I'm curious, what voltage do you measure across the resistor when the bulbs are burning too bright? If you're getting 29 volts across the bulbs, I'd expect next to nothing across the resistor. Perhaps the resistor lowers its resistance as it heats up.

The bulbs are in the 25 volt part of the circuit probably as a cost savings.
Top Score Bulbs (resized).jpg
To get the bulbs into the 6 volt circuit they'd have to replicate a bunch of stuff on the 6 volt side including A, FS and Motor 1C switches and the Conservative/Liberal adjustment jones plug. This way they just need to add a resistor.

If you decide to replace the resistor be sure to get one that has the right power rating. I think that 85 ohm resistor needs to be at least a 10 watt resistor.

Another option would be to use a 28 volt bulb like an 1819.

/Mark

I'm just speculating about the cost savings, but it seems plausible. Compare what they did with the alternative:
Top Score Bulbs 2 (resized).jpg
The top circuit is what was built. Below it are the two circuits that would be required to get rid of the big resistor, and to use the bulbs on the 6 volt supply instead of the 25 volt supply. The resistor in the upper left red box eliminates the need for all of the redundant stuff in the lower right red box.

I would take PBR's advice on which resistor to use. 47 bulbs are lower power, and therefore higher resistance than 44 bulbs. To get the appropriate voltage across the bulb you're using you'd need to size the resistor appropriately. The game would have been designed with 44 bulbs so if you switch to 47s the resistor too should be changed ideally. I'm sure 47 bulbs would work with the 75 ohm resistor for a while but they'll burn hotter than intended and burn out sooner than they should.

But to answer the original question, it sounds like the 85 ohm resistor is drifting in value so replacing it should do the trick as @currieddog suggested.

Hi pinballman3 +
I am not good in using meters measuring Volts, Amperes, Ohms. I do have a cheap, simple meter - I measured Ohms on a 44 bulb and on a 47 bulb - same reading. I do not know if it is 1.6 Ohms or 16 Ohms or 160 Ohms or what. currieddog wrote in his post-7 about PBR asking about using an 44 or 47 bulb meaning 75 or 150 Ohms resistor. I searched the net - got some german site advertizing an variable resistor (up to 200 Ohms) 25 Watt --- see the JPG. The prize 6.87 Euro roughly is around 8 dollars - does Radio Shack sells such resistors of "many Watts" ?

I am interested to learn about the secrets of such wiring --- how many Ohms, how many Watts, does it matter if we use an 44 or 47 bulb - can we use an 455 flashing bulb - how does the resistor cut - does it cut Amperes and the bulb does not care about Volts ? I know close to nothing about electronics. pinballman3 : Are You interested in these questions and do ask in here https://pinside.com/pinball/forum/forum/technical-help ? Greetings Rolf

0Variable-Resistor (resized).jpg

Hi currieddog
Yes, I read MarkG 's post-8, second paragraph. I agree with his guesses but I would like to have facts and (simplified) theory for a dumb EM-man (me). It is pinballman3 's topic - so asked him to take the problem to the experts (tech-help). If he does not want to - I will do.
You wrote in post-7 "have Mayfair and Central Park" - and posted a snippet from probably "Mayfair". I do not have the schematics to Mayfair but I do have the schematics to Central Park - I looked it up - an 35 Ohm 10 Watt resistor after the bulb. I do have a pinside snippet of "Melody" schematics --- 75 Ohm 10 Watt resistor. "Bronco" also has 75 Ohm, 10 Watt resistors. "Dancing Lady also has "75 Ohm, 10 Watt. "Ice Revue" has 35 Ohm, 10 Watt. "Orbit" has 85 Ohm, 10 Watt. "Jet Spin" has 75 Ohm, 10 Watt. "King Rock" has 85 Ohm, 10 Watt. (Now I stopped looking through different Gottlieb-Schematics)

The examples agree on "10 Watt" (Top Score does not tell Watts) - many times 75 Ohms are mentioned -- but sometimes 85 Ohms - but sometimes 35 Ohms. How comes ? Greetings Rolf

I made the same mistake. Then I was reminded recently that you can't reliably measure the resistance of an incandescent bulb when it's cold because the resistance of the bulb filament increases as the bulb heats up. So for bulbs it's more reliable to use their ratings to figure out their operating resistance.

A 44 bulb is rated at .25 amps when running at 6.3 volts while a 47 bulb is rated at .15 amps.

Using Ohm's law (V = I * R or, voltage = current * resistance) the resistance of the bulbs at temperature is:
- about 25 ohms for a 44 bulb (6.3 volts = .25 amps * 25 ohms)
- about 42 ohms for a 47 bulb (6.3 volts = .15 amps * 42 ohms)

The purpose of the large resistor in series with the bulb is to drop the voltage from the 25-30 volt relay supply voltage down to the 6.3 volts that the bulb is designed for. How much voltage is dropped across the large resistor depends on the resistor value and on the current that passes through it (V = I * R again).

An 85 ohm resistor would have a voltage drop across it of 21.25 volts when .25 amps pass through it. The same resistor would drop only 12.75 volts across it when .15 amps pass through it, leaving the 47 bulb with more than double the voltage it was designed for.

So while 75-85 ohms is about right to drop the voltage to 6.3 volts for a #44 bulb that draws .25 amps, a larger resistance of 140-150 ohms or so is required to drop the voltage to 6.3 volts for a #47 bulb that draws .15 amps. (LED fans beware: an even larger resistance would be needed if using an LED bulb in this circuit.)

Note that none of these figures is precise. That's not an oversight. Supply voltages vary depending on the transformer and on the voltage supplied by your utility. Resistances can vary 5-10% unless otherwise specified. And I'm sure there's variability in the bulbs too due to manufacturing and other differences. The good news is that 5-10% is usually good enough unless you're dealing with precision equipment (test equipment, audio, etc.).

Hi MarkG
thanks for the 44-bulb and 47-bulb ratings and the insights ... I did some search in the net and have found https://en.wikipedia.org/wiki/Voltage_divider , on the bottom of the page are links to calculators - in here https://learn.sparkfun.com/tutorials/voltage-dividers/ just below the "calculator" comes "simplification" ... "if R2 and R1 are equal then the output voltage is half of the input". AND THIS is the theory behind ( I havent realized until now) when I advertize "make an SteveFury-Testlight for an Gottlieb or Williams pin (24VAC) using two 12Volt car bulbs in series (one behind the other)" - the two car bulbs are two equal resistors in an Voltage-Divider. Greetings Rolf

Interesting topic. Thanks guys for the technical overview.

Another way to look at it is that two resistors or bulbs in series (or, wired end to end) will have the same current flowing through them. Given that, they divide the total voltage between them proportional to their relative resistances. The larger of the two resistors will have a larger voltage drop across it. Equal resistors will divide the voltage equally. If one resistor is replaced with a wire (no resistance) the other resistor will have all of the voltage across it.

Hi pinballman3
I did learn some "new to me" stuff in Your topic - I learned so much that I do not have to ask in tech-generic. I believe that Your posts -16/-17 are some kind of "final words - how You fixed the problem". Greetings Rolf