My 1997 Mercury outboard motor is acting up. Really strange, it will not generate any spark on any cylinder, then it will on/off. When it works it seems to work for several days to even months. When it doesn't work, I "mess" with it for a while then it suddenly starts working again. When it doesn't work it is always in the slip connected to my home made shore power. It has always started on the lake (not connected to shore power). My current theory, based on a forum post I found googling, is that the battery maintainer "may" be to blame. The last time it didn't work, I simply unplugged the battery maintainer and it started right up. However I'm suspicious because looking at the electrical diagram I don't see how the battery voltage is even connected to the ignition circuit. There are separate magneto windings for the alternator and the ignition module (switch box). The ignition module is completely self powered through the magneto windings. My only guess is that an over voltage on the battery maintainer might generate a bad electric field in the alternator windings that affects the magneto windings; the windings are in the same physical structure.
For troubleshooting the ignition module, Mercury specifies all the voltages as peak voltages. It took me a while to understand this, but it makes perfect sense to me now. The whole ignition module is based on pulsed voltages. The waveforms probably do not resemble sine waves in any way. Your basic meter assumes a 50-60Hz sine wave. Even true RMS meters assume a sine wave. These meters will register a voltage for pulsed A/C but there is no accurate way to relate that reading to the actual voltage waveform. Ignore posts that talk about multiplying by 1.4, that also assumes a sine wave. After I thought about it for a while I realized that the only mathematically correct way to specify the magnitude of a pulsed A/C waveform using only one value is to specify the peak voltage. I think that all capacitor discharge ignitions (CDI) share this measurement problem.
Assuming that my ignition problem is not solved, I want to be ready the next time it comes back. I started thinking about how to build a peak adapter, reading about others that have tried (google again), and thinking about the features of the commercial modules. All of the commercial modules I've seen are self powered with no range settings. This implies that there are only passive components. There can not be any type of amplifier (operation amp) because any type of amp will require a primary voltage source greater than the output of the module; up to 400VDC. The only way I can imagine to build a completely passive peak converter is to use a simple rectifier and capacitor. This turns out to be rather simple because of the extremely large input impedance of modern A/C volt meters, typically in the 10s of megaohms. Just build a high voltage (e.g. 600V) full wave bridge, add a high voltage capacitor, and throw in a series resistor or two for safety. Probe jacks make it even better.
I'm guessing at a lot of this because I do not know the waveform shape or frequency. If someone wants to loan me an oscilloscope, I'll measure my Mercury engine and then build a spice model for this. It's been 10+ years since I've used spice but this should be a very easy circuit to simulate.
Diode Voltage Drop
Diodes have a forward voltage drop, typically about .6V but varies based on diode type and forward current. This voltage drop limits the lowest voltage that can be measured. However, I think that voltages down to just less than 1V can be measured if the forward voltage drop of the diodes is known and added to the reading.
Diode Switching Speed
Diodes also have a switching speed that might come into play if the voltage pulses are relatively short. The right way to determine the required switching speed is to hook an oscillator scope up to the signals and measure the pulse width.
Input Current Limit
Good capacitors tend to have extremely low impedance; this will look like a short to the device under test when first connecting the adapter. Since the pulses are repetitive, a series resistor can be used to limit the input current without changing the final reading. Once the capacitor voltage approaches the pulse voltage, the voltage drop at the resistor will approach zero.
Good capacitors also tend to hold a charge and can deliver that charge in a very short period (again low impedance). Once the adapter is disconnected from the meter there will be two safety problems. First shorting the probes could produce a very large current pulse. Another series resistor will limit this current to something relatively safe. This resistor will not affect the meter reading because the A/C volt meters have a very small input current. Second the adapter will hold a relatively dangerous voltage. A shunt resistor across the capacitor will guarantee that the adapter discharges in a reasonable time. The value of this resistor must be carefully selected by considering the pulse frequency and pulse width to keep from partially discharging or limiting the full charge of the capacitor between pulses. However it must also be sufficiently low to discharge the capacitor to a safe voltage in a short time.
Accommodating 200V DC Meters
Some meters do not measure DC over 200V. Since the peak adapter will generate up to 400VDC, some may need a voltage divider. The resulting meter reading will need to be multiplied by the divider ratio. This is an area of active consideration for me. I'm torn between including the divider at the input vs. the output. In the input it will lower the voltage requirements for both the diodes and the capacitor but it will increase the time required to charge the capacitor. In the output, the divider its self will suffice for both the safety current limit and safety shunt but will not help with the voltage requirements. I'm leaning toward the output side. I really do not know the frequency or pulse shape of the signal being measured. This will make it hard to select a divider that does not introduce an RC time constant too large for the signal being measured.