Incandescence bulbs as dummy load, few questions ... - Page 1
Forum section: electronics / beginners

author: BravoV
author_rank: Super Contributor
timestamp: January 13, 2020, 02:48:55 am
content: Recently I scored an adjustable DC bench PS capable of 350V and max at 0.2A (70W).My current adjustable DC dummy load is capable only for much lower voltage, at 350V DC definitely will kill it instantly.So in order to test the PS, the dummy load that capable at 350V with max current @0.2A, I've been thinking to use two 220V ordinary incandescence bulbs "in series", which are capable of voltage max at 220V x 2 bulbs = 440V, which is still above the PS max output, cmiiw.The ideal load resistance is -> 350V / 0.2A = 1750 OhmAs two bulbs in series -> 1750 Ohm / 2 bulbs = 875 Ohm at "on" resistance for each bulb.Single bulb when connected to 220V power source -> 220V / 875 Ohm = 0.25ASo the right bulbs to use are rated at -> 220V x 0.25A = 55 Watt type.Cmiiw please.Aware that incandescence bulb has much-much lower resistance when cold, those above calculations on the bulb's resistance was at ON state at rated 220V voltage.While in series, both bulbs at 350V, each one of them only will be powered at 350V / 2 bulbs = 175V each, clearly the bulb's resistance will be lower than at fully rated 220V.Locally available bulbs which are close to above calculations are 40W or 60W, also others like 15, 25 or 75 Watt.Questions :1. Did I do it right ?2. So the safest bulbs to use is 40W type instead of 60W for this particular PS load test ? It doesn't need to be precisely loading the PS at exact 0.2A, say tad lower is fine for me. Do not want to overload the PS than the rated current capability at 0.2A.3. Any other matters that I missed or other considerations ?TIA

author: IanB
author_rank: Super Contributor
timestamp: January 13, 2020, 03:22:18 am
content: It seems to me that you want to seek out 120 V 25 W bulbs, since their operating current would be 25/120 = 0.21 A. If you put three of these in series you would have a nominal load that would draw 0.21 A at 360 V, and therefore approximately 0.20 A at 350 V.

author: BravoV
author_rank: Super Contributor
timestamp: January 13, 2020, 03:28:33 am
content: Quote from: IanB on January 13, 2020, 03:22:18 amYeah, that is an ideal situation, but mains here is at 220VAC, and as Dave love to say, the 120V bulb is rare as hen's teeth.

author: schmitt trigger
author_rank: Super Contributor
timestamp: January 13, 2020, 03:51:15 am
content: You did right.You are absolutely correct that the resistance of an incandescent bulb is far lower when cold.For instance a 65W, 120V bulb I measured a cold resistance of 14.5 ohm, and hot with a little help from Ohm calculates at 221.5 ohm.

author: BravoV
author_rank: Super Contributor
timestamp: January 13, 2020, 04:02:51 am
content: Quote from: schmitt trigger on January 13, 2020, 03:51:15 amThanks.As expected, your bulb there, has approx. 15 times ( 221.5 (hot) /14.5 (cold) ) higher resistance when powered on, this ratio is similar to most numbers I've read on the net.Fyi, I do not need to be precisely loading the PS at 0.2A@350V, say even at 0.15A or better 0.175A, I will be happy, hence I don't need to drill down the complexity finding the exact perfect/ideal bulb, beside the volt vs current consumption for incandescence bulb is not linear anyway.

author: RoGeorge
author_rank: Super Contributor
timestamp: January 13, 2020, 04:12:52 am
content: The inrush current for incandescent lightbulbs is about 10 times the nominal current. Once I measured 3 different types of incandescent lightbulbs:To avoid an overload, maybe charge an external capacitor first (through a current limiting resistor), then connect the capacitor straight to the source output, then connect the lightbulbs. The idea is to accumulate enough energy in the external capacitor, so it will help to "start" the filament(s) of the lightbulb(s). Once the filament is heated to red, the lightbulb tends to act like a constant current regulator.

author: BravoV
author_rank: Super Contributor
timestamp: January 13, 2020, 04:25:03 am
content: @RoGeorge, thanks for sharing the inrush condition at various bulbs there at power up.At my situation, inrush current is not an issue, as my plan is to slowly crank up the voltage, say from zero up to 350V, while slowly watch the current consumption, and once settled, I will bake both PSU and the bulbs, while I do measurement on the PS output quality overtime.This is the PS , it has CV and CC feature, both knobs V & I are 10 turns pots.PS : Hell, your post give me the idea to test how good the CC mode does this PSU have.

author: fourfathom
author_rank: Super Contributor
timestamp: January 13, 2020, 06:44:38 am
content: An incandescent bulb (or bulbs) can be a very useful dummy load, but when calculating the filament resistance be aware that this is highly nonlinear, even at steady-state. I can't find the numbers now, but I've seen approximations where the current/voltage relationship (resistance) varies somewhere between the square root and the cube root of the applied voltage. It's definitely not a straight-line function.

author: calzap
author_rank: Frequent Contributor
timestamp: January 13, 2020, 08:23:04 am
content: Why not use high wattage wire-wound resistors? They're not that expensive. And they're more compact and tougher than light bulbs, and their resistance is more stable. You can parallel them if need be to achieve the required wattage. I try not to exceed more than 1/2 the rated wattage, and even then they'll get hot.Mike in California

author: BravoV
author_rank: Super Contributor
timestamp: January 14, 2020, 12:53:34 am
content: @Mike , here at brick n's mortar shops, incand. bulbs are still available, they're so dirt cheap as these are practically hard to sell old stock piling up for years in their shop, single bulb is less than a buck, and if I purchase more, probably will come with greater discount.Beside, these bulbs will still be useful for future PS test.

author: schmitt trigger
author_rank: Super Contributor
timestamp: January 14, 2020, 01:20:11 am
content: Well, this thread piqued my curiosity and decided to measure the lamp's resistance vs voltage, using a Variac and a pair of DMMs measuring voltage and current.The lamp is a Philips 120V, 65W nominal, outside reflector.An Excel spreadsheet completes the analysis. Horizontal axis is voltage, vertical is resistance.Of course, the "0" volt resistance was measured with a DMM in ohms mode.Perhaps you would like to repeat this experiment with a couple of lamps. It only takes a few minutes.

author: vu2nan
author_rank: Contributor
timestamp: January 14, 2020, 03:36:31 am
content: Quote from: IanB on January 13, 2020, 03:22:18 amIndeed workable!Regards,Nandu.

author: BravoV
author_rank: Super Contributor
timestamp: January 14, 2020, 04:19:43 am
content: Quote from: schmitt trigger on January 14, 2020, 01:20:11 amThanks a lot for the chart, really appreciate your effort here.Yeah, I think I will do some profiling work on various incand. bulbs, now the problem is I forgot where I put my variac, as its rarely used, must be drowned somewhere in the piles of junks.Also suddenly pop out in my mind about DC vs AC, does anyone know if there is significant difference when an incandescence bulb is powered AC vs DC ?Of course, AC is say at 220V or 110V AC RMS@50/60Hz ... vs ... pure DC at 220V or 110V.

author: IanB
author_rank: Super Contributor
timestamp: January 14, 2020, 06:09:04 am
content: Quote from: BravoV on January 14, 2020, 04:19:43 amNot in the short term for testing a power supply as planned here.In real world use for lighting purposes AC lamps should not be used with DC. They can fail sooner and if they do fail the DC supply will sustain an arc much more readily than AC leading to potentially energetic consequences.

author: Gyro
author_rank: Super Contributor
timestamp: January 14, 2020, 01:48:12 pm
content: Quote from: calzap on January 13, 2020, 08:23:04 amIncandescent lamp filaments are so hot that they dissipate most of their power as thermal radiation rather than conducted or convected heat (in fact only a small percentage comes out as light). Radiant heat dissipation is way more efficient than conducted or convected. You don't have the same problems with hot casings and heatsinks, the bulb radiates its heat directly to objects within line of site - the walls, ceiling, stuff around them, you. All get warmer, but the large surface area involved means that nothing heats up very much.The same applies to vacuum tubes, their Anode dissipation is pretty much all by radiation. A power anplifier tube can quite happily dissipate 50W upwards where in the same situation, a transistor would need large heatsinks and possibly fans too.If you can screen yourself from the annoying light output, a light bulb is an excellent high power load.

author: schmitt trigger
author_rank: Super Contributor
timestamp: January 14, 2020, 02:13:23 pm
content: Something I found very appealing from the curve above, was that from about 1/3 of its rated voltage (40 volts in this case) the resistance increase is quite linear.Which means simple interpolation will allow you to dial in any resistance value within the range, quite accurately.