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need to build vss signal divider


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oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: September 05, 2013 at 9:22 PM / IP Logged  
Like I said, image names CAN NOT be that long.
Going high-voltage has little to do with the alternator age - it often has more to do with vehicle age (ie, ground integrity), though yours may have a maximum output voltage.
(I have found a few Jap alternators that won't exceed ~15.5V - IMO a design feature in case the S/Sense terminal is disconnected - not that I have tested them with a bad engine-body GND.)
You're sure you have the ECM/speedo taking its signal from 4017 output 1 (or 2 or 3)?
To play safe, connect your output LED to another output (ie, 1-3, but not the one the ECM uses).
If the ECM/speedo expects a square wave input, I can't see what the problem is. The 4017 output will switch to within 0.05V of either rail which is as good as an OC output or "+V" output.   
So if the 4017 is advancing its output, the ECM/speedo should be happy.
Unless it expects an OC ouput and hence not above +5V etc...
Or if it expects a sinewave...
At first when I saw your optical VSS buffer, I thought it was a zero crossing detector for a reluctor circuit (since it has ~6 semiconductors).
It could be an output squarer for a "raw" optic or Hall signal, but that is rare - usually optic & Hall sensors come as "modules" which include the square-wave output circuity.   
But I assume that circuit is feeding the 4017 etc? (That isn't after the 4017?)
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 05, 2013 at 9:39 PM / IP Logged  
Actually it is an optical signal. That module in the picture is no longer in the car. It uses optics to detect the cable driven speedometer spinning and converts that into a 2kppm squarewave.
Since the new cluster is digital and uses an electronic (Magnetic) VSS mounted at the transmission the stock speedometer head is removed rendering the optical VSS module pictured useless.
The way I hooked up the 4017 was to splice into the signal wire coming from the magnetic VSS (Which according to the manufacturer outputs 8kppm squarewave) and run that to the CLOCK pin. On the test bench when I pulse 12VDC to the clock pin it is properly advancing and resets to Q0 when Q5 goes HIGH. Signal wire to ECM has been verified at output Q3 but there is something I;m missing on the vehicle end.
I have contacted someone who is very knowledgeable in that area so I'll report back when I hear from them.
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 05, 2013 at 9:41 PM / IP Logged  
BTW when you said the behavior of electrolytic capacitors breaks my theory, what were you referring to?
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: September 05, 2013 at 9:53 PM / IP Logged  
An electrolytic cap (ie, 100uF etc) does NOT provide the same protection as a 0.1 or 0.01uF cap - you need the smaller caps as well - even if next to the 100uF - though the 0.1uF etc decouplers are usually placed close to the chip +V pins.
You'll see many digital circuits like that, ie with large 100uF filter caps wherever, but the smaller decouplers strategically located around the PCB at chip +V pins or after/at every few chips etc.
The big caps are for power filtering, the decouplers are to remove chip-caused transients.
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 06, 2013 at 9:37 AM / IP Logged  
Spark, i have just received a response:
[QUOTE]The 4000 series CMOS, even with buffered outputs, doesn't have enough 
drive capacity. Further buffer the output with a 2N3904 transistor. 
Output of 4017b to a 10K 1/W, then to the base. Ground the emitter and 
tie the collector to the ECM input.
I do the same when using either LS or HC logic. You may be able to 
drive both the cruise and the ECM from the one transistor. Inside the 
ECM is a 5.1K resistor as a pull up to +IGN volts.[/QUOTE]
I will try this and report back.
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 09, 2013 at 9:51 AM / IP Logged  
Oldspark,
For the sake of doing my first ever circuit the right way I went to town with this thing and it is nearly complete. All I have to do is add the transistor to power the output pulse LED since its 20mA draw will max out the Q4 output.
I did want to add those protection diodes you mentioned. Could you explain to me one more time how they work I didn't fully understand.
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: September 09, 2013 at 10:59 AM / IP Logged  
I assume you mean the "clamping" diodes to ensure the signal does not exceed +V & 0V by more than (say) 0.6V - 0.7V for silicon dioess, or ~0.3V for Schottky diodes.
It's probably easiest understood graphically (or pictorially?) - viz: Uses for a Double Diode? or Testing Protection/Clamp Diodes with .... or Design femtoampere circuits....
EG - the first http://electronics.stackexchange.com link shows...
need to build vss signal divider - Page 2 -- posted image.
Forget the "labels" etc, but maybe you can see that if the input (sclk) is higher than (say) +Vs plus 0.7V (or whatever the diode's fwd voltage drop at that current is), then the upper diode "shorts" the input to +Vs so the input cannot exceed - or is "clamped" to a max - of +Vs plus Vd (Vd = diode's fwd Voltage drop).
Likewise if the input Vin (sclk) is much lower than 0V, the lower diode shorts Vin to 0V, hence Vin cannot drop lower than 0V - Vd or about -0.6V (for silicon diodes; ~-0.3 for Schottky diodes, or ~-0.2 for "ancient" Germanium diodes).
Ooops - I should have said "lower than GND" instead of "lower than 0V", but it's too late now... (Hey man, some systems use a +ve GND - eg - very old vehicles, and many telcos.)
BTW (IMO...) - Ain't the diodes "arrow heads" great? Provided the voltage across that arrow head (is +ve to -ve and) exceeds the diode's forward voltage drop (aka Vf), that diode clamps aka shorts the voltage to its line end aka Cathode (or Kathode of you like the "symbolic" --K--- = ---|<--- analogy) - but adding or allowing for that Vd diode voltage drop difference.
Does that make sense?
I've added a bit of Kathode = Cathode = "line end" and that current (+ve to -ve) flows THRU the arrow head (into the line = Cathode end) to try to demystify and make basic diode behavior easy to remember.
That voltage clamping technique is common (and reliable and simple). EG - it is used to make FETs and MOSFETs "static proof" as shown in the 3rd link above. (Before that, FETs like CMOS had to be handled VERY carefully!)
If your input is OC (open collector) etc, then such diodes may be overkill. But the way I look at it, two 5c diodes (common 1N400x or maybe 1N914 etc) should ensure downstream circuitry does not see more than +V + Vd or less than 0V - Vd even if a relay coil spike or fan motor transient or lightning strike enters that circuit. (Ok, maybe not a lightning strike!)    
That sort of protection is typical for "automotive hardened" electronics.
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 09, 2013 at 2:34 PM / IP Logged  
Thank you I will look into all this once I have finished with the transistor circuit to power the output pulse LED.
Now please bear with me, but I am trying to understand how to determine what current to drive the transistor base at and am getting lost throughout google searches. What I have determined so far and please correct me if I am wrong:
HFE = minimum current gain
I am using an MPS2222A transistor only because I had it laying around and figured why not vs going out and buying a transistor. The jedec data sheet for this transistor states the current gain for the 2222A @ 10mA load @ 10V is 75. However the next step is 150 @ 150mA @ 10V. How would I determine the current gain with a 20mA load? Also, at full alternator output the load is going to be driven at 14V and not 10.
I also have this 20mA @ 3V LED regulated through a 470ohm resistor which is perfect for a 12V load however at 14V the resistor value should be a bit higher to 550ohm but as far as I'm concerned all this will do is shorten the LED life a bit which is not a concern because it will have no use once off the test bench. I just want to throw the LEDs in there so I can watch the pulses being divided on the test bench "for fun".
I have the input clock LED powered directly from the CLOCK input wire but since these LEDs draw 20mA nominal this will max out the Q4 outout of the CMOS and not have enough drive capacity left to pulse the ECM and cruise control. So, I have decided to tie the transistor base to the Q4 output and use the transistor to power the output LED.
Also, understand that since the resistor value is a bit off for the LED load this will result in the LED pulling slightly more current and throw our base current calculation off a tad.
oldspark 
Gold - Posts: 4,913
Gold spacespace
Joined: November 03, 2008
Location: Australia
Posted: September 09, 2013 at 11:36 PM / IP Logged  
Ah (or arghhh!) - hence why I like using FETs or MOSFETs. There are no "base" (gate) resistors to calculate - as long as the ON voltage exceeds the Vgs "gate turn on voltage" (typically round 5V) it turns fully on.
But usually 2 resistors are used - namely a resistor from Gate to Source (ie GND or 0V for an N-channel MOSFET which is equivalent to an NPN transistor) to ensure it turns off, and a series Gate resistor (like a transistor's Base resistor) to limit Gate current.
Since the 4017 pulls low (0V) when an input is not on, the Rgs is not needed. But may as well include one to ensure it's off if the 4017 is unpowered or not connected etc. Rgs is typically 1M but is not that critical - eg, 10k or 100k etc. (FETs only need nA or uA to turn on unlike transistors that may need mA or nearly mA to turn on.)
And the Rg "series" current limiting resistor is not to protect the Gate, but the feeding circuit in case of high Drain thru Gate currents (which I think is a fault or breakdown situation).
Hence the minimum Rg value would be (say) 16V/1mA = 16k => 18k to limit a 16V supply thru the (Drain to) Gate to 1mA which is easy for the 4017 to handle.
I chose 16V as a typical max automotive design voltage, and 1mA because - from memory - the 4017 will sink that current on a low/off output.
Since probably less than 1uA is required to turn the gate on (1000 times less than 1mA), that 18k could be up to 18M, but too big means low current which is susceptible to noise injection, hence keep it under 1M or less.
I'd suggest for a "one size fits all" resistor size, maybe 100k for both the series 4017 to Gate Rg resistor, and the Rgs resistor from the Gate to Source.
FYI - I omitted that Rg has to be small enough to allow enough Gate ON current for (say) a 4017 output voltage of 10V into a Vg-on of ~5V, but 1M or 100k will ensure that - ie, a "worst case" 4017 output voltage of 10V into a Vg of 5V means 5V across Rg. 5V/1M or 5V/100k = 5uA or 50uA respectively which should be more than enough to turn on a MOSFET.
The next calc is the current limiting series LED resistor. That is relevant for both FET/MOSFET and transistor solutions.
EG - assume a MAX of 20mA thru the LED at a supply max of 16V. Ignore LED and FET/transistor voltage drops.
Hence R-LED = 16V/20mA = 0.8k = 800 Ohms, hence 820 Ohms to be less than 20mA (using the common "preferred" resistor values).
Allowing (say) 2V for red LED, it's (16V-2V)/.02A = 14/.02 = 700Ohms, hence again 820R (R = Ohms) else 680R. 680 Ohms should be fine since we are assuming a 16V supply, and we are ignoring the FET/transistor voltage drop.
Plug in "real" values and you might find 560 Ohms is ok.   
In practice, the LED resistor is not too critical accuracy wise since whether it's 10mA or 15mA or 20mA may not have much impact, and although a "20mA LED" may mean infinite or 5,000,000 hour life at 20mA, it may NOT mean it blows instantaneously above 20mA. Maybe 25mA reduces its life 1000-fold - ie, it only lasts 5,000 hours (about 8 months full-time operation). NOTE that I'm only guessing and IMO being pessimistic for the sake of illustration.
I'll leave your transistor deign as is but offer the following:
- HFE should not change much with voltage. (It might increase a bit with higher voltage, but HFEs vary anyhow. And since you are using its minimum HFE you should be safe.)
- design for a 4017 output of 10V (or maybe 8V?). Hence it should work whilst cranking or totally flattening a battery whilst getting home without an alternator.
You want the transistor (or FET) to supply AT LEAST the desired LED current (ie, 20mA; maybe 25mA). The LED resistor then limits that current to a max of 20mA etc.
And you probably want the transistor (or FET/MOSFET) to be "fully on" so that it has a minimal voltage drop (Vce or Vds) to limit its power dissipation (that voltage drop times the current thru it).
CAVEAT - I am very rusty when it comes to transistor design, but straight "on & off" situations are reasonably easy.
BTW, if you are using (say) a hex Schmitt inverter to clean up the input source (clock), maybe you can use unused inverters to drive the LED if they handle 20mA etc. Or if they are OC (open collector) outputs, 2 or more paralleled to sink the 20mA LED current.
Insert other unused inverters to un-invert if inversion is not desired.
Sorry for the long reply, but I thought I'd cover various points or options as well as show how often design values are not that critical (it's more important allowing for worst-case situations).
And I've made it an "automotive design example" rather than for a test bench (ie, an 8V or 10V to 16V range).
firedemonsic 
Member - Posts: 23
Member spacespace
Joined: September 09, 2011
Location: Virginia, United States
Posted: September 10, 2013 at 9:53 AM / IP Logged  
That's all useful info. I'll be sure to re-read once I have had a GOOD night's sleep because I'm running on fumes ATM and having trouble taking stuff in.
However last might I discovered a new problem. At this point I am inclined to just remove the LEDs but if it is a simple fix I will. Do that.
What is happening is I noted that the input pulse from the VSS has now dropped to 5V along with the output on Q4 while the supply voltage is still 12. I have the input pulse LED powered off the CLOCK signal wire and grounded with a 480ohm resistor.
With the output transistor driven LED hooked up in the manner I indicated above (Base of transistor to same output Q4 as signal wire through 54kohm then emitter to ground and collector to outout LED ground through another 480ohm resistor).
As long as the transistor is hooked up the 4017 will ceases to count. The input pulses from the vss are still coming but the cmos does nothing. Also the output pulse LED remains constantly lit (At least I got all the resistance values right through all that transistor math!).
I suspect current is taking a ground path it should particularly through the LED resistors but how do I correct this?
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