Resistance on Q14
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Resistance on Q14
I'm on step 72 of a MS1 V3 build, and I'm getting zero resistance between two of the pins on Q14. I removed the transistor and checked the contacts, and still zero ohms. I cleaned the contacts, and still zero. Is it supposed to be that way, or did I bridge them somehow and I haven't unbridged them yet?
#3
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I am following your writeup, but I like the step-by-step soldering instruction given in the megamanual.
Believe me, I'd be clueless without this forum and some of the instruction on DIY.
Believe me, I'd be clueless without this forum and some of the instruction on DIY.
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Depends on which two pins you're measuring.
If R37 is installed (Step 71a) then you will measure zero ohms between pin 2 (base) and pin 1 (emitter to ground). Actually, the measurement you'd get if you owned a super-expensive milliohmmeter would be 0.05 ohms (five one-hundredths of an ohm) but we're talking test equipment in the $500 and up range. A normal meter will read that as zero.
What prompted you to make this measurement?
If R37 is installed (Step 71a) then you will measure zero ohms between pin 2 (base) and pin 1 (emitter to ground). Actually, the measurement you'd get if you owned a super-expensive milliohmmeter would be 0.05 ohms (five one-hundredths of an ohm) but we're talking test equipment in the $500 and up range. A normal meter will read that as zero.
What prompted you to make this measurement?
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Depends on which two pins you're measuring.
If R37 is installed (Step 71a) then you will measure zero ohms between pin 2 (base) and pin 1 (emitter to ground). Actually, the measurement you'd get if you owned a super-expensive milliohmmeter would be 0.05 ohms (five one-hundredths of an ohm) but we're talking test equipment in the $500 and up range. A normal meter will read that as zero.
What prompted you to make this measurement?
If R37 is installed (Step 71a) then you will measure zero ohms between pin 2 (base) and pin 1 (emitter to ground). Actually, the measurement you'd get if you owned a super-expensive milliohmmeter would be 0.05 ohms (five one-hundredths of an ohm) but we're talking test equipment in the $500 and up range. A normal meter will read that as zero.
What prompted you to make this measurement?
I was taking the measurement to see if I'd bridged a couple pins with solder. When the measurement came up zero, I pulled the transistor out to check the front side, and everything looked clean so I came here to ask questions. One of the steps previously had mentioned checking a transistor with a multimeter for just such a problem, but never mentioned that it might be zero on purpose.
I have a VERY cheap multimeter. Got it for free in one of my classes. Does everything I need a multimeter to do.
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Not really, just how to read a schematic.
Taken individually, all of the various circuits that make up the MS are actually pretty simple. Let's look specifically at one of the injector drivers. I've deleted from this clip some parts of the circuit that are not relevant.
So here we have the whole primary INJ1 driver, minus the flyback circuit. The injector has +12 applied to one side, and the other comes into this diagram at the point marked INJ-1. The green line illustrates the primary current path.
To make the injector turn on, the CPU applies a logic high to pin 2 of the IXDI404, which in turn applies +5 to the output on pin 7. This causes the gate of FET Q1 to turn on, permitting current to flow through it and through R37 to ground. But what's with R37? It's only a .05 ohm resistor, so it can't be doing much. More on that later.
Take a look at the placement of Q14. Its collector-emitter junction (the primary flow path through a transistor) goes from the gate of Q1 to ground. If it were ever to turn on, it would short the voltage being supplied by U4 to ground, turning off Q1, and preventing the injector from working. On the other hand, the base of Q14 is more or less directly connected to ground (through the aforementioned .05 ohm resistor) so it can't possibly turn on, right?
Enter ohm's law. E = I x R. Voltage = Current x Resistance.
As current flows through a resistor, there is a voltage "drop" across the resistor. In other words, a difference in the voltage on one end of the resistor vs. the other. Now the bottom end of R37 is connected quite solidly to ground, so the voltage on that end is always going to be zero. But as current flows through it, a very small voltage will begin to appear on the top end of that resistor, which is where the base of Q14 is actually connected.
Say that 2 amps is passing through R37. That's about the amount of current that will flow through a pair of high impedance injectors with 13 volts supplied to them. Two amps through .05 ohms will cause a voltage difference of about .1 volt (100 millivolts). Not much.
Figure a pair of low impedance (3 ohm) injectors- 8.7 amps. That much current flowing through R37 will generate .435 volts. More, still not a ton.
Figure 14 amps. That'll produce about .7 volts.
Time to look at the datasheet for the 2N3904, aka Q14. The Base-Emitter saturation point (basically the amount of voltage required to turn the transistor on) varies slightly with temperature and collector current, but it's in the range of about .7 volts. Aha!
With 2 amps passing through R37, there's not enough voltage to turn Q14 on. Ditto with 8.7 amps. But at 14 amps, the voltage on the top of R37 becomes high enough to turn on Q14! When this happens, Q14 will cause Q1 to turn off, thereby stopping the flow of current through the injector circuit.
It's a safety feature. Q1 can only handle so much current without being destroyed, so if some douchebag connects four lo-z injectors to a single injector channel, this prevents the circuit from blowing up. Technically, none of it is necessary. You can eliminate Q14 and R37, with a jumper across the R37 holes (and do the same on INJ2) and it'll work just fine so long as you don't overload it. In fact, I didn't install them on mine to free up a pair of spaces on the heatsink where some of my mods live.
Taken individually, all of the various circuits that make up the MS are actually pretty simple. Let's look specifically at one of the injector drivers. I've deleted from this clip some parts of the circuit that are not relevant.
So here we have the whole primary INJ1 driver, minus the flyback circuit. The injector has +12 applied to one side, and the other comes into this diagram at the point marked INJ-1. The green line illustrates the primary current path.
To make the injector turn on, the CPU applies a logic high to pin 2 of the IXDI404, which in turn applies +5 to the output on pin 7. This causes the gate of FET Q1 to turn on, permitting current to flow through it and through R37 to ground. But what's with R37? It's only a .05 ohm resistor, so it can't be doing much. More on that later.
Take a look at the placement of Q14. Its collector-emitter junction (the primary flow path through a transistor) goes from the gate of Q1 to ground. If it were ever to turn on, it would short the voltage being supplied by U4 to ground, turning off Q1, and preventing the injector from working. On the other hand, the base of Q14 is more or less directly connected to ground (through the aforementioned .05 ohm resistor) so it can't possibly turn on, right?
Enter ohm's law. E = I x R. Voltage = Current x Resistance.
As current flows through a resistor, there is a voltage "drop" across the resistor. In other words, a difference in the voltage on one end of the resistor vs. the other. Now the bottom end of R37 is connected quite solidly to ground, so the voltage on that end is always going to be zero. But as current flows through it, a very small voltage will begin to appear on the top end of that resistor, which is where the base of Q14 is actually connected.
Say that 2 amps is passing through R37. That's about the amount of current that will flow through a pair of high impedance injectors with 13 volts supplied to them. Two amps through .05 ohms will cause a voltage difference of about .1 volt (100 millivolts). Not much.
Figure a pair of low impedance (3 ohm) injectors- 8.7 amps. That much current flowing through R37 will generate .435 volts. More, still not a ton.
Figure 14 amps. That'll produce about .7 volts.
Time to look at the datasheet for the 2N3904, aka Q14. The Base-Emitter saturation point (basically the amount of voltage required to turn the transistor on) varies slightly with temperature and collector current, but it's in the range of about .7 volts. Aha!
With 2 amps passing through R37, there's not enough voltage to turn Q14 on. Ditto with 8.7 amps. But at 14 amps, the voltage on the top of R37 becomes high enough to turn on Q14! When this happens, Q14 will cause Q1 to turn off, thereby stopping the flow of current through the injector circuit.
It's a safety feature. Q1 can only handle so much current without being destroyed, so if some douchebag connects four lo-z injectors to a single injector channel, this prevents the circuit from blowing up. Technically, none of it is necessary. You can eliminate Q14 and R37, with a jumper across the R37 holes (and do the same on INJ2) and it'll work just fine so long as you don't overload it. In fact, I didn't install them on mine to free up a pair of spaces on the heatsink where some of my mods live.
Last edited by Joe Perez; 07-10-2008 at 12:45 AM. Reason: Whoops- Q14, not Q4.
#7
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Wow, that is pretty cool. I never would have had any idea what any of that meant on my own but I actually understood what you were saying all the way through.
so I could free up a space on my heatsink by getting rid of R37 but then I just have to make sure I stick with high-z injectors connected correctly? Sounds pretty simple. I don't know if I'll need any more heat sink locations but I'll keep it tucked away in the back of my mind.
Thanks!
so I could free up a space on my heatsink by getting rid of R37 but then I just have to make sure I stick with high-z injectors connected correctly? Sounds pretty simple. I don't know if I'll need any more heat sink locations but I'll keep it tucked away in the back of my mind.
Thanks!
#8
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Without the PWM circuit and VR circuit, the board is bare of almost any components. removing the PWM circuit frees two spots on the heatsink. leaving you with 3 spots open for whatever you need.
#9
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Ok, new question, but still regarding resistance.
I'm modifying my board as in the build your megasquirt thread, but my pullup kit didn't come with any 470ohm resistors. Am I supposed to get this somewhere else, was it missing from my kit, or am I probably just not seeing it in some other portion of the kit?
I'm modifying my board as in the build your megasquirt thread, but my pullup kit didn't come with any 470ohm resistors. Am I supposed to get this somewhere else, was it missing from my kit, or am I probably just not seeing it in some other portion of the kit?
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is it possible that there is a typo on the bag? the part number is 470QBK-ND but the description says "Resistor, 270ohm,5%,.5w,axial"
I can't see the color bands on the resistors in the thread because they're blocked by the shrink tube. Google, here I come!
I can't see the color bands on the resistors in the thread because they're blocked by the shrink tube. Google, here I come!
#13
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How about this capacitor for running high-res code? It's not supposed to be included with the pullup kit, and the little baggie of .1 microfarad caps is empty. Do I really have to order a $1 capacitor and have it shipped here, or is there something else I can do?
#16
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Alright. I'll take a look at radioshack. I don't have high hopes though because the one around here doesn't have a lot of little components like that. They moved more to things like cell phones, ipods and televisions. They still have the little drawers in the back with the switches and buttons and such things, so I'll give radioshack a try tonight.
#17
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Eliminating these parts does not have any effect upon PWM or Lo-Z injector drive. They are an overcurrent protect, and nothing more.
Lo-Z injector drive is made possible by the pair of circuits marked as "Note 3" on the schametic, consisting of Q9/Q10 (INJ1) and Q12/Q13 (INJ2) plus a couple of resistors. In particular, the removal of Q9 and Q12 will free up an additional pair of heatsink spaces, should they be needed.
A caveat: since you have an MS1, you are encouraged to run the Hi-Rez code. This software does not directly support Lo-Z (PWM) injector drive regardless of how you have your hardware set up. Despair not, however. The company that brought us the JimStim also builds an outboard PWM injector controller: http://www.jbperf.com/p&h_board/index.html This board will allow you to run Lo-Z injectors regardless of what software build or internal MS hardware configuration you have. In fact, if I had Lo-Z injecotrs, even on an MS2, I'd probably still run that board since it's a much cleaner design with isolated grounds for the injectors- less noise inside the MS case.
It's a standard stock item that's still part of the current store layout configuration. You'll find it in the pull-out drawer section in the back. P/N 272-1069 or 272-1053.
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