You could. But to do it safely it would be better to use a step-down transformer (to provide mains isolation) with a 4.5W resistive load. To get 4.5W from the LED bulbs they could simply be connected in parallel: not series.

There isn't enough current from the power source to connect them in parallel, but there is enough voltage hence why they are connected in series. It seems it isn't as trivial (for a novice) as I first thought.

 

There isn't enough current from the power source to connect them in parallel, but there is enough voltage hence why they are connected in series. It seems it isn't as trivial (for a novice) as I first thought.

So simplify it.
Assume that the supply is 230V DC not 230V rms AC.
The capacitor has an impedance at 50Hz of about 6.8k, so assume it is a resistor of 6.8k.
You no longer need the bridge rectifiers.
Now can you see what's going on?

 

Connecting them the way you propose would mean that you had 10 capacitors in series which means you one tenth of the capacitance. This means that you have 10 times the reactance. In the existing lamps there are effectively 6 LEDs in series which (Assuming white LEDs) would give a voltage drop of about 20 volts. So with a 240 for a single lamp you would have 220 volts across the capacitor. (I am ignoring the small voltage drop across the bridge rectifier and the 330 ohm resistor for simplicity.)
When you have the 10 lamps in series there will be about 200 volts across the LEDs. So now you only have 40 volts across the capacitor. ("The capacitor" in this context is the 10 original ones in series.) So the ratio of voltage across the capacitor is 40/220. But now the reactance of the 10 capacitors in series is 10 times the reactance of a single capacitor. So the ratio of the new current is 40/220 x 1/10 = 40/2200 = 2/110 of the original current. So the light output of each lamp will be just under 2% of the original. so the total light output for the 10 lamps will be about 20% of the original output of one lamp. (This is assuming a linear relationship between LED current and light output.)

Les.

 

When you have the 10 lamps in series there will be about 200 volts across the LEDs. So now you only have 40 volts across the capacitor.

If there is 200V across the LEDs and the supply is 240V then there is 132V across the capacitor = SQRT(240^2 - 200^2)
because the capacitor voltage is 90 degrees out of phase with the LED voltage.
Or am I just being picky, and spoiling the point you are trying to make?

 

Hi Ian, Seeing your comment in post #22 about not needing bridge rectifier are you proposing having two groups of 5 of the original LED configuration connected in anti phase parallel so each group is on for half a cycle ? So you would just be using the LEDs from the original lamps together with a suitable value capacitor ?
Bhope691, Can you explain why the power supply can't supply the lamps connected in parallel ?

Les.

 

Hi Ian, Seeing your comment in post #22 about not needing bridge rectifier are you proposing having two groups of 5 of the original LED configuration connected in anti phase parallel so each group is on for half a cycle ? So you would just be using the LEDs from the original lamps together with a suitable value capacitor ?
Bhope691, Can you explain why the power supply can't supply the lamps connected in parallel ?

Les.

I was suggesting that the TS should start by trying to understand a similar system with a DC supply.

(I did suggest earlier that he could increase the power by putting more LEDs in series on the same capacitor.)

 

Ian, I have just read post #24. You are correct I forgot to do vector addition. I normally ignore it when most of the voltage drop is across the capacitor but ignoring it as I have with such a large number of LEDs in series gives a significant error.
Edit. Making the point will make the TS realise that is even more complicated than he thought.
Les.

 

Bhope691, Can you explain why the power supply can't supply the lamps connected in parallel ?

It's a bespoke portable power supply which can only supply a small amount of current - approx 7mA max but high voltages.

I was suggesting that the TS should start by trying to understand a similar system with a DC supply.

So if I consider the 10 LEDs as being powered by a DC supply I can then work out what a single circuit (combination of the 10 LEDs) would look like under the same conditions, and what is being dropped across each of the components? Or am I getting more confused?

 

It's a bespoke portable power supply which can only supply a small amount of current - approx 7mA max but high voltages.

Then there's no way you can get 4.5W out of it using those LED bulbs. 180 LEDs x 3V (Vfwd) x 7mA = 3.78W.

 

So if I consider the 10 LEDs as being powered by a DC supply I can then work out what a single circuit (combination of the 10 LEDs) would look like under the same conditions, and what is being dropped across each of the components? Or am I getting more confused?

Yes - that's the idea.
That's where we get RMS from - it's the amount of AC voltage that produces the same power in a resistor as DC voltage.

 

If your power supply can only give 7mA, how did you measure 33mA in your post #1?

 

Short-circuit output current for 470nF should be 33mA. So where did you get 7mA from?

 

In post #28 the TS specified that the maximum output from the power source was 7 mA.

Les.

 

Then there's no way you can get 4.5W out of it using those LED bulbs. 180 LEDs x 3V (Vfwd) x 7mA = 3.78W.

The total drop measured is over the 10 LED capsules, is there not power lost on the other components?

Short-circuit output current for 470nF should be 33mA.

The output current can only be 33mA, it can't be less?

 

1
It is just a power calculation. 180 x 3 = 540 volts. The maximum current you can supply is 7 mA so 540 volts x 0.007 amps = 3.78 watts.
BUT you have 240 volts available so your AVAILABLE power is 240 volts x 0.007 amps = 1.68 watts.
2
Not if it has 230 volts at 50hz connected across it.
Why do you want to use this mysterious power source ? It seems totally unsuitable for your purpose.

Les.

 

The output current can only be 33mA, it can't be less?

It can be less. 33mA is when the full 230V is across the capacitor. With more voltage across the LEDs, there's less across the capacitor, so less current flows.

 

Yes - that's the idea.
That's where we get RMS from - it's the amount of AC voltage that produces the same power in a resistor as DC voltage.

Then "one" circuit would essentially be a summation of the capacitors and resistors in the "ten" circuits?

I did suggest earlier that he could increase the power by putting more LEDs in series on the same capacitor.

Which capacitor would I put more LEDs in series on?

 

Yes it will be less when in series with a number of LEDs
Is your power source 240 volts AC 50 hz and a PURE sine wave ?

Les.

 

Is your power source 240 volts AC 50 hz and a PURE sine wave ?

No

Why do you want to use this mysterious power source ? It seems totally unsuitable for your purpose.

I wanted to understand the power source and how it works with LEDs compared to purely resistive loads.

 

I wanted to understand the power source and how it works with LEDs compared to purely resistive loads.

With a purely resistive load, it is a simple capacitor and resistor in series problem.
I = V/SQRT(R^2+1/(2*PI*F*C)^2)
you can ignore the bridge rectifier.

With the LEDs it gets more complicated.