AEM WIDEBAND 30-4900 WITH M4 (M2R) VOLTAGE DIVIDER

[chick0]

Hi guys, I know its kind of been answered before, but im seeking some clarification if anyone can chime in and give an answer.

(This aem gauge does not have the inbuilt divider option where you can change the voltage output at the back of the gauge itself.)

So to begin with, I have an aem 30-4900 wideband gauge... Runs the lsu4.2 bosch o2 sensor.... Now the gauge has 2 analog outputs + and -, gives a 0-5 v output.... the wires are + white - brown...

http://aemelectronics.com/files/instruc ... 0Gauge.pdf - PAGE 4 IS THE WIRING DIAGRAM.



Now question is: Can i successfully run an inline voltage divider between the gauge and the m4?
one of these: http://wbo2.com/faqs/autronic.htm

So it would be connected as: + and - from aem gauge, then voltage divider, then connect + and - pin 31 & 32 on m4.


The reason i ask is, through searching on the net people have done this with other widebang gauges and different ecus mainly autronic. However I emailed AEM and they told me you can not run a voltage divider to change the voltage out. Which confuses me as the output is after the gauge itself so it does not affect the gauge from what i can tell... Here is their email:


Unfortunately a voltage divider will not allow you to change the output signal from 0-5 volts to 0-1 volt on the Wideband Fail-Safe Gauge. Also the Wideband Fail-safe gauge will no longer read properly within the software calibration.

[David Ferguson]

I think you should be able to do this, but be sure to use a high-resistance divider. For example, I would use a precision 80k ohm and 20k ohm resistors to convert the voltage range from 0-5V to 0-1V, the gauge would see a 100k load, which it could surely handle.

If you own the device, give it a try.

[chick0]

Thanks for your input. Appreciate it!!

With what you said david, although the wires of the aem gauge are an output wire, putting a resistor in there will put load on the gauge then?

Basically the 02 sensor itself runs to the gauge with its own loom... Then there is a second loom which has the output wires to run to an ecu....and the ground and power to power on the gauge...

[chick0]

David, i've just found the actual voltages of the aem, I think the resistors may need to be different? Please advise if you can. Here are the voltages:


volts afr lambda
0.5 8.50 0.58
1.0 9.69 0.66
1.5 10.88 0.74
2.0 12.06 0.82
2.5 13.25 0.91
3.0 14.44 0.99
3.5 15.63 1.07
4.0 16.81 1.15
4.5 18.00 1.23

[David Ferguson]

With what you said david, although the wires of the aem gauge are an output wire, putting a resistor in there will put load on the gauge then?

The gauge that is supplying the output signal is suppling a voltage -- the hardware that does this can only support a maximum amount of current (the higher the load, the more current is required). By using a large resistance value (100k ohms), very little current should be required and I would expect the gauge hardware could handle that. Normally the inputs the ECUs and Loggers also present a high-impedance (resistance), and so they also don't require any real current.

By using a voltage divider with values that are similar to the inputs the the ECU / Logger I can't imagine that the gauge won't work.

The voltage divider using 80k / 20k will divide the input voltage by 5 -- so for your table:

.1V = 0.58 Lambda (8.5 AFR for your fuel)
.9V = 1.23 Lambda (18.0 AFR)

[chick0]

Thanks David! Ill try this today shops havent opened yet only 8 am haha...

Just online i can see 20k ohm 1/4watt %1 resistors, but i cannot find 80k resistors, only 82k ones...

Should i just buy 4x20k wire them in series to = 80k?

Or can i scale down instead of divide by 5, go down and divide by 10 like they did here: http://wbo2.com/faqs/autronic.htm

[David Ferguson]

My suggestion is to buy the 82k ohm resistor.

Measure the resistance of the actual resistors you are using, and your scale factor will be determined by this equation

Scale = R1 / (R1 + R2)

R1 would be the 20k resistor, and R2 the 80k (approx).

So the scale = .2 = 20,000 / (20,000 + 80,000).

Your measured voltage in the M4 would be InputVoltage * scale. (5V x .2 = 1v)

Here's how that would change if you had a resistor that measured 20,000 and 82,000

Scale = 20,000 / 102000 = 0.196

So for a 5V input, you would measure .98 volts.

The reason you don't want to divide by 10, is that then you will only be using half of the resolution of the 1V input on the M4 (since 5V input would only be .5V output).

[chick0]

Thanks for the info.again david! My cousina been helping me with it as well from the states, hes an elec. Engineer. He said i can possibly also try a 120k and 30k res. He said:

if you want an exact 5:1 ratio and not too concerned on the low current, why dont you use 30K and 120k?

Both 30K and 120K are standard values so that would give you 1V output from a 5V inputt he output current for a 30K and 120K will be a bit too low though around 33uA and am not sure if your ECU can still recognize that small amount of current

actual delay with an increased resistance is at microsecond range so it might be negligible from your application.

- - - - -

From your experience is it also worth trying those resistors david? Just as a secondary option?

[David Ferguson]

Quit typing and start soldering.

I re-read the original question -- the AEM folks were answering the wrong question -- they thought you wanted to use the voltage divider between the wideband controller and the gauge. arghh...

I suspect you would probably be fine with a 10k voltage divider.

Just try something!!! It will work.

[chick0]

Done and soldered...

If i ground the wire output from the gauge to chassis i get my .98 volts...

If i connect that ground output to my LA - (pin 32)input im getting .43 volt reading...

So for wiring is it

LA+ out from gauge connects to LA + input pin 31

Ground output from gauge to chassis ground

LA - input (pin 31) to chassis ground