Which do you all think I should try first?

Whichever one you can build from components you already have?

 

I only need a minimum of 5 VAC (RMS) to send to the computer. Which ever schematic most easily accomplishes this is what I will go with.

You rather imply that's a low voltage but its 14Vpp and is marginal to generate from the vehicle battery directly.

But you could try the circuit in #6 and, if that doesn't provide sufficient voltage, you could add the 555 negative voltage supply shown in post #17 to nearly double its maximum output voltage.

 

You rather imply that's a low voltage but its 14Vpp and is marginal to generate from the vehicle battery directly.

But you could try the circuit in #6 and, if that doesn't provide sufficient voltage, you could add the 555 negative voltage supply shown in post #17 to nearly double its maximum output voltage.

Thanks for everything. This gives me lot's of hope. I knew I could count on you guys !!!

 

As I stated you can't get 5Vac RMS with a 13.5Vdc supply.
The max would be about 4.8V RMS.

Below is my take on what's required to get at least 5.5Vrms from a 1.1Vrms signal.
It uses a 555 oscillator in a negative peak doubler circut to generate -12V from the +13.5V battery supply to power the opamp U2.
The opamp has a gain of +5 to amplify the 1.1Vrms signal to 5.5Vrms (15.4Vpp) for the computer.
A 3Vrms signal will cause clipping of the sine wave output (at about 25Vpp) but that shouldn't be a problem for the computer.

View attachment 151737

This one confuses me a bit. Is it two separate circuits on two perfboards? Do I take the voltage from the car as V1 for the schematic on the left, and then that output into the circuit on the right? Could you explain a bit further? Thanks.

 

his one confuses me a bit. Is it two separate circuits on two perfboards? Do I take the voltage from the car as V1 for the schematic on the left, and then that output into the circuit on the right?

You can put it all on one perfboard.
Yes, V1 is the car battery voltage.
V2 is the signal from your transmission.
Sig_Out goes to the computer input.

 

I didn't include any filtering on the battery voltage, which is needed.
That's shown below with the addition of R7 and C5.

 

I didn't include any filtering on the battery voltage, which is needed.
That's shown below with the addition of R7 and C5.

View attachment 151782

Maybe an explanation for all of the inputs and outputs and how the three circuits are joined together. I'm not fully understanding it. I want to have a perfect understanding of it, as I may attempt it this weekend. And thanks again !

 

I didn't include any filtering on the battery voltage, which is needed.
That's shown below with the addition of R7 and C5.

Another reason to test with two 9V batteries. Again, just sayin'. I'm not convinced we know what signal is required. It sounds like we need an AC signal centered at ground, but I'm not super confident in that.

 

how the three circuits are joined together.

What three circuits?
My circuit contains the amp in post #6 and a 555 circuit to generate -12V.
That's all you need.

 

What three circuits?
My circuit contains the amp in post #6 and a 555 circuit to generate -12V.
That's all you need.

I was confused on how to make it all one unit, with voltage in and the gained voltage out. That's all.

 

I was confused on how to make it all one unit

Sorry, I don't quite understand your confusion.
The circuit has connections for the battery voltage, ground, an input from the transmission, and the output to the computer.
Are you okay now?

 

Sorry, I don't quite understand your confusion.
The circuit has connections for the battery voltage, ground, an input from the transmission, and the output to the computer.
Are you okay now?

I think so......I'll give it a try this weekend and let you know if I get stuck. Thanks !

 

Here's a simple amplifier to try:
View attachment 151751
This will give at least 5V peak output (albeit uni-polar) from a sensor signal as low as 1V peak (~0.7V RMS) and can tolerate a sensor signal >30V. R3 not only determines gain but also provides an impedance typical of an inductive tacho sensor, in case the computer does a check at power-up to see if the sensor is connected.

Another question Alec: is there any way of changing the resistor values to get a steady 7-9 volts output? I want to make real sure I have a steady voltage above 5 so I don't have any other problems. Thanks, sorry for the hassle.....

 

Alec: is there any way of changing the resistor values to get a steady 7-9 volts output?

Reducing R1 (to, say, 100 Ohms) will increase the output amplitude. Like the original sender, it won't produce a 'steady' output (if, by that, you mean a constant amplitude independent of shaft speed).

 

is there any way of changing the resistor values to get a steady 7-9 volts output? I want to make real sure I have a steady voltage above 5

You seem to be ignoring the difference between a peak-to-peak voltage and an RMS voltage.
Why is that?

 

You seem to be ignoring the difference between a peak-to-peak voltage and an RMS voltage.
Why is that?

Not sure I understand the difference. My basic understanding is RMS is AC. What is the significance?

 

Not sure I understand the difference. My basic understanding is RMS is AC. What is the significance?

The RMS voltage is not just any AC, it is the value of AC voltage that generates identical power as the same value DC voltage.
Calculating the RMS AC voltage for this equivalent gives a √2 (1.414) difference between the peak sine-wave signal and the RMS value.
So the difference is that a 5V peak to peak AC signal is only 5Vpp/(2*1.414) = 1.77Vrms.

Thus if you really need a 5Vrms signal, then the circuit must generate a 5Vrms*2*1.414 = 14.1Vpp output.
This is barely obtainable with an amp operating from the vehicle battery if the charging battery voltage is a least 14.1V.
That's why I proposed a circuit that adds a negative voltage to increase the maximum pp output of the amp if 14Vpp is not sufficient.

 

The RMS voltage is not just any AC, it is the value of AC voltage that generates identical power as the same value DC voltage.
Calculating the RMS AC voltage for this equivalent gives a √2 (1.414) difference between the peak sine-wave signal and the RMS value.
So the difference is that a 5V peak to peak AC signal is only 5Vpp/(2*1.414) = 1.77Vrms.

Thus if you really need a 5Vrms signal, then the circuit must generate a 5Vrms*2*1.414 = 14.1Vpp output.
This is barely obtainable with an amp operating from the vehicle battery if the charging battery voltage is a least 14.1V.
That's why I proposed a circuit that adds a negative voltage to increase the maximum pp output of the amp if 14Vpp is not sufficient.

Is this why there is why I see V- and V+ symbols going from opposite rails into opposite sides of an op amp? I always wondered about this. I have seen schematics of two nine volt batteries one feeding from the negative into one rail, one feeding from the positive into the other rail and then into the op amp. So your saying this swing produces greater amplitude and gain? That seems easy enough to build if need to go that route. Thanks for the explanation; I'm still a novice.

 

Is this why there is why I see V- and V+ symbols going from opposite rails into opposite sides of an op amp?

Yes.
The op amp peak-to-peak output voltage can be no greater than the supply rail voltages.
Thus for a +9V and - 9V supplies the maximum output you could generate would be 18Vpp or 6,36Vrms.

 

Yes.
The op amp peak-to-peak output voltage can be no greater than the supply rail voltages.
Thus for a +9V and - 9V supplies the maximum output you could generate would be 18Vpp or 6,36Vrms.

Thanks for the clarification. Bearing this in mind, if I wanted to use an op amp with those two 9 volt batteries only and simply reduce it down to 7-8 volts output with a square wave, and NOT use my vehicles sensor as an input at all, could I do that? The reason I'm asking is that one guy years ago said he used "only a battery powered op amp with vcc+ and vcc- and bypassed the sensor, with output directly to the PCM" . This was his quote, and I can no longer reach him, nor did he say how he rigged it nor did he show a diagram/schematic. I am try to re-construct his thoughts based on the information from his post. He said this: "by using a simple LM741 op amp I created an amplifier with +vcc and -vcc. Parts are 741 op amp, two batteries, a 1k ohm resistor and a 100K ohm resistor onto a circuit board". That was it, but did confirm it worked and shut off the "check engine light" by sending the require 5V or more to the computer. I am trying to wrap my head around what a schematic would look like. Thanks for helping me, I just really want to solve this problem.