MS build thread
#42
Boost Czar
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its just a way to activate the fuel pump relay. you either add the ground to it in place of the AFM (no additional circuit, but extra wiring) or backfeed 12v to the relay to activate it.
back-feeding the 12 volts would be giving it 12v to the spot labeled ECU STA on the above diagram, the ground to the AFM isn't needed that way. make sense? you dont want to just give it 12v from the harness at 1B because then the pump will be running anytime the key is on.
techincally you can do the same by jumping F/P and GND in the diagnostics box, but it's not the right way.
back-feeding the 12 volts would be giving it 12v to the spot labeled ECU STA on the above diagram, the ground to the AFM isn't needed that way. make sense? you dont want to just give it 12v from the harness at 1B because then the pump will be running anytime the key is on.
techincally you can do the same by jumping F/P and GND in the diagnostics box, but it's not the right way.
#43
#46
ok, well anyways I got it done and threw it into the car but seems trying to crank it I see no rpms, I checked my CAS pickup wiring on the harness I built to see if it was that, seems thats fine, so it must be the mods most likely. That or the pins are wired internally to a different pin for pickup compared to the harness.
#48
Heres a log file if anyone wants to check it out that may have more helpful insight, all I can think of off hand is its not picking up the CAS or there maybe more settings I forgot about, but I figure it should still be able to show the rpms with the correct trigger wheel settings.
thoughts? maybe it won't output unless some circuit is connected that I missed?
thoughts? maybe it won't output unless some circuit is connected that I missed?
#50
Boost Pope
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It's a safety feature. If you crash and rupture a fuel line, you don't want the pump to keep running. Burning to death is one of the shittier ways to go from what I understand.
The official, according-to-Hoyle nomenclature for the ground wires on the 1.6 ECU is as follows:
2A: Injector ground (goes to ground common JC-03)
2B: Output ground (goes to ground common JC-03)
2C: CPU ground (goes to ground common JC-02)
2D: Input (sensor) ground (goes to ground common JC-02)
Eventually, JC-02 and JC-03 wind up getting commoned together, and terminate at the ring lugs on the back of the intake manifold.
They key point is that the sensitive stuff (sensor and CPU ground) is kept separate from the noisy stuff (injector and output drivers) for as long as possible, right up until they common at the end.
From the log, I can tell you that it is seeing the second (CMP) signal at pin 11 of the ECU (PTC4, which is the 16's position, is toggling, as the decimal value of portc transitions between 3 and 19 while cranking) however it does not appear to be seeing the primary trigger.
another useless question, why on the 1.6l standalone harness there is noted chassis ground and sensor ground and on the boomslange its just know as black ground. For my grounds I only put two to the chassis grounds and left sensor ground unconnected
1 cable to 3 grounds on the MS
1 cable to 3 grounds on the MS
2A: Injector ground (goes to ground common JC-03)
2B: Output ground (goes to ground common JC-03)
2C: CPU ground (goes to ground common JC-02)
2D: Input (sensor) ground (goes to ground common JC-02)
Eventually, JC-02 and JC-03 wind up getting commoned together, and terminate at the ring lugs on the back of the intake manifold.
They key point is that the sensitive stuff (sensor and CPU ground) is kept separate from the noisy stuff (injector and output drivers) for as long as possible, right up until they common at the end.
Heres a log file if anyone wants to check it out that may have more helpful insight, all I can think of off hand is its not picking up the CAS or there maybe more settings I forgot about, but I figure it should still be able to show the rpms with the correct trigger wheel settings.
#51
I did a contunity test on it and my multimeter shows 1800~ on the cas line, measureing from where I put the resister on the diodes 5 and 9 to the pin on the db37 connector on the ecu, maybe this has to do with it?
though from pin 11 on the cpu to the db37 connector cas pin I get 0 so that is good
EDIT: nvm this crap, figured what that was about.
Still thinking, I need to monitor to see if 12v is coming though to the cas pickup
I was wondering if portc was that pickup, so is the other pickup another port to check for?
though from pin 11 on the cpu to the db37 connector cas pin I get 0 so that is good
EDIT: nvm this crap, figured what that was about.
Still thinking, I need to monitor to see if 12v is coming though to the cas pickup
I was wondering if portc was that pickup, so is the other pickup another port to check for?
Last edited by Techsalvager; 08-02-2010 at 09:45 PM.
#52
Boost Pope
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You should have a ~470 ohm resistor from the striped side of D5 to a source of +12 (the striped side of D9 is an acceptable source), and that's it. If you measure from the striped side of D5 to pin 24 of the DB-37, you should read ~330 ohms (the value of R10.) And from the striped side of D9 to pin 28 of the DB37 (the +12 source), you should see ~470 ohms. Take these measurements twice, with the probes in both orientations, to be sure you're not measuring through a semiconductor or some such BS.
though from pin 11 on the cpu to the db37 connector cas pin I get 0 so that is good
EDIT: nvm this crap, figured what that was about.
I was wondering if portc was that pickup, so is the other pickup another port to check for?
Portc is a combination of five different pins at the CPU, one of which happens to be the pin that we use for CMP. In CS-speak, it can be thought of as a register. In order for it to make sense, you have to decode the decimal value in the log to binary, and then align the bit positions in the resultant binary number with the pins of the CPU that they correspond to.
On the "C" port, you are seeing pins 7 through 11, where pin 7 is PTC0 and pin 11 is PTC4.
During the binary-to-decimal conversion, PTC0 has a value of 1, PTC1 has a value of 2, PTC2 has a value of 4, PTC3 has a value of 8, and PTC4 has a value of 16.
So, for most of the log, portc is equal to 3. Decoded out, that's 0000011, where PTC0 is all the way on the right (least significant bit) and PTC7 (which doesn't exist) is all the way on the left (most significant bit.) So, PTC0 and PTC1 were high, and all others low for most of the log.
(For beginners, binary is most easily read from right to left, FYI. You start at a value of 1, and then double the value for each successive bit to the left. If there's a 1 in that bit position, you add its value to the total.)
Occasionally, while you were cranking, portc changed from 3 to 19. Decoded again that's 00010011, telling me that PTC0, PTC1 and PTC4 were high, and all others low. Since PTC4 is pin 11 of the CPU, and that's where CMP is connected, we know that the CPU was seeing activity from CMP.
Sadly, the primary trigger connects to the IRQ pin, which is not part of a register that we can see directly, however since RPM never deviated from 0 even a little bit, we can surmise that the CPU saw no activity on that pin.
Have I mentioned yet that you're nuts for building a 1.01 board?
#55
Where are you measuring, exactly?
You should have a ~470 ohm resistor from the striped side of D5 to a source of +12 (the striped side of D9 is an acceptable source), and that's it. If you measure from the striped side of D5 to pin 24 of the DB-37, you should read ~330 ohms (the value of R10.) And from the striped side of D9 to pin 28 of the DB37 (the +12 source), you should see ~470 ohms. Take these measurements twice, with the probes in both orientations, to be sure you're not measuring through a semiconductor or some such BS.
You should have a ~470 ohm resistor from the striped side of D5 to a source of +12 (the striped side of D9 is an acceptable source), and that's it. If you measure from the striped side of D5 to pin 24 of the DB-37, you should read ~330 ohms (the value of R10.) And from the striped side of D9 to pin 28 of the DB37 (the +12 source), you should see ~470 ohms. Take these measurements twice, with the probes in both orientations, to be sure you're not measuring through a semiconductor or some such BS.
It's more complicated than that.
Portc is a combination of five different pins at the CPU, one of which happens to be the pin that we use for CMP. In CS-speak, it can be thought of as a register. In order for it to make sense, you have to decode the decimal value in the log to binary, and then align the bit positions in the resultant binary number with the pins of the CPU that they correspond to.
On the "C" port, you are seeing pins 7 through 11, where pin 7 is PTC0 and pin 11 is PTC4.
During the binary-to-decimal conversion, PTC0 has a value of 1, PTC1 has a value of 2, PTC2 has a value of 4, PTC3 has a value of 8, and PTC4 has a value of 16.
So, for most of the log, portc is equal to 3. Decoded out, that's 0000011, where PTC0 is all the way on the right (least significant bit) and PTC7 (which doesn't exist) is all the way on the left (most significant bit.) So, PTC0 and PTC1 were high, and all others low for most of the log.
(For beginners, binary is most easily read from right to left, FYI. You start at a value of 1, and then double the value for each successive bit to the left. If there's a 1 in that bit position, you add its value to the total.)
Occasionally, while you were cranking, portc changed from 3 to 19. Decoded again that's 00010011, telling me that PTC0, PTC1 and PTC4 were high, and all others low. Since PTC4 is pin 11 of the CPU, and that's where CMP is connected, we know that the CPU was seeing activity from CMP.
Sadly, the primary trigger connects to the IRQ pin, which is not part of a register that we can see directly, however since RPM never deviated from 0 even a little bit, we can surmise that the CPU saw no activity on that pin.
Have I mentioned yet that you're nuts for building a 1.01 board?
Portc is a combination of five different pins at the CPU, one of which happens to be the pin that we use for CMP. In CS-speak, it can be thought of as a register. In order for it to make sense, you have to decode the decimal value in the log to binary, and then align the bit positions in the resultant binary number with the pins of the CPU that they correspond to.
On the "C" port, you are seeing pins 7 through 11, where pin 7 is PTC0 and pin 11 is PTC4.
During the binary-to-decimal conversion, PTC0 has a value of 1, PTC1 has a value of 2, PTC2 has a value of 4, PTC3 has a value of 8, and PTC4 has a value of 16.
So, for most of the log, portc is equal to 3. Decoded out, that's 0000011, where PTC0 is all the way on the right (least significant bit) and PTC7 (which doesn't exist) is all the way on the left (most significant bit.) So, PTC0 and PTC1 were high, and all others low for most of the log.
(For beginners, binary is most easily read from right to left, FYI. You start at a value of 1, and then double the value for each successive bit to the left. If there's a 1 in that bit position, you add its value to the total.)
Occasionally, while you were cranking, portc changed from 3 to 19. Decoded again that's 00010011, telling me that PTC0, PTC1 and PTC4 were high, and all others low. Since PTC4 is pin 11 of the CPU, and that's where CMP is connected, we know that the CPU was seeing activity from CMP.
Sadly, the primary trigger connects to the IRQ pin, which is not part of a register that we can see directly, however since RPM never deviated from 0 even a little bit, we can surmise that the CPU saw no activity on that pin.
Have I mentioned yet that you're nuts for building a 1.01 board?
Also I liked that spark output trigger setup you came up with to allow flashing the ecu with the igintor and coilpacks still in, think I will add that in later, first thing is first, get rpms\cas to pickup then fuel circuit.
#56
Boost Pope
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Actually, I screwed that up. You don't want to go to D5, you need to go to the side of R10 which is away from D5:
If you can't figure out which side is which, it'll be the side of R10 which reads 0 ohms (full continuity) to pin 24.
The way I drew it originally wouldn't work because the CAS would be trying to pull through R10.
If you can't figure out which side is which, it'll be the side of R10 which reads 0 ohms (full continuity) to pin 24.
The way I drew it originally wouldn't work because the CAS would be trying to pull through R10.
#58
Boost Pope
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Background:
The CAS provides what are known as open-collector outputs. That means that it doe not output a voltage, rather it provides a closure to ground when active. It either conducts to ground or it does not.
So, we take +12 (such as from D9) and run it to the point indicated above on R10. When the CAS is inactive (not conducting to ground) the current flows through the 470 ohm resistor, then through R10, and then through the LED section of U4. This causes U4 to turn on.
When the CAS becomes active, it provides a short-circuit path to ground for the current flowing through the 470 ohm resistor. Instead of flowing through R10 and then U4, it takes the much easier path through the CAS. With no current flowing through U4, it turns off.
Coincidentally, this is how things are working on the CPU side of U4 as well. When the CAS is conducting to ground (and thus, U4 is "off"), current flows from Vcc, through R11, and shows up at IRQ-1, which is pin 14 of the CPU. It sees this as a logic "high."
When the CAS becomes inactive (no longer conducting) thus causing U4 to turn on, the current flowing through R11 instead takes a short-circuit path to ground through the transistor portion of U4, which causes the voltage at IRQ-1 to fall to zero (well, not exactly zero, but it's a long story involving semiconductors and forward voltage drop, and suffice to say that it's close enough.)
C11 can be ignored in this simplistic explanation.
The CAS provides what are known as open-collector outputs. That means that it doe not output a voltage, rather it provides a closure to ground when active. It either conducts to ground or it does not.
So, we take +12 (such as from D9) and run it to the point indicated above on R10. When the CAS is inactive (not conducting to ground) the current flows through the 470 ohm resistor, then through R10, and then through the LED section of U4. This causes U4 to turn on.
When the CAS becomes active, it provides a short-circuit path to ground for the current flowing through the 470 ohm resistor. Instead of flowing through R10 and then U4, it takes the much easier path through the CAS. With no current flowing through U4, it turns off.
Coincidentally, this is how things are working on the CPU side of U4 as well. When the CAS is conducting to ground (and thus, U4 is "off"), current flows from Vcc, through R11, and shows up at IRQ-1, which is pin 14 of the CPU. It sees this as a logic "high."
When the CAS becomes inactive (no longer conducting) thus causing U4 to turn on, the current flowing through R11 instead takes a short-circuit path to ground through the transistor portion of U4, which causes the voltage at IRQ-1 to fall to zero (well, not exactly zero, but it's a long story involving semiconductors and forward voltage drop, and suffice to say that it's close enough.)
C11 can be ignored in this simplistic explanation.