resistance. I would start with the sliding contact in the middle, and would be highly confident that I would be able to set the voltage across the load to the desired value with no real difficulty.

complacentsee

Yep, I would still switch to a pair of resistors once I had found my "perfect" values. Don't know how a pot would hold up to the conditions of the engine bay. But then again, that's probably more of a personal preferance thing.

Fusion4u2nv

sounds like it was reversed

complacentsee

Nope, just kept dropping below the forward bias voltage of the diode.

perpetualmotion

I re-read some of that tech article again, and although it makes perfect sense overall, some of it is patent nonsense. In particular:

"The first mistake people often make is trying to trick the ECM by using a resistor inline on the O2 sensor. ... The problem is our ECM uses a mega-ohm measurement device, meaning just about no matter what kind of resistance you put in, the full amount of voltage will still be seen. Remember, resistance does not drop voltage, it limits current. If the ECM were to apply a decent load, then resistance would inadvertently drop the voltage it saw. So this is the right idea, but the wrong method."

This is patent nonsense. The basic notion of a voltage divider, which is the universal method to lower the voltage across a load, is that of putting an appropriate resistor in series with the load. By example, if there is no resistance in the circuit except for the load and the resistor the you add, and if the resistor you add has twice the resistance of the load, then the voltage across the load will drop to 1/3 of what it was previously, because the voltage across the resistor must be twice the voltage across the load (because the current is the same through both and because the resistor has twice as much resistance), and because the combined voltage drop must be the same as what it previously was across the load itself. The total series resistance is now three times what it was previously, so the current will drop to 1/3 of what it was previously. It is no coincidence that the voltage across the load, and the current through the load, have both dropped by the same fraction, 1/3.

No matter how large the input resistance of the circuit in the ECU, as long as it is reasonably stable, which it certainly should be, a simple voltage divider is the standard solution. It is just a matter of finding the right value of series resistance to add.

Whether or not this will actually work depends on whether the ECU does actually sense the current. If so, there is probably no way to do this trick at all. When I first read "internal resistance too high", my assumption was that it refers to the resistance of the heater element. But I am not at all certain of this, and if this detected fault does in fact apply to the sensor itself, as opposed to the heating element, then it all likelihood it simply will not be possible to perform this trick at all. If the person who wrote the tech article was successful, this suggests that the ECU does not sense the resistance of the circuit involving the sensor itself.

Which brings me to this: "The average rectifier diode is a one-way valve. It has a voltage drop when current passes through it, which for the sake of discussion is usually around 0.5 volts."

An ideal diode has zero resistance when D.C. voltage is applied to it in one direction, and infinite resistance when D.C. voltage is applied in the other direction. But real diodes are different. In the "forward" direction, the internal resistance remains quite high until the voltage drop across the diode reaches a threshold around .6 to .7 V (assuming a common silicon diode). This is the "forward bias" voltage. Once that threshold is reached, the voltage drop across the diode remains fairly constant for a fairly wide range of current values, and this means that the diode does not behave like a simple resistor, where voltage drop and current remain in a fixed ratio (the resistance). But the potential problem (there's a pun, intentional or otherwise), is that it is not guaranteed that the voltage generated by the O2 sensor will be adequate to achieve the required forward bias voltage across the diode. The person who wrote the tech article is claiming, in effect, that the electrical characteristics of the sensor and the circuit in the ECU are just right to yield a .5 V voltage drop across the diode, i.e., slightly less than what is needed for the diode to be properly biased, but the right amount to achieve the desired effect. I'll take his word for that, but it bugs me that what he wrote implies that a voltage drop of .5 volts is assured even if the voltage drop across the diode is less than that. The annoying thing about diodes and such (it applies to transistors as well), is that it makes no sense to talk of the "supplied voltage", because the only thing that is measurable and meaningful is the voltage drop across the component. So by necessity, we talk about how the current varies with the voltage drop across the diode. The diode is in effect a resistor having very high resistance when the forward voltage is below the bias voltage, and once that voltage threshold is reached, the effective resistance of the diode decreases as the voltage across the diode increases.

The bottom line is that if a diode works here, it is because the characteristics of the sensor and the circuit in the ECU are such that the voltage drop across the diode is just right to achieve the desired effect. It should be possible to put an ohmmeter in series with this arrangement to determine the current, and assuming that the voltage drop is indeed .5 volts, it should be possible to determine the effective resistance of the diode, and to replace the diode with a resistor having that same value and yielding the same net effect.

But from my perspective, the potential issue with the voltage drop across the diode not being adequate to properly bias the diode is of concern. It seems to me that it is easily possible that slight variances in the input resistance of the circuit in the ECU, or in the sensor, could throw off the voltage drop across the diode. I am not convinced that this solution is highly repeatable. If I were messing with this, I would use a simple potentiometer as I described, and once the optimal position of the **** on the potentiometer is found, the resistance can be measured and, if desired, the potentiometer replaced with a simple resistor. The diode might work, but whereas the person who wrote the tech article clearly believes that it is fool proof and that there is no reliable way to accomplish the desired effect using simple resistors, the truth is much the opposite.