Check the electronic board, with lamp and variac

[ME]

Does it matter if the bulb goes in series with the variac  input , also can a variac be powered from a pure sinewave inverter?.

[CharlotteSwiss]

I used a single 75W light bulb (but I have a second bulb socket somewhere if I wanted), and a space heater as the DUT.
[...]
But if I set it to medium temperature mode (I don't know how many watts, I think 1200...), as soon as I turn it on, the heater doesn't even start, and the lamp stays on, I think, at maximum brightness.

I only use the dim bulb method for things expected to draw relatively little current. Hifi stuff, electronic gadgets, etc. If I know they will draw thousands of Watts in normal operation, I don't. At least not with the heater/motor/... running. I don't have any 3000W bulbs around

For those things, I have an amp meter with a current transformer and an override switch in my overengineered dim bulb box.

Sometimes I wish I had some kind of programmable AC fuse that I could set to arbitrary current limits like in my DC power supply.

The heater was just a test, allowing me to have three different power settings. I'll do further experiments with much lower-power devices. As for the lamps, the lowest I have is 40W; the highest is 75W. I'll use a second lamp holder and experiment with two bulbs (maximum power I can get up to 150W for now). I imagine I'll have to connect them in parallel.

Okay, I'll try with lower-power loads. I was also thinking that, in addition to the lamp, when I test a board, I always have the current visible with the clamp meter. Lamp on or off, if there's a short, I think we'll have an abnormal current in the test circuit...

IMO you're overthinking this.

Dim bulb testers are so simple once you get your head around how they react with a working DUT and we can change their visual feedback just by using a lower or higher wattage bulb for any given DUT.

We know how much current a DUT should draw therefore with bulb selection and how much each bulb glows we have a rough idea of the DUT current draw.

For the unknown DUT a dim bulb is an excellent 1st test power up but after that it can be put away until it's needed again after major PSU repair for example or the next DUT.

The dim bulb tester is a simple and elegant solution costing just cents and time if you have a good box of scrap devices for the parts to make one from.

Here's one I knocked together from scrap a couple decades back and linked from an old thread.

I basically use the lamp to prevent additional components on the board from burning out. However, I'd like it to alert me when there's a short circuit in progress.

[CharlotteSwiss]

I've started studying and experimenting, and I think I've made progress.
My DUT was an old radio, with a maximum power of 6W. In this case, having no problems, the light bulb didn't turn on at all (both 60W and 40W), and this told me there was no short on its circuit board. The supply voltage remained constant at 240V.
Let's say I had a serious short on the radio circuit board: the light bulb would certainly turn on, but the AC input voltage to the DUT would certainly drop, indicating an abnormal current draw. I hope I've understood correctly.
Now, however, I'd like to understand how to proceed when analyzing a device with a maximum power of 500W. Let's assume this device doesn't have any shorts, but if I connect the light bulb (say, a 60W one), the same light bulb will always be on with a nice glow, despite not having any shorts. So, it won't be of any help in evaluating any anomalies. In these cases, could the solution come from the variac, using a lower supply voltage? (But then we'd have the problem of potential problems with the switching part of the DUT...).

But I'm making progress.

And since the forum is a nest of nerds, my future idea would be to build a junction box with: two lamp holders; a switch for each lamp; a switch to activate/deactivate the lamp protection; and a display that measures the voltage and current involved 

[nonius_]

the same light bulb will always be on with a nice glow, despite not having any shorts. So, it won't be of any help in evaluating any anomalies. In these cases, could the solution come from the variac, using a lower supply voltage?

It will not 'glow nicely', it will burn at near full brightness! The lamp will be seeing roughly 500/(500+60) = 0.89 = 89% of line voltage, i.e. 205Vac. (making quite a few assumptions here; it's just a quick 'back-of-the-enveloppe' calculation).

The variac won't help: if the devices are purely resistive loads, the dim-bulb will still see 89% of line voltage, just at a lower line voltage.... It won't burn as bright but that's because you've lowered its supply voltage.

The only real solution here is to use a dim-bulb of higher power-rating; a bulb that is in proportion to the power of the device being tested.

[tautech]

I've started studying and experimenting, and I think I've made progress.
My DUT was an old radio, with a maximum power of 6W. In this case, having no problems, the light bulb didn't turn on at all (both 60W and 40W), and this told me there was no short on its circuit board. The supply voltage remained constant at 240V.
Let's say I had a serious short on the radio circuit board: the light bulb would certainly turn on, but the AC input voltage to the DUT would certainly drop, indicating an abnormal current draw. I hope I've understood correctly.
Now, however, I'd like to understand how to proceed when analyzing a device with a maximum power of 500W. Let's assume this device doesn't have any shorts, but if I connect the light bulb (say, a 60W one), the same light bulb will always be on with a nice glow, despite not having any shorts. So, it won't be of any help in evaluating any anomalies. In these cases, could the solution come from the variac, using a lower supply voltage? (But then we'd have the problem of potential problems with the switching part of the DUT...).

But I'm making progress.

And since the forum is a nest of nerds, my future idea would be to build a junction box with: two lamp holders; a switch for each lamp; a switch to activate/deactivate the lamp protection; and a display that measures the voltage and current involved 

250W heat lamps are available but they don't offer the same visual clues as an incandescent light bulb.
There are also 15W bulbs available too like those used in a fridge and oven lights.

But most of what we often repair is low wattage so it's easiest to select a bulb based on the expected current/wattage draw plus a little more and if it glows dim all is well.

Still we expect some initial extra brightness at power On but dimming immediately once the magnetizing and cap charging currents are satisfied.

Keep playing with 100% operational DUT's and different wattage bulbs to get a proper handle on what you should expect to see.
No need to make it more complex than is needed.....

[CharlotteSwiss]

the same light bulb will always be on with a nice glow, despite not having any shorts. So, it won't be of any help in evaluating any anomalies. In these cases, could the solution come from the variac, using a lower supply voltage?

It will not 'glow nicely', it will burn at near full brightness! The lamp will be seeing roughly 500/(500+60) = 0.89 = 89% of line voltage, i.e. 205Vac. (making quite a few assumptions here; it's just a quick 'back-of-the-enveloppe' calculation).

The variac won't help: if the devices are purely resistive loads, the dim-bulb will still see 89% of line voltage, just at a lower line voltage.... It won't burn as bright but that's because you've lowered its supply voltage.

The only real solution here is to use a dim-bulb of higher power-rating; a bulb that is in proportion to the power of the device being tested.

What formula is used to calculate the percentage of voltage that will be applied exclusively to the lamp? (rated power/rated power + lamp power)?
The problem is that I still haven't figured out how to calculate the ideal lamp power based on the device's power...

I've started studying and experimenting, and I think I've made progress.
My DUT was an old radio, with a maximum power of 6W. In this case, having no problems, the light bulb didn't turn on at all (both 60W and 40W), and this told me there was no short on its circuit board. The supply voltage remained constant at 240V.
Let's say I had a serious short on the radio circuit board: the light bulb would certainly turn on, but the AC input voltage to the DUT would certainly drop, indicating an abnormal current draw. I hope I've understood correctly.
Now, however, I'd like to understand how to proceed when analyzing a device with a maximum power of 500W. Let's assume this device doesn't have any shorts, but if I connect the light bulb (say, a 60W one), the same light bulb will always be on with a nice glow, despite not having any shorts. So, it won't be of any help in evaluating any anomalies. In these cases, could the solution come from the variac, using a lower supply voltage? (But then we'd have the problem of potential problems with the switching part of the DUT...).

But I'm making progress.

And since the forum is a nest of nerds, my future idea would be to build a junction box with: two lamp holders; a switch for each lamp; a switch to activate/deactivate the lamp protection; and a display that measures the voltage and current involved 

250W heat lamps are available but they don't offer the same visual clues as an incandescent light bulb.
There are also 15W bulbs available too like those used in a fridge and oven lights.

But most of what we often repair is low wattage so it's easiest to select a bulb based on the expected current/wattage draw plus a little more and if it glows dim all is well.

Still we expect some initial extra brightness at power On but dimming immediately once the magnetizing and cap charging currents are satisfied.

Keep playing with 100% operational DUT's and different wattage bulbs to get a proper handle on what you should expect to see.
No need to make it more complex than is needed.....

I'm not complicating things, but I can't figure out what the ideal lamp wattage would be. I imagine there's a calculation to be made.
I mean, if I had a device that draws a maximum of 500W, there's a calculation to say that an xxW lamp would be better than an xxW one.
I'm missing something. Obviously, I'm doing the tests...

[tautech]

500W is where I would instead rely on fast blow fusing instead of a dim bulb tester as you can't cover every need with one.

It's all about the diagnostic principles you might chose to use.
So let's imagine you have a fault with a DUT that draws 500W where it would make better sense to disconnect its PSU from the rest of the circuit load to first ensure the PSU wasn't the problem.
Then you might do resistance readings on each rail to see if one has a short.....this is where a bench PSU and a thermal camera comes into its own.

There are many ways as they say to skin a cat and a dim bulb tester is just one of them but it's not suitable for all jobs.

I often park projects and sleep on them when often the next day see them in a different (and sometimes smarter) light.
There are no hard and fast DUT repair methods other than those with known common faults however experience counts but these days with the internet freely available to all so we get to benefit from the experience of others.

A mentor decades back said if he was to have just one instrument it would be a DSO which always intrigued me but with some application of what lies between our ears he was right of course.

[pcprogrammer]

What formula is used to calculate the percentage of voltage that will be applied exclusively to the lamp? (rated power/rated power + lamp power)?
The problem is that I still haven't figured out how to calculate the ideal lamp power based on the device's power...

I can all be done with Ohms law and the rules on series networks. The back of the envelope calculation of nonius_ makes use of the fact that power is based on current times voltage and according to Ohms law current is voltage divided by resistance. Take into account the formula to work on a voltage divider, where the voltage is split based on the two resistance values, and you have the back of the envelope calculation.

Vr2 = Vin * (r2 / (r1 + r2))

If r1 is the resistance of the lamp and r2 the resistance of the DUT, the percentage of voltage on the DUT is calculated with r2 / (r1 + r2).

As an example r1 = 100 Ohm and r2 = 900 Ohm, would mean that 90% of the voltage is over r2. ( 900 / (100 + 900) = 900 / 1000 = 90 / 100 = 90% )

The current in this case is based on I = Vin / (r1 + r2) and the total power is then P = Vin * I. With the voltage Vin being the sum of the voltages across the resistors this can be reworked to P = (Vr1 + Vr2) * I.

You can work out the rest with not to complex mathematics.

I'm not complicating things, but I can't figure out what the ideal lamp wattage would be. I imagine there's a calculation to be made.
I mean, if I had a device that draws a maximum of 500W, there's a calculation to say that an xxW lamp would be better than an xxW one.
I'm missing something. Obviously, I'm doing the tests...

Based on the above explanation about the back of the envelope calculation, you can now conclude that to check a DUT with a high power rating, you need a lamp with a much higher rating to allow the DUT to work as intended.

But like tautech, and I previously, indicated, there is no one clear cut solution to hardware repair. Take for instance a personal computer, it can have a power supply with a 1000W rating, but that means it can deliver a maximum of 1000W into the motherboard and peripherals. It does not mean it is always consuming a 1000W from the mains. It depends on the parts used in the personal computer and the task it is performing.

If you start to work on a board, and you suspect it to have a short, the first thing to check is if there is an actual short with the use of a continuity tester, like what most digital multi meters have build in.

[CharlotteSwiss]

Thanks, Tautech and pcprogrammer, for the explanation. I'd add that I use the lamp when I have a device to test, or after I've changed some components, to prevent other parts of the board from burning out. Once I've verified with the lamp that there's a short, I obviously do a voltage-free scan on the board to find the problem.

From the formulas above on the divider, we can therefore say that if I have a 500W DUT, using a lower-wattage lamp will result in a higher voltage on the DUT, while if I use a higher-wattage lamp, I will have a lower voltage percentage on the DUT.
Example: with a 25W lamp: 500/525 = 95% of the voltage on the DUT.
Example: with a 200W lamp: 500/700 = 71% of the voltage on the DUT.
This means that in the event of a short circuit, the 200W bulb will provide more protection for the DUT, as it will draw less current.

But since this is all complex     , the solution would be to use an intermediate bulb (e.g., 60W), the purpose of which would be to prevent further damage to the board in the event of a short circuit. If its light stays on for a long time, I'll try to figure out where the problem lies. If, however, the bulb stays off, or almost, I can connect the DUT to the mains without the bulb   

[pcprogrammer]

Thanks, Tautech and pcprogrammer, for the explanation. I'd add that I use the lamp when I have a device to test, or after I've changed some components, to prevent other parts of the board from burning out. Once I've verified with the lamp that there's a short, I obviously do a voltage-free scan on the board to find the problem.

From the formulas above on the divider, we can therefore say that if I have a 500W DUT, using a lower-wattage lamp will result in a higher voltage on the DUT, while if I use a higher-wattage lamp, I will have a lower voltage percentage on the DUT.
Example: with a 25W lamp: 500/525 = 95% of the voltage on the DUT.
Example: with a 200W lamp: 500/700 = 71% of the voltage on the DUT.
This means that in the event of a short circuit, the 200W bulb will provide more protection for the DUT, as it will draw less current.

But since this is all complex     , the solution would be to use an intermediate bulb (e.g., 60W), the purpose of which would be to prevent further damage to the board in the event of a short circuit. If its light stays on for a long time, I'll try to figure out where the problem lies. If, however, the bulb stays off, or almost, I can connect the DUT to the mains without the bulb   

It is the other way round. Example: with a 25W lamp: 500/525 = 95% of the voltage on the lamp. Why because the lamp has a higher resistance than the DUT. Resistance is an inverse function in respect to power. The higher the resistance in series with the same voltage source means a lower current and thus less power.

If the bulb does not light up, and the bulb is in proper working condition, the DUT either does not draw current and is still defective or it works as intended and can then be safely connected to the mains. Not every defect results in a short. It can also be an open circuit.

Keep in mind that the bulb will only stay of or very dim if the current through it is lower than the nominal current. So testing a resistive device of 500W with a 25W bulb will always make the bulb light up rather brightly, short or not.

[CharlotteSwiss]

I knew I wouldn't get it right, but then again, I said it, it's not easy to understand 

Maybe I need to think differently to understand better:

500W DUT, with 3 different lamps (60W; 500W; 700W)

Lamp with a wattage lower than the DUT: 500/560 = 90% of the current to the lamp.

Lamp with a wattage equal to the DUT: 500/1000 = 50% of the current to the lamp.

Lamp with a wattage higher than the DUT: 500/1200 = 41% of the current to the lamp.

This tells us that the more the lamp's wattage is equal to or higher than that of the DUT, the less bright its light will be, and probably, if I don't have any short circuits, it will remain off.

Perhaps at this point I should calculate a lamp wattage that can, however, allow at least a nominal current to the DUT for normal operation.

I hope I've made at least one step forward now... 

[pcprogrammer]

You are mixing up your currents and voltages. The percentage is about voltages and not current. The current is equal through both the lamp and the DUT.

Take your example of 700W lamp and 500W DUT, assuming the DUT is a normal resistive device and for the sake of the argument that the lamp is too, will result in the lamp seeing 41% of the voltage, which means that it will glow at only 41% of the brightness. The DUT gets 59% of the voltage, which means that it will not go up to its full potential.

Lets say the DUT is a normal resistive heating element rated 500W at 240V. This means it has a resistance of ~115 Ohm. The lamp rated 700W at 240V has a resistance of ~82 Ohm. The total resistance of the series network is 197 Ohm. So the current through this series network is 240 / 197 = ~1.22A.

The voltage across the heater is 115 * 1.22 = ~140V, the voltage across the lamp is 82 * 1.22 = ~100. This shows that the back of the envelope calculation is correct. 100 / 240 = ~41%

But the power consumed is only 240 * 1.22 = ~293W, which is no way near the 1200W sum of the device wattage's. The heater is only generating 140 * 1.22 = ~171W of heat.

As I wrote earlier the real world of lamps is a bit different due to the fact that it has a positive temperature coefficient, which means that the resistance goes up as the temperature rises. So initially the 700W lamp will have a lower resistance and will only be around that calculated 82 Ohm when fully heated. This changes the above scenario and the heater will get a larger slice of the voltage, but one needs the resistance vs temperature vs current curve of the lamp to determine the actual value.

You should evaluate the need for and what wattage to use on a per case basis. A transistor radio requires a different approach then for instance a cathode ray tube television. In a more modern light, a lot of devices make use of a switching power supply that is separate from the actual device itself. This allows for working on the device at a lower voltage by using a current limited bench power supply. This way the current limiting lamp indicator on the supply can take on the function of the lamp.

If the problem is with the switching power supply it becomes a different story for which the lamp can be of use.

As I'm not a real repair man, I don't have a dim bulb tester device, and tend to stay clear of the mains voltage. Have had enough shocks throughout my life. 

One anecdote, I once worked on a simple transformer based linear power supply that had an issue. I constantly removed the power cord to make some changes, right up to when it was working again and needed to be put back together. I never knew I could jump that high. 

[CharlotteSwiss]

Thanks, pcprogrammer, for your patience. I'd be a terrible student!
True, I was confused. The current flowing through the circuit is the same, but the voltage changes.
So, going back to your example, if we had a short on the DUT (which is 500W) and we used a 700W lamp, the current flowing would only be 1.22A, much less than the maximum allowable current (500/240 = approximately 2A), and this would prevent further damage.
Perhaps what I'm having trouble understanding is how the lamp behaves when there's no short on the DUT; that is, its filament should have a higher resistance.

I also find it difficult to experiment, both because of the lack of high-resistance lamps and the lack of shorted DUTs.

As for electric shocks, fortunately I've never had one, although as a child I sometimes enjoyed pinching the wires to feel the line voltage along my body... 

[pcprogrammer]

Thanks, pcprogrammer, for your patience. I'd be a terrible student!
True, I was confused. The current flowing through the circuit is the same, but the voltage changes.
So, going back to your example, if we had a short on the DUT (which is 500W) and we used a 700W lamp, the current flowing would only be 1.22A, much less than the maximum allowable current (500/240 = approximately 2A), and this would prevent further damage.
Perhaps what I'm having trouble understanding is how the lamp behaves when there's no short on the DUT; that is, its filament should have a higher resistance.

I also find it difficult to experiment, both because of the lack of high-resistance lamps and the lack of shorted DUTs.

If there was a short in the 500W DUT the current would be higher than the 1.22A. It would be based on just the resistance of the lamp which is calculated to be 82 Ohm, so the current would be 240 / 82 = ~2,93A. This is higher than the nominal rated current of the DUT, but the DUT is defective because it has a short. Without the lamp the mains fuse would blow because the current would only be limited by the wire resistance. Even if that adds up to lets say 1 Ohm, the current could go up to 240A, which is enough to fry thin copper wires.

A 60W lamp, when on, has a high enough resistance. Only when it is cold the resistance is relatively low. But that is the whole point of the dim bulb tester. Low resistance when cold so not a lot of voltage drop across the lamp, but high resistance when hot, limiting the current through the DUT.

The question is how much do you know about things like Ohms law and simple electrical networks? If not much, than that is something to look into.

As for electric shocks, fortunately I've never had one, although as a child I sometimes enjoyed pinching the wires to feel the line voltage along my body... 

You were then lucky that there was no direct path to ground, or you have incredible high skin resistance, otherwise you would have jumped too. 

But I know the sensation. A slight tingling that makes you think, something is not right here.

[CharlotteSwiss]

I know Ohm's Law, as well as some other formulas. Of course, I don't study this topic every day. For example, a few years ago I was more knowledgeable, now I'm less so. But I do have a passion for this world of electrons 
In fact, I was also confused about the lamp's resistance, which, as you said, is the logic behind this test.

I don't remember having major problems sensing mains voltage as a child. Once, with my father, at a small company, I received a higher shock (about 400V), and I remember having a wrench in my hand, which I used to touch the inside of some live conductors, and we found the same wrench 10 meters away... 

[pcprogrammer]

I know Ohm's Law, as well as some other formulas. Of course, I don't study this topic every day. For example, a few years ago I was more knowledgeable, now I'm less so. But I do have a passion for this world of electrons 
In fact, I was also confused about the lamp's resistance, which, as you said, is the logic behind this test.

As I do not use them daily, I do find that I have to look them up from time to time when in need. What is important is the understanding of the principle. This is what a teacher once told me. Formulas can be looked up, but if you don't know how to use them they are worthless.

Yes, the dynamic behavior of the lamp is what it is all about. Once you understand that, it should all become clear.

I don't remember having major problems sensing mains voltage as a child. Once, with my father, at a small company, I received a higher shock (about 400V), and I remember having a wrench in my hand, which I used to touch the inside of some live conductors, and we found the same wrench 10 meters away... 

You have been living dangerously 

[MarkF]

But I know the sensation. A slight tingling that makes you think, something is not right here.

I once got shocked by aircraft 400 Hz power.
There is NO tingling at all.
It reaches right out and grabs you HARD.
Can't say I recommend the experience.   


Another time I burnt 1/4" off the end of a pair of needle nose pliers.
Metal handles.  No insulation.  Pretty sparks...


Charlotte,
  You may benefit by not worrying how much current the DUT needs.
And look at it from the point of how much current do you want to initially give the DUT and
seeing the what happens.  You can always repeat with more current.
  Guess at how much current the DUT needs and decide how much you want to provide.
Your initial concern should be avoiding fire and melting.  Not DUT operation.

[CharlotteSwiss]

You have been living dangerously 

My whole life has been lived dangerously, and will continue to be so (obviously not only for electricity-related reasons) 

Charlotte,
  You may benefit by not worrying how much current the DUT needs.
And look at it from the point of how much current do you want to initially give the DUT and
seeing the what happens.  You can always repeat with more current.
  Guess at how much current the DUT needs and decide how much you want to provide.
Your initial concern should be avoiding fire and melting.  Not DUT operation.

Okay. If I use a 60W lamp, it will have a standard light output when 0.2A flows through the circuit; if the DUT requires less current, the lamp will have a more moderate light output, and if the current is very low, the lamp won't even turn on.
Can we say that for DUTs with a nominal current greater than 0.2A, a 60W lamp isn't the optimal solution?

I feel like I keep going around in circles, and I always end up banging my head against a wall... 

[MarkF]

You're getting a lot of varying answers and there is no correct one.
Each case is unique.  It depends on the DUT.

If the DUT has a transformer, then 0.2A @ 240V would provide a lot of power on the secondary side.
Which may be a good starting point or even high for devices that aren't power hungry.

For DUTs that have a motor or heating element, 0.2A wouldn't be near high enough.

I don't look for the Dim Bulb tester to prevent damage. 
I want to stop the DUT from turning into an arc welder.
Here in the USA the breakers for outlets can be as high as 20A @ 120V.
Just limiting an unknown DUT to 1A is a game changer.  (100 W vs 2400 W)

I never even try to find the optimal solution.  It's not the way I use it.

Start low and work your way up. 
You will get the feel for it after trying a few different devices.

[CharlotteSwiss]

It really depends on the scenarios involved.
I'm starting to think that perhaps, instead of using a light bulb, it would be simpler to use a fuse, ready to trip as soon as the current exceeds the rated power of the DUT. Sure, I'd have a few dead fuses, but the situation would be more under control.

[MarkF]

It really depends on the scenarios involved.
I'm starting to think that perhaps, instead of using a light bulb, it would be simpler to use a fuse, ready to trip as soon as the current exceeds the rated power of the DUT. Sure, I'd have a few dead fuses, but the situation would be more under control.

The light bulb will react much much faster than a fuse.
And the fuse current might be 200% of its rated value.

By-the-way, I also have a fuse in series with my light bulbs in my tester.
It is what I expect as a maximum current for the dim bulb tester (fallback protection).

[nonius_]

I'm starting to think that perhaps, instead of using a light bulb, it would be simpler to use a fuse, ready to trip as soon as the current exceeds the rated power of the DUT. Sure, I'd have a few dead fuses, but the situation would be more under control.

Simpler to use a fuse? Yes. But I'd hardly call it 'more under control': bulbs have a PTC (positive temperature coefficient) characteristic, whilst fuses hardly have this (in comparison to an incandescent lamp). E.g. in the case of the large inrush-current of my welding machine (which I've added a permanent dim-bulb to; see image earlier in this thread) a fuse would do literally nothing to lower the inrush-current; instead, it would just blow (which is what the house breakers always did with this machine). The dim-bulb solved that problem.

Besides, when using fuses you'd still have to correctly dimension the fuses (AND their speed-rating), just as with a dim-bulb....    And better make sure you have plenty of fuses at hand, things can get expensive pretty quickly. Not to mention the (extra) damage you may cause to the DUT.

The nice thing of a dim-bulb is that it acts as cheap auto-resetting fuse, the rating of which isn't very critical due to the PTC character.

As others have said, I think you're way overcomplicating things. A dim-bulb is a simple effective tool, not a panacea for anything that could possibly go wrong. Usage isn't very critical or difficult (for me, with 40+ years of electronics experience), but you need to have a mental picture of how it works. Judging from your previous remark about the relation of bulb dimensions vs. load rating, you're not there yet. Just play around a bit with 'normal' dim-bulb dimensions (40W, 60W, 100W) and normal apparatus (laptop, computer monitor, soldering iron, radio, TV).

Hint: the HIGHER the power rating of a light bulb, the LOWER its resistance. Maybe that's what was causing your confusion earlier in this thread?

[pcprogrammer]

But I know the sensation. A slight tingling that makes you think, something is not right here.

I once got shocked by aircraft 400 Hz power.
There is NO tingling at all.
It reaches right out and grabs you HARD.
Can't say I recommend the experience.   

Another time I burnt 1/4" off the end of a pair of needle nose pliers.
Metal handles.  No insulation.  Pretty sparks...

Wearing shoes with rubber soles or standing on a wooden floor and only touching the life wire, there won't be a lot of current and all one feels is a tingling. 

Having a spark jump from a charged color cathode ray tube into you, that is also no fun at all.  Hurts like hell.

[tautech]

But I know the sensation. A slight tingling that makes you think, something is not right here.

I once got shocked by aircraft 400 Hz power.
There is NO tingling at all.
It reaches right out and grabs you HARD.
Can't say I recommend the experience.   

Another time I burnt 1/4" off the end of a pair of needle nose pliers.
Metal handles.  No insulation.  Pretty sparks...

Wearing shoes with rubber soles or standing on a wooden floor and only touching the life wire, there won't be a lot of current and all one feels is a tingling. 

Having a spark jump from a charged color cathode ray tube into you, that is also no fun at all.  Hurts like hell.

It's never wise to offer advice about HV, mains or otherwise without preaching safety first and last otherwise you might be a statistic !
Shocks, and I've had a good few are to be avoided ........maybe that's what's wrong with me. 

When one has to work with HV the best advice is to only use one hand while the other resides in a pocket so to eliminate a discharge path across the chest.

[CharlotteSwiss]

To properly test the use of the series lamp, I'd need to experiment with a short circuit. I don't currently have any short circuit boards available. Could I create a simple, safe short circuit to test with the series lamp? This might help me better understand the behavior of the lamp (of different wattages) in the presence of a short circuit.