[rusEfi] Clio Mk2 172 Ph1 Race Car

[kb1gtt]

The 1N4007 is probably fine. If not it will break and then you'll see some odd behaviors with the VVT, but should not cause any catastrophic issues. If the diode fails, you would simply replace the diode and it would start working properly.

I'm ignorant about the nitty grittys of VVT. I believe it's about 1A continuous and the 1N4007 is typically rated for 1A continuous. The diode should not be on 100% of the time, so you should be well under it's ratings. Making a better prediction is kind of complicated, as there are things like recovery time, and other factors to consider when predicting how hot a diode will get. In this case because a failed diode is not expected to cause any permanent damage, just short term performance issues, I say just install it and see how hot it gets. It will probably be cold enough, and will probably work just fine.

[jalava]

This VVT solenoid is of on/off type with current draw of 1.4A when active. I see that low channels are named with 1A so I'm bit uncertain if I really can draw that much current through low driver.

Slowness is of course relative, but VVT cam action itself is quite slow as when VVT solenoid activates it takes a while for oil pressure to enter camshaft and turn the dephaser to new timing.

[kb1gtt]

The key with the Frankenso's ability to drive amps is it's ability to stay cool. If it overheats, there is a thermal shutdown which will turn it off. I laid lots of copper to help get the heat away, so hopefully it will typically stay cold enough for most applications. If it gets to hot, then adding a heat sink will be the fist line of defense. I expect the traces can easily handle 5A, and with some good heat sinking should be able to handle 10A. Most of the heat generated at the injector drivers is not from the constant amps, but from the inductive energy which is absorbed by the MOSFET and turned into waste heat. You should be fine with the 1.4A by a large margin, especially with only being activated once every couple seconds or less.

So far no one has had a thermal shutdown, so the thermal theories seem to be holding their ground. The 1A comment is because I feel significantly confident it will work at 1A, it is unknown how much above 1A you can go before you start to hit thermal problems.

[jalava]

Moving to ignition wiring, we are using Bosch 0 227 100 203 I'm 3.2 ignition amplifier's two channels for driving our Sagem wasted spark coil pack.

Connecting the amp gnd to power earth (EFI GND connects to this), and 5V high drivers to base pins of transistors, do I need to connect gnd_output to gnd, or are they internally connected?

[kb1gtt]

Hmmm, this module wants 20mA drive, and 33 ohms drive to GND. I think that instead of using a wire to make the hi/lo driver to 5V, you want that wire to be a 150ohm resistor. So if we assume your using P604, you want 150 ohms for the jumper that selects your drive voltage. Then you want a R601 or R604 to be a 33ohm resistor. If you do that, it appears it would work well.

[kb1gtt]

Oh, about the GND, I would think you want that wire connected direct to the battery, or at least a solid connection to the frame. That wire does not need to get back to Frankenso. Your Frankenso ground should be fine. Keep in mind your ignition and dwell currents are going to create the most voltage drop. The 20mA drive from the Frankenso is small compared to other currents.

[jalava]

Hmmm, this module wants 20mA drive, and 33 ohms drive to GND. I think that instead of using a wire to make the hi/lo driver to 5V, you want that wire to be a 150ohm resistor. So if we assume your using P608, you want 150 ohms for the jumper that selects your drive voltage. Then you want a R601 or R604 to be a 33ohm resistor. If you do that, it appears it would work well.

I presume you mean P604 (v0.4) if using HL 1/2 ad ignition channels?

[kb1gtt]

Yes you are correct, I meant P604, not P608. I just updated my prior post to prevent confusion of those who read this thread at a later date. Thanks for the note.

[jalava]

If I understand the datasheet correctly, it requires minimum of 10mA of Ib and it should be in range of 10-23mA.

If I put 150 Ohm resistor to P604, and keep the 100 Ohm resistors like they are now, I'm getting 20mA of Ib drive (5V / 250 Ohm = 20mA), which should be enough. So based on this soldering just the 150 Ohm resistor to the P604 should give correct current to the ignition amplifier. Desoldering and swapping the R604 & R601 seems tedious with my tools as I don't want to burn the led next to it.

Going to test the setup tomorrow to see if it responds as expected.

I'm bit worried about using this for wasted spark though as it is rated for 8000rpm with one cylinder per spark plug. Now as we go with wasted spark setup, we are basically triggering those two transistors inside it twice as often.

[kb1gtt]

It appears they want 33 ohms pulling to GND when you are off. I think you are neglecting this. Leaving the 100 ohms may be OK, but you may also experience delayed off times. Or under certain conditions you could get a run away condition. Keep in mind that the channel where current flows in the piece of glass, in the module is acting like a resistor. You have high voltage and current at the top, and 0V at the bottom, with a gradient in between. Some where in the middle of this voltage drop gradient, is your activation point. Your external circuit is driving the to about 1V to turn this on, which requires about 20mA. If you voltage gradient provides say 5mA, when you turn off you'll turn off slowly. If the voltage gradient can provide 15mA, then you will not shut off at all. The common solution is to include a pull down resistor. Basically to keep the module on, you need 10mA, and to get the 1V, with a 33 ohm parallel resistor to GND, you would need 30mA for the resistor, as well as 10mA for the transistor. So you would need a total of 40mA to be provided by the gradient, not just the 10mA. These pull down resistors can be really important. Even more important with MOSFET's, as those are voltage driven, not current driven.

Keep in mind the ignition module is very similar to a diode. At around 10mA to 20mA, it probably drops around 1V. So you probably have more like 4V / 250 ohms = 16mA from your drive.

Instead of removing the resistor, and replacing with a 33 ohm, you could add a 49 ohm resistor on top of the existing 100 ohm resistor. The combined resistance would equal about 33 ohms.