An electronics question

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Greg_Lewis
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An electronics question

Post by Greg_Lewis »

I need a little electronics help — my knowledge of such things is measured in negative numbers. I’ve got a dimmer like the one in the photos below. Taking the guts out of the case, I need to replace that potentiometer with the smallest one I can get. Space is critical and I’ll be taking off the terminal block as well and soldering on the connections. I’ve seen pots on circuit boards that are extremely low profile. I don’t need to have the replacement mounted on a board; I can wire it to the existing one. The replacement will only be adjusted once so I do not need a knob; something with a small screwdriver slot will be fine.

So I have two questions:
1. What should I look for? I see scores of various ones in the Newark catalog but I have no idea of what I should get. I can see that this is a 1k ohm unit but other than that I’m clueless.
2. How can I tell the wattage rating of the existing one so I can get a proper value replacement?

Or..... do you think I could cut off the stem on the existing one without it falling apart?

Thanks!
s-l1600.jpg
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IMG_0701.JPG
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Greg Lewis, Prop.
Eyeball Engineering — Home of the dull toolbit.
Our motto: "That looks about right."
Celebrating 35 years of turning perfectly good metal into bits of useless scrap.
choprboy
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Re: An electronics question

Post by choprboy »

Well... that all depends on what you want to use it for... Potentiometers are available in many different sizes/shapes for mounting and are rated by 5 aspects:
1) Total resistance (or full scale resistance) - resistance of the potentiometer at full scale.
2) Resistance scale - typically either "linear taper" or "audio taper":
linear taper - where the resistance increases from 0 to full scale in a linear fashion (i.e. 10% turn equal 100ohm, 20% = 200ohm, 30% = 300ohm, ...100% = 1000ohm)
audio taper - where the resistance increases logarithmically (i.e. 10% turn equal 20ohm, 20% = 50ohm, 30% = 100ohm, 40% = 200ohm, ...100% = 1000ohm)
3) Number of turns - how many turns of the shaft to go from 0 to full scale (i.e. 1 turn, usually about 300deg rotation,, or many turns, 5, 10, 20 revolutions, whether it has detent positions, or in some cases a linear slide action instead of a turn)
4) Power rating - how much power the potentiometer can dissipate at max current and voltage
5) Terminal configuration - where the terminals connect to the resistive element (as shown above it is a standard 3 terminal with one on either end of the resistor and a center tap which varies in resistance to the other two terminals based on where the potentiometer wiper is currently located on the resistive element).

The above with the "B1K" marking appears to be a cheap, generic potentiometer: 1Kohm, linear taper, single turn, 1/8W to 1/2W, 3-terminal potentiometer. There seems to be a large range of power dissipation specs in identically marked B1K parts, though I would tend to believe the values at the low end of the range as being realistic.
https://www.protechtrader.com/manuals/P ... asheet.pdf
https://www.jameco.com/z/RV24AF-10-18F- ... 33675.html

You can replace it with any similarly rated potentiometer assuming the original part was actually spec'd correctly to begin with, though its hard to determine a "correct" value without knowing what the circuit is actually doing. From the first picture I would have expected it to be a power chopper dimmer circuit where very little current flows thru the potentiometer. Though from the second picture of the circuit board it looks like it is a bare bones cheap design where all the current flows thru the potentiometer. If the "12-24V 0.8A" rating is to be believed the potentiometer is dissipating a minimum of 0.6W and potentially much more... so it was likely vastly under-spec'd to their claimed dimmer performance. If this is acting as a full rated voltage motor speed controller or light dimmer it has a short outlook on life. If it is a low level mic input volume control then there likely isn't any issue...

You can likely cut the stem off, all the way down to the start of the threads, it is normally just a solid shaft, though then the question becomes how do you turn it. And depending on how you cut it you can damage the wiper. From the picture I'd guess the shaft is just a cheap aluminim/zinc "pot metal" alloy.

Newark, Digikey, and Mouser are all major electronic suppliers, though they cater to major purchasers, so single piece prices can be quite expensive. Though their catalogs are large and can give you a great range of possibilities to look at and compare. For one-off purchases if you don't need an exact manufacturer part number or form factor you may want to look at the many surplus/hobbyist suppliers that sell in small quantities. Though their variety is far more limited. Some examples:
All Electronics
Electronic Goldmine
Elliott Electronic Supply
Jameco
SparkFun

From standard:
https://www.allelectronics.com/item/nlt ... aft/1.html
https://www.jameco.com/z/RV24AF-10-18F- ... 33675.html
To horizontal wheel:
https://www.jameco.com/z/3352T-1-102-91 ... 91695.html
To tiny:
https://www.allelectronics.com/item/tpr ... mer/1.html
https://www.sparkfun.com/products/9806
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Bill Shields
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Re: An electronics question

Post by Bill Shields »

What are you doing with the end result?

What actual controlled voltage and current is required?
Too many things going on to bother listing them.
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Greg_Lewis
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Re: An electronics question

Post by Greg_Lewis »

THANK YOU Choprboy for your thorough explanation and education on these things. Yes, this is a cheap PWM dimmer that is sold on Ebay by various Chinese sources for about $5. So if that’s the retail price, we can imagine the manufacturing cost of the components.

And I didn’t notice the decimal place on the current rating. What’s funny is that I’m using these to dim the headlights on my locomotives. I’ve already put one on my gas/hydraulic engine. The bulbs in the lights are the little halogen 12-volt 20-watt peanut bulbs, so if my math is right they should be drawing 1.6 amps. I haven’t smoked the dimmer on the diesel yet. Now I don’t run them dimmed except when there is approaching traffic on the railroad so they aren’t running for more than a minute at most (Headlight wiring bypasses the dimmer when on full.). But nevertheless, with electronic components I’m surprised that the one on the diesel hasn’t had a coronary.

So, Bill, your questions should be answered above. The reason I bought these dimmers is that they were cheap and easy to work with and I could set the pot to whatever level to get the dimming I wanted. Then I wire the headlight to a double-throw switch so I can flip from high (bypassing the dimmer), to dim.

Now all this discussion moves my brain cell into another possibility. Waking up in the middle of the night I realized that I could put whatever dimming device needed under the tender deck and just run two wires to the cab instead of one — one full voltage and one dimmed. That eliminates the size issue of parking a package of dimming circuitry in the locomotive cab where it would be ugly and not look prototypical.

So the next step is what about a fixed resistor? And if so, would I just go for the 20 watts or a little more for a fudge factor? Keep in mind that it is only in use for perhaps 60 seconds at most when passing traffic on the railroad. The question would then be how to determine the resistance needed to get the dimming I want. Just for the halibut I hooked it up on the bench and took a reading with my meter. You can see that I got about 1.6 volts at the dimming level I think I want. But am I getting a legal reading considering that a PWM dimmer doesn’t really cut the voltage but cuts the on-time of the full voltage?

And if my meter is giving the right reading, then how do I calculate the values of the needed resistor? I went to an online ohms law calculator but I don’t see where I’d put in both the input voltage and the output voltage to get an answer.

Thanks again for all your help!


The alleged circuit board:
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Halogen 20w peanut bulb:
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IMG_0706.JPG (164.18 KiB) Viewed 828 times

Bench test showing 1.6 volts at dimmed level:
IMG_0707.JPG
IMG_0707.JPG (140.01 KiB) Viewed 828 times
Greg Lewis, Prop.
Eyeball Engineering — Home of the dull toolbit.
Our motto: "That looks about right."
Celebrating 35 years of turning perfectly good metal into bits of useless scrap.
choprboy
Posts: 322
Joined: Sat Oct 29, 2011 11:23 pm

Re: An electronics question

Post by choprboy »

Ah yes.. .a PWM circuit, much better than just trying to run the currently directly thru a potentiometer, what I was expecting based on the completely bare backside of the board and no apparent thru-holes or components. And yes a PWM circuit is a type of chopper circuit that turns on and off very rapidly with different lengths of on time depending on how much power is set to be passed. The potentiometer in that case just sets a relative voltage level that relates to the on time (disipating very little power), setting a relative average output voltage that the circuit will try to track (really on at full voltage for a period, then off for a second period, averaged over time).

From the picture I can see a 78L05 voltage regulator that powers the other components at a constant 5V regardless of the 12-24V input. This feeds a NE555 single timer circuit that is creating a continuous chain of output pulse at high frequency (this might be configured to output a ramping voltage instead of a simple on/off pulse, probably at somewhere around 10kHz, but I would have to draw out the circuit, look at spec sheets, and remember how to do the math to verify...). The pulse output is feed to a LM358 op-amp acting as a comparator between the output of the 555 and a voltage reference from the potentiometer. The 358 then turns the output power transistor in the lower right corner (probably a mosfet, but I don't recognized the part number) on and off in sync with the timer. The transistor turning on and off rapidly creates an average output voltage to the powered device. The circuit doesn't appear to have any feedback monitoring, so the average output voltage will vary a bit with input voltage.

As you know your source voltage (a 12V battery), lamp rated voltage and power, and the average voltage at your low power setting, you can calculate the average equivalent resistance. A light bulb is not a perfect resistor (resistance varies greatly with filament temperature), but for the purposes of a back-of-the-napkin calculation, and the way it is used here, we can assume it is perfect. First we start with the simple voltage/current/power equations and calculate all the other values needed:
  • 1) V = I * R
  • 2) P = V * I
  • 3) P = I^2 * R
where V = voltage, I = current, R = resistance, and P = power. The 12V battery will have a nominal voltage of somewhere between 10V (completely flat) and 14.5V (on charger at full charge). For a fully charged lead-acid battery discharging under load, the voltage will typically stabilize at about 12.5V (slowly dropping to about 12.0V for the majority of its life). So we will use that as "source voltage". The lamp at rated voltage can therefore be calculated to have a current at full power of:
  • 2) 20W = 12V * I
where I = 1.67A. The resistance of the bulb is therefore:
  • 1) 12V = 1.67A * R
where R = 7.20ohm. (At 12.5V this lamp will be drawing 1.73A at 21.7W). At a reduced voltage of 1.62V the current thru the bulb will be:
  • 1) 1.62V = I * 7.20ohm
where I = 0.23A. So at low power the drop across the bulb will be 1.62V with 0.23A of current (0.36W). We need to drop or "waste" the rest of the 12.5V supply voltage across a load resister to get that output. So 12.5V - 1.62V = 10.88V across the resistor:
  • 1) 10.88V = 0.23A * R
  • 2) P = 10.88V * 0.23A
where R = 47.3ohm resistance and P = 2.50W of power dissipated by the resistor. Now, this is a fairly large amount of power for a standard resistor, though power resistors can be found up to the 100s and 1000s of W power range. So a typical resistor value that would meet this would be a 50ohm resistor with a minimum power rating of 2.5W (note1: 5W+ would be better) (note2: 47ohm is also a standard value resistor value, though I couldn't immediately find a 47ohm power resistor in a very quick look) (note3: this will get quite hot so don't place it next to anything burnable, if possible get a resistor with a heatsink). So something like this, though there are many others as well:
https://www.jameco.com/z/CR10-50-RC-Xic ... 46629.html

Another option, instead of completely throwing away the dimmer circuit, would be to cut the line between the potentiometer wiper terminal and the 358 op-amp, and install a switch to flip between the wiper output and the full voltage output at one of the potentiometer ends. That way you could easily switch between full output and low output with a separate switch, but still have the ability to adjust the low power setting if you needed to. I can't tell which end of the potentiometer at first glance, depends on how the op-amp was configured as a comparator. It looks like the potentiometer feeds a variable voltage between 0.41V and 4.59V to the op-amp, so full output may be at 0.41V or 4.59V relative potentiometer input. You also need to be careful as a switch moving from one position to another momentary goes open, which could caused "bad things" to happen to the op-amp. That is probably easy to fix with a single additional small resistor on the circuit, but where that additional resistor goes depends on the op-amp configuration. I can go into that with additional measurements of the live circuit... but this post is already too long...
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Greg_Lewis
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Re: An electronics question

Post by Greg_Lewis »

Thank you so much, Choprboy, for laying this out for me. I appreciate it. I've always struggled with electronics as you can't see what is happening. Give me something with levers and gears and I'm much happier.

As to your final comment about cutting into the circuit for a full-power switch, on my other locomotive, where I have room for the entire unit including the plastic case, I just wired a bypass for the entire dimmer. That setup has worked fine. But here with the steamer, space is an issue so I need to get creative. I see some 47 ohm, 5w power resistors on Ebay for five bucks and change including shipping and they don't look too big that I can't find a place to hide one including a heat sink. So I think I'll conjure up some of those and see what happens.

(From one of the Chinese Ebay sellers I see the 47 ohm and 51 ohm. Perhaps I should get some of each. They're only a buck each.)

Thanks again for all your helpful explanations!
Greg Lewis, Prop.
Eyeball Engineering — Home of the dull toolbit.
Our motto: "That looks about right."
Celebrating 35 years of turning perfectly good metal into bits of useless scrap.
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