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  #1  
Old 09-04-2002, 07:03 PM
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Question Modifying a MB dual coolant sensor to trigger an electric fan?

I was wondering if the electrical wizards here can help me.
I want to replace the SINGLE coolant sensor with a MB *double* sensor,
that way I don't have to use the sensor that sticks into the radiator
fins to trigger my electric puller fan that replaces my viscous fan.

Pins 1 & 4 act just like the "blue" coolant sensor that triggers the
Aux.
fans.

However, Pins 2 & 3 have a smaller value.
The Derale adjustable thermostat uses a sensor that gives higher values.

Here's a brief table (yes I know, it's incomplete) in OHMS:

Double-sensor
Celsius Pins 1/4 Derale
44.4 970 5525
48.9 810
54.4 730
60.0 570
65.6 470
71.1 396
74.4 361
76.7 345
82.2 309 2924
85.6 268
87.8 252
91.1 228
93.3 215
95.6 195
97.8 190 2750
98.9 188 2460
100.0 186 2240


If I want to trigger the Derale to come on at 93.3-Celsius how can I get
the MB sensor on pins 1/4 to mimick the values of the Derale (note- it's
adjustable to a point)?

I was thinking of adding in series a resistor (I think resistors in
series you add them together) that would get me in the "magic range" but
the resistor value cannot be high-enough to trigger it on until
93.3-Celsius.


Thanks in advance,
:-) neil
1988 360TE AMG
1993 500E
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  #2  
Old 09-04-2002, 07:44 PM
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If you are talking the Benz 4 prong sensor , the values are the same , they are just isolated from one another as they feed different systems.
The resistor in series will not change the cut in. It has to be in parallel. Reason is the total resistance of the sensor will now drop.
[ parallel resistors total value is always less than the smallest resistor]
So, using an ohm meter , measure the resistance of the sensor compared to the gauge at the temp you are looking for and
compare to your chart for the desired ohms.
you are looking for 215 at 93 - correct?

Plug what you have for sensor read at temp and using RT formula for par. resistance, the unknown R can be fiqured..
[ look at google for a par R plug the number data computer"
That makes it easy..
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  #3  
Old 09-04-2002, 07:56 PM
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Thanks Arthur. I read the values while boiling water TWICE with my Fluke VOM, and got different values from Pins 1/4 vs. Pins 2/3. I got NOTHING when reading Pines 1/2 or 3/4.

BTW- I'm using MB sensor 011.542.51.17
What 4-prong dual MB sensor are you using (part number please).


Pins 2/3 seem to mimick the stock sensor.

C Pins 1/4 Pins 2/3 Derale STOCK
44.4 970 2150 2120
48.9 810 1810 1850
54.4 730 1350 1450
60.0 570 1230 1160
65.6 470 1040 1030
71.1 396 870 858
74.4 361 750 743
76.7 345 710 710
82.2 309 630 2924 600
85.6 268 540 540
87.8 252 510 501
91.1 228 464 455
93.3 215 431 430
95.6 195 391 391
97.8 190 376 2750 380
98.9 188 367 2460 370
100.0 186 358 2240 340

:-) neil
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  #4  
Old 09-04-2002, 08:20 PM
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BTW- IF my readings of Pins 1/4 are correct (970-ohms @ 44C; and 186-ohms @ 100C), then per the Parallel Resistor circuit formula, 970-OHMS is the most resistance I can get from a parallel resistor.

Perhaps I need to increase the resistance value of the circuit (by a series resistor) to a value that I can drop with an additional resistor in parallel.

BTW- the target is to have a reading of 3.35K-Ohms at 93.3C, however, the reading needs to be under 2.9K-Ohms under 93.3C.


Does that make any sense?
I am so confused . .

:-( neil

Last edited by ke6dcj; 09-04-2002 at 08:30 PM.
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  #5  
Old 09-04-2002, 08:42 PM
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< >>

Ok , Now I see

I was mis-reading the first chart and thought the target was 215...

Problem now is the Benz sens is an NTC type, so as you heat up, the R. drops...

So parallel or series are both going in the wrong direction for your set-up...
Let me think on it ...looks like you need a positive sens. [ R increase w/temp]
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  #6  
Old 09-04-2002, 08:57 PM
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Arthur- the 4-prong MB sensor I'm using decreases resistance as the temperature increases (approx. 970-ohms @ 44C and 186-ohms @ 100C;
sensor range: 870-ohms;

The Derale's radiator probe also decreases resistance as the temperature increase (apprx. 5.5Kohms @ 44C and 2.24Kohms @ 100C;
sensor range: 3.26-ohms
Derale preferred trigger point: 3.3kohms @ 93.3C

What next?

BTW- what 4-prong MB sensor are you using?

:-) neil
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  #7  
Old 09-04-2002, 09:13 PM
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I am not using a 4 prong , I am just looking at charts..

OK
Your sens is Neg , but your chart shows your sens at
430 ohms at 93.3 C

So why is your target different ? And why are youooking for less ohms at lower temp ?
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  #8  
Old 09-05-2002, 01:14 AM
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I did another measurement with the Derale sensor. Here are the results:

F C Derale/Target (Ohms)
106 41 25200
120 49 17000
130 54 13300
140 60 10000
150 66 8600
160 71 6400
170 77 5500
180 82 4500
185 85 4300
190 88 4000
195 91 3500
200 93 3300
206 97 2900
210 99 2700
212 100 2600

Here are the values of Mercedes sensor:

C MB Sensor (Ohms)
44.4 970
48.9 810
54.4 730
60.0 570
65.6 470
71.1 396
74.4 361
76.7 345
82.2 309
85.6 268
87.8 252
91.1 228
93.3 215
95.6 195
97.8 190
98.9 188
100.0 186

Note- I'm using the MB Sensor to feed into the DERALE adjustable relay. In short, I'd like to the DERALE relay to trigger the relay between 3500 -ohms (91C) and 3300-ohms (93 C).

The MB Sensor at 91C is approx 228-ohms; and at 93.3C approx. 215-ohms.

So- what can I do to use the MB Sensor with the Derale relay?

:-) neil
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  #9  
Old 09-05-2002, 02:29 AM
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Sorry, still can't . . .

read those tables or columns!

Neil, it only takes a few minutes more to HARD format them as I requestedl. Besides, later when somebody else reads this thread, it will mean something.

Art, I see your not having much luck either! Have at it, man! The key is a parallel resistor??
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  #10  
Old 09-05-2002, 02:59 AM
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Jim- I typed them out with SPACES (removed all TABS), and it still came out like above!

How about this link:
http://www.silcom.com/~neilv/sportline/pages/Derale-table.htm

:-) neil
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  #11  
Old 09-05-2002, 02:59 AM
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OK, against my principles

I read those s@%tty charts and, bottom line, you can't!

The MB sensors trigger around at (200-300 ohms) low ohms and it appears from what you said, the Derale triggers at 3.xKohms, high ohms.

You CAN'T add a series resistor that's an order of magnitude greater than the desired trigger point, since that resistor will swamp the resistance trigger point.

Now if you want to build an 'amp' whose trigger point is, say 93 ohms that in turns kicks in a relay that will turn on the fans, that can be done. But it's a start-from-scratch-circuit unless we can find something from Radio Shack (as an example) that can be modified to do what you want using the MB sensor.

It can be done but not with what you currently have!
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Last edited by JimF; 09-05-2002 at 03:07 AM.
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  #12  
Old 09-05-2002, 04:02 PM
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This will work and. . .

it's the simplest circuit that will do what you want.



The key part(s) is the R2 potentiometer. It's a 3watt RS part of 25 ohms variable. This will set the trigger point to any point around 200 - 250 ohms from the Mercedes 4 term sensor. The diode across the relay prevents spikes from wiping the transistor. Any NPN transistor with a TO-220 case will work.


The graph shows the relay pull-in point (12V relays typically pull-in at about 8V or greater) based on the resistance of the CTS. Note that at 225 ohm point (adjustable by R2), the relay will fire and start the fan.

It's not elegant but it'll work. A 'better' solution involves use of an OP-AMP and transistor plus more parts.

PS: A really neat solution is to forget the relay and use a "heavy enough" transistor and fire the fan directly. In this version, the fan's speed will be proportionally controlled and come on gradually as the resistor changes in the area of 250 to 200 ohms. Cool!
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Last edited by JimF; 09-05-2002 at 04:08 PM.
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  #13  
Old 09-05-2002, 05:27 PM
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Jim- I just knew you had circuit brewing in your head, all good Ham Radio guys/gals do!

I'm not very good in reading schematics can you clarify the following:

1) the schematic show R2 with a value of 15.5-ohms, is that the set-point for your simulation of the 25-ohm potentiometer?

2) what does Rf at 100K do? I see it's in parallel with the MB CTS. Does it bring the values of the CTS into a range to work with R2?

If my SPAL fan is rated at 23Amps @ 12VDC, is it really feasible to drive through a large transistor?

Can I contract you to make me two of these?

Thanks in advance,

:-) neil
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  #14  
Old 09-05-2002, 07:00 PM
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'94 S500: only 793 sold!
 
Join Date: Mar 2001
Location: San Diego, CA
Posts: 1,925
Circuit clarification. . .

It's not the Trump Plaza but it'll work!

1) Yes, the 15.5 ohm point is the 'center' of the simulation but since the pot is 25 ohms, you can set it 'exactly' based on the final circuit values. The value of R2 determines where the relay will turn on and thus at what temperature the circuit fires.

2) Rf is called a "feedback' resistor giving the circuit some positive feedback to help the transistor switch. It does nothing to the MB CTS values.

3) The SPAL fan does require lots of amps so the circuit as shown would need some 'beefing' to withstand the power dissipation. For example, with the fan 'half-on' the power dissipation in the transistor would be 6 - 7Volts x 22 amps = 143 watts!!! That would reguire a substanial heatsink and a TO-3 case style transistor. What comes to mind is the blower regulator module on my car. See the pic in this thread:
A/C blower's output is low . . .

BTW, did I use the right values for the MB CTS that you are considering?

I really don't want to build this circuit but I'll look around for some 'device' that you can easily modify for your needs. There's got to be someting that's just right.

PS: On second thought, what's wrong with using that blower module??? Other than it's fairly expensive ($170)??? I think it could be used!!!!!
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Last edited by JimF; 09-05-2002 at 07:07 PM.
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  #15  
Old 09-05-2002, 07:36 PM
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Using the MB blower regulator does sound like a good design. I wonder if the W124 blower regulator (124.820.27.10) might be better. It has a giant heatsink (resembles a Pentium IV heatsink), and MB engineers were smart to place it in the air-path of the blower so in a sense, it self-cooled itself!

Unfortunately, its a royal PITA to replace (they should have made an access port!) and also expensive ($274).

You did get the MB CTS values correct.

BTW- why would an op-amp be a better design?
From the "simple" circuit you posted, it sure looks like very few parts to go bad; and probably a very HIGH MTBF.

Perhaps you can make me TWO of these "simple" circuits OR TWO of the op-amp version.

Thanks again,

:-) neil
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