Night Light

AnalogKid

Joined Aug 1, 2013
12,128
(2) the power dissipation is so low you don't really need a day/night detection mechanism. That's not a bad idea though, but it would take a bit more circuitry to add that feature, like a photocell and transistor or two and some resistors and maybe a capacitor.
Or not. Day/night detection already exists in post #1.

ak
 

MrAl

Joined Jun 17, 2014
13,703
Or not. Day/night detection already exists in post #1.

ak
Hi,

Oh that's a possibility then, depending on the retrofit work/parts required.

Personally, I'd just go out and buy one or more likely order one online :)
Cheap way out. Just checked Amazon, 2 for $6 USD. Probably cheaper ones out there too.

Indicator lamps are easy to make with LEDs though. A resistor and LED with a diode. The resistor can be high value like 50k or something for a 120vac line system so not too much heat. The high brightness LEDs don't mind just a little current for a little indicator brightness.
 

AnalogKid

Joined Aug 1, 2013
12,128
You might be able to reduce power dissipation by adding a resistor of the right value in series with D2.
That's the idea behind increasing R2 - it returns both the D2 current and the C2 peak voltage to the values of the original 120 V application.

Of course a bridge rectifier is always better.
Disagree. It would reduce flicker to the point that C2 might not be necessary, but it is larger, more expensive, and takes up more board space.

ak
 

MrAl

Joined Jun 17, 2014
13,703
Thinking, if I take the photo cell part out of the circuit, would that make any difference has to runing a led bulb ?
To reply again, if the circuitry will work with an LED too but it would take some modifications. You could trace out the circuit and show it here so we can figure this out.

You know if you buy one they are quite cheap online. You will miss out on the hobby value that way though of course.
 

AnalogKid

Joined Aug 1, 2013
12,128
Re-thinking the #12 circuit . . .

At 50 Hz, R1 || C1 = 8.8 K, while at 60 Hz it is 7.44K. Increasing R2 by 1 K was a seat-of-the-pants estimate of what it would take to adapt the circuit to twice the original input voltage. Accounting for frequency, it looks like the correct increase is 1.3 K. But . . .

Wait. R1-C1 is higher (than the rest of the circuit) enough in impedance that the circuit front end looks more like a current source than a voltage source. At 60 Hz, everything to the left of D2 looks like 8.44K, for a peak D2 current of 19 mA. At 240 V and 50 Hz, the correct equivalent front-end impedance for the same D2 current is 17.8 K. Thus, R2 should be increased from 1.0 K to 9.1 K.

Or leave R2 at 1.0 K, and decrease the value of C1 to 0.15 uF. This is the better solution; C1 becomes smaller and cheaper, and everything else is unchanged.

Or something like that.

ak
 
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Tonyr1084

Joined Sep 24, 2015
9,744
So I take it, that there is nothing I can do to this circuit, to be able to run a led bulb !
If the halogen bulb worked before and you want to replace it with LED, simply find an LED that is designed to work on that voltage and install it somehow. I don't know what sort of bulb base the existing unit is so you may have to do some modifications. On the other hand, LED night lights are so cheap you can probably get one for less cost than getting an LED that runs on the voltage the halogen bulb used to work on. Sometimes it's cheaper to toss something than it is to actually fix it. But I understand the desire to fix. It's ingrained in most of us here.
 

MrAl

Joined Jun 17, 2014
13,703
There is no guarantee that an off the shelf 240-volt LED will work with that circuit.
Yes, unfortunately that's true too.
I still think it's better to start from scratch that way you set all the parameters too, no guess work. This has been done many times in the past. You also get all brand new parts in there. Complete control over power dissipation, heat, illumination levels, color(s) of the LED, etc. There are some night lights now that change color too but I do not think I would want that.
 
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