MERC - V. REG. OUTPUT

[VARISTORS]

THE V. REG. OUTPUT IS 5 V from my MERC 40 HP outboard. I was expecting 14-16 V. WHY SO LOW?
THE INPUT TO my BATTERY ISOLATOR IS 5 V. THE
OUTPUT IS 13.6 V TO EACH BATTERY. HOW DOES THE VOLTAGE INCREASE ? WHAT HAPPENED TO THE 0.7 V DROP?

[rstofer]

I would put it back in the factory configuration and then measure the voltage with the motor running.<p>If I had to guess, I would guess the alternator has run without a battery and the alternator diodes are toast. But that is just a guess.

[Chris Smith]

Run everything hooked up and check the voltages, both in the AC and DC mode.

[Robert Reed]

Jack
Need more info. Was motor running at time of measurment?
Terminal voltage on your batterys should be 12.6 VDC
13.6 VDC indicates they are being charged. The charging wave form may have a low duty cycle thereby producing a lo DC reading on your volt meter. It may be performing its job ok. Is there a problem or were you just curious?

[dyarker]

The output of a voltage regulator feeds the alternator field winding; it will be lower when the battery is charged. Since you probably wouldn't have posted the question unless there was a problem ... I'll repeat "more info" please.

[Robert Reed]

One other thing Jack-
Small motors usually dont have an altrenator as such. They carrie 3 or 4 coils arranged in magneto fashion, thereby producing 3 or 4 pulses of charging current for each revolution, again relating to former reference of erroneous volt meter readings

[VARISTORS]

READINGS WERE TAKEN WITH MOTOR RUNNING BETWEEN BATTERY GROUND AND ISOLATOR INPUT. NO ALTERNATOR, STATTOR OUTPUTS TO V.. REG. TO ISOLATOR. BATTERIES ARE BEING CHARGED.. GUESS ITS A CURIOSITY QUESTION -
ONLY A 5 V OUTPUT???? THEN GOING THROUGH THE ISOLATOR (PRESUMABLY JUST DIODES) THE V. JUMPS TO 13.6 V???

[rstofer]

No, the voltage doesn't jump. I would look at the measurements again. How do you know the batteries are charging? Just because they have voltage? Or are you measuring the current and direction?

[Gorgon]

Hi Jack,
To check the charging, measure the voltage over the battery poles when the motor is stopped and then when it is running. If the voltage is higher when running, the battery is charging.<p>You said that the unit in question is an 'isolator'. If it is a true galvanic isolator, you can not measure the input voltage with reference to the battery. You need to take both + and - from the same side of the isolator, either the input(regulator) or the output(battery).<p>TOK

[Robert Reed]

Jack
It still appears as a waveform/duty cycle situation.
Crude example--bursts of energy at say a duty cycle of 10% and peak voltage of 16 volts.When peak exceeds batt. term. voltage, charging current flows breifly and possibly limited by source impedance. The isolator you refer to is probably nothing more than a diode or bank of diodes (one for each pickup coil). these keep the battery from discharging back into the "Charger" between charging pulses, when its voltage is lo or nonexistant. Or when the motor is not running. The primary purpose of small outboards is to propel a craft and charging systems here are almost an after thought which are of the simplest design possible to do the job.
If you have access to a 'scope, I'm sure the presrentation would clear up any questions in your mind. This would give a clear picture of what is taking place and not the average voltage that your DMM is displaying.

[VARISTORS]

THE ISOLATOR IS DESIGNED TO KEEP THE MOTOR BAT. & HOUSE BATS. SEPARATED. IT CONSISTS OF TWO DIODES WHICH BOTH TEST GOOD (REVERSING OHM METER LEADS GIVES VERY DIFFERENT READINGS) ONE FEEDS THE MOTOR BATTERY THE OTHER THE HOUSE BATS. & V.
REG. FEEDS THEM BOTH. THE MOTOR WAS ADVERTISED AS PUTTING OUT A MAX. OF 16 A. WILL TEST FURTHER & GET BACK. THANKS

[Robert Reed]

Okay Jack
Now I get a clearer picture of the "isolator". This device is allowing two batterys to be charged simultaneously without ever allowing the weaker battery to drain the stronger one.This is probably is an after market device-installed by you or former owner?
But there is still problly a series diode involved in OEM charging circuit somewhere. Measurements you are taking are at the juntion of these two diodes (isolator and OEM).
The DMM readings would still be similar at this point (as dicussed in previous posts) as Batt. DC is prevented from back feeding thru diodes.
I am surprised to hear that charging rate is 16 amps--A pretty good output considering the size of your motor.

[haklesup]

Just a coulple of thoughts<p>Could you be trying to measure an AC voltage with a DC setting on your meter? If truly DC, AC voltage should be 0V<p>Could you be changing ground references? For instance comparing a differnetial voltage measurement at one point to a ground referenced one.<p>Try removing one of the batteries and see if it makes more sense.