Wow, I am simply blown out by (and grateful for) all the support so freely given by all the members here!

Yes, this does make sense now. Looks like I was missing the connection to ground at the junction of the SCR cathode and R3. I will recheck my circuit and validate this and also post back on whether I find any other faulty components. As far as I can remember this was never plugged into anything but a 120V AC wall outlet.

Thank again to everyone!

 

Depending on what other loads may have been on the same circuit, there coud have been distortion products present. Things such as a light dimmer or a variable speed furnace blower, or a TV or computer with a switcher power supply. There are lots of sources of electrical noise that we might not even realize.

 

The bulb looks like a neon lamp. Did it glow orange?

NE-2: https://www.jacksonelectricsupply.com/A1A-NE-2-Neon-Glow-Bulb-Wire-Terminal-Base-p/a1a.htm

If it's indeed a neon lamp, then shunting the lamp off makes sense. Note you have to have a 100K resistor in series for 120 operation foran NE-2 which was a common bulb. With Neon, the on and off voltages are different. They burn out.

R4 needs to shunt > 1uA around the lamp to keep the triac/scr off. i.e minimum load for the triac/SCR.

 

The bulb looks like a neon lamp. Did it glow orange?

NE-2: https://www.jacksonelectricsupply.com/A1A-NE-2-Neon-Glow-Bulb-Wire-Terminal-Base-p/a1a.htm

If it's indeed a neon lamp, then shunting the lamp off makes sense. Note you have to have a 100K resistor in series for 120 operation foran NE-2 which was a common bulb. With Neon, the on and off voltages are different. They burn out.

R4 needs to shunt > 1uA around the lamp to keep the triac/scr off. i.e minimum load for the triac/SCR.

The lamp did not look at all like a neon lamp The internals had supported a single filament close to the axis of the glass bulb. This is typical of bulbs intended to be used in a parabolic reflector, and optimized for that sort of application.

 

If it were neon it would have a current limiting resistor. It doesn't. So it's not a neon.

 

If it were neon it would have a current limiting resistor. It doesn't. So it's not a neon.

Certainly correct, Tony, although I have seen neon bulbs with the resistor built onto one lead.

 

Yeah @MisterBill2, but they still would be shown in the circuit diagram. Which is the point I was making. There is no limiting resistor on the lamp.

 

Yes, probably. But as soon as I saw the photo of the bulb it was obvious that it had been an incandescent bulb with a single section of filament along the center axis. A neon or other gas filled bulb has electrodesand a fairly narrow gap between them.

 

Guys/Gals

Updates:

This is definitely not neon. In fact, the reason I don’t like the night lights available now is that they almost all use LEDs which give off an unpleasant bluish light, whereas this older night light has a warm incandescent glow – which is why I’d like to restore it if it makes sense.

The circuit as drawn by @LesJones is correct (thank you!), I had missed the connection to ground (DC negative) from the junction of the SCR cathode and R3 (220 K ohm resistor)

I removed C1 from the circuit. With my multimeter it measures 845nF. I have no way of checking ESR, on the resistance setting with my multimeter it exhibits the normal capacitor behavior – starts off with high current/low resistance and then settles to no current/infinite resistance (open). It definitely does not appear to be shorted.

I have yet to test the SCR, that will be next.

Calculations

The impedance (capacitive reactance) of C1 would be 1/(2*pi*f*C) = 1/(2*3.14159*60*845*10^-9) = 3139 ohms. In parallel with the 330K resistor (R1) it would be about 3K ohms impedance.

If we redraw the circuit in a simplified block diagram (as below), with V being the voltage dropped across the bulb and R its resistance and P the power used by the bulb, I get these scenarios for various values of R.




The total circuit impedance is Z=√(R squared+3000 squared). (The 3000 is the impedance of R1/C1)

The current in the circuit is I=120/Z

Then the voltage across the bulb is V= I*R and power used by the bulb is P=V*I.

I am ignoring the voltage dropped across the diodes, and also the effect of the SCR circuit in parallel to the bulb, since its impedance would be about 60K when light and about 300K when dark, so minimal impact to the net impedance. And the shunt itself would be open when the lamp is lit.

R Z I V P Ohms Ohms Amps Volts Watts 1 3000 0.04 0.04 0.00 10 3000 0.04 0.40 0.02 20 3000 0.04 0.80 0.03 30 3000 0.04 1.20 0.05 50 3000 0.04 2.00 0.08 100 3002 0.04 4.00 0.16 1000 3162 0.04 37.95 1.44 2000 3606 0.03 66.56 2.22 3000 4243 0.03 84.85 2.40 5000 5831 0.02 102.90 2.12

Questions

• @LesJones – you have stated that the shunt would not waste power because the current would be 90 degrees out of phase with the voltage. Isn’t that the same situation when the power flows through the lamp rather than the shunt? How does any power get used by the bulb?
• Do my calculations make sense? To get any meaningful wattage it seems the bulb would have to be very high resistance/high voltage, and even then, it seems about 2 watts or so max. Doesn’t seem right to me, so I feel I am doing something wrong.

Sorry if these are stupid questions, I have not touched electronics for 50 years so this is a bit rusty!

 

In an incandescent lite the load is resistive, and thus the current and voltage are in phase and the power is real. So power is used in the bulb. producing light.

Now for LED colors, there are also "warm white" LEDs but they cost more. But they are certainly available from the larger distributors and home improvement stores. Home Depot has a wide selection, look at "warm white" lights. They are specified as having a lower color temperature. You could have three of the lower color tep LEDs in series with a variable resistor running off a stable 12 volt supply and get just the light that you want.

 

In an incandescent lite the load is resistive, and thus the current and voltage are in phase and the power is real. So power is used in the bulb. producing light.

The filiment resistance is also about 15x lower when cold than when hot assuming tungsten.

 

The filiment resistance is also about 15x lower when cold than when hot assuming tungsten.

Certainly the filament resistance is much lower cold than hot. This resistance varies in a most non-linear manner as the temperature increases, no question about that. The result is that bulbs light up quickly. Of course the load is still resistive as the bulb lights. There is indeed a current inrush as a filament type of bulb is driven by a constant voltage source. How is this all relevant to the voltage and current phase relationship?

 

A few points.
1 / The diagram in post #29 would not work as the capacitor does not provide a DC path.
2 / MisterBill2 has answered your question about power in the bulb. (Post #30)
3 / You would need to test the capacitor with a much higher voltage than that used on a DMM. As the capacitor is rated at 250 volts AC you would need to test it for leakage with with least 250 * 1.414 (Root 2) = 354 volts DC.
4 / C1 should be a Y rated capacitor and could be a self healing design. This means that it could have broken down for long enough to have caused the damage but repaired itself and could now test OK. Google X and Y capacitors and self healing capacitors.

Les.

 

Hi,
Can someone help me understand how this circuit works? It is for a night light which plugs into a wall outlet and is supposed to come on in the dark, and it’s not working.

I believe I have traced it accurately and have checked 3 times, but it just seems wrong! I can attach pictures of the actual circuit board (both sides) if it would help.

I would expect that after the bridge rectifier I should have 120x1.414 = 170V DC. This then goes straight to the lamp, which seems crazy high, plus then what is the purpose of the rest of the circuit? And I am measuring only 105V DC at the lamp. The lamp itself is burnt out and I suspect that it is a low voltage (and of course low wattage) lamp that got zapped by high voltage.

Here are a few questions:

• What is the purpose of R1 and C1?
• R4 is burnt (open) and the markings are also burnt. I assume the circuit failure is because R4 has burnt out. What would be the appropriate value here, assuming that the output to the lamp should probably be 12 volts DC max, and most likely lower, maybe 3 to 6 volts. There are no markings visible on the lamp to help determine its rating.
• How does this circuit control (step down) the voltage to the lamp, the way it is drawn? Or, if I have drawn it incorrectly what is the more likely way it should be drawn?

Thanks, I would really appreciate any insights.


View attachment 219806

Yes the original circuit does not look right because there is a cap in series with the SCR on the DC side. That is unless that SCR is really a triac, which still wouldnt make sense though.

 

1 / The diagram in post #29 would not work as the capacitor does not provide a DC path.

I'm confused by this -- this is the same circuit as drawn by you in Post #19, just put into a block daigram. There is a resistor in parallel with both capacitors, so there is always a DC path.

3 / You would need to test the capacitor with a much higher voltage than that used on a DMM. As the capacitor is rated at 250 volts AC you would need to test it for leakage with with least 250 * 1.414 (Root 2) = 354 volts DC.
4 / C1 should be a Y rated capacitor and could be a self healing design. This means that it could have broken down for long enough to have caused the damage but repaired itself and could now test OK. Google X and Y capacitors and self healing capacitors.

Good point. I guess I have no way of testing this.

In an incandescent lite the load is resistive, and thus the current and voltage are in phase and the power is real. So power is used in the bulb. producing light.

Let me try my question the other way around. Yes, the bulb is resistive, but so is the resistor in the shunt, so then why won't the shunt also draw (waste) power when the SCR is triggered during the day?

Any comments on the calculations and how does one get more than about 2 watts through the bulb? The resistance used would of course be the "steady state" resistance after the bulb has warmed up, not the cold resistance.

And the bulb was not a 12 volt bulb, based on the length of the filament.

Any way to determine/"guesstimate" from the photo what kind of bulb this is -- voltage/watt rating?

Thanks again to everyone for patience with my continuing questions!

 

I'm confused by this -- this is the same circuit as drawn by you in Post #19, just put into a block daigram. There is a resistor in parallel with both capacitors, so there is always a DC path.


Good point. I guess I have no way of testing this.


Let me try my question the other way around. Yes, the bulb is resistive, but so is the resistor in the shunt, so then why won't the shunt also draw (waste) power when the SCR is triggered during the day?

Any comments on the calculations and how does one get more than about 2 watts through the bulb? The resistance used would of course be the "steady state" resistance after the bulb has warmed up, not the cold resistance.


Any way to determine/"guesstimate" from the photo what kind of bulb this is -- voltage/watt rating?

Thanks again to everyone for patience with my continuing questions!

My guess is a mains voltage bulb of a quite low wattage, possibly one or two watts. With a really good magnifier and a good light it should be possible to see the remaining ends of the filament and verify that it was really very thin. Given the small size of the bulb body, one choice would be to go to the Graingers website and look through their listings of bulbs, which should include sizes, and see what bulbs are available in a similar size and appearance. That should help reduce the guesswork. Probably the costof a replacement bulb will be about what the original assembly cost.

 

1 / In all the circuits before post #29 a bridge rectifier is used with the capacitor on the AC side. In post #29 you show only half wave rectification.

Shunt resistor question. The shunt (R4) resistor will dissipate some power during the day. If it is a low value the amount of power will be low.
Your table of power dissipated in the load (The lamp) shows that the maximum power is when the load resistance is equal to the reactance of the capacitor. (You have demonstrated the maximum power transfer theorem.)
I see no way to work out the rating of the original bulb (Or the value of R4). It' current rating must be less than 40 mA. I would guess that it was probably 12 or 24 volts. To get the maximum light output it would need to be 85 volt and 30 mA rating.

Les.

 

Thanks guys!

1 / In all the circuits before post #29 a bridge rectifier is used with the capacitor on the AC side. In post #29 you show only half wave rectification.

I showed only 2 diodes for simplicity in understanding the current path and doing the calculations, of course the 2 other diodes are still there for the other half of the cycle.

I am glad that my calculations seem accurate, and to repair this I would probably have to find a suitable high voltage lamp and also replace C1-- certainly both not worth doing from a cost perspective.

Thanks to all who pitched in to help me through this process!