Dial Lamp Puzzle.

[Robert Darwent]

Of course! 50Hz AC is 3.14 x RMS Voltage yes? So half wave (zero voltage on the negative going half cycle) is half of 3.14 = 1.57.

No I don't believe so David, I think you're confusing pi with the square root of 2 there.

I'm confused, can you expand on that David.

For the sake of this explanation, assume the mains AC voltage is a pure sine wave, then the following is correct;

V[average] = V[peak] x 0.636

and

V[rms] = V[peak] x 0.707 (0.707 being 1 over the square root of 2)

rearranging the above

V[peak] = V[rms] / 0.707

or

V[peak] = V[rms] x 1.414 (1.414 being the square root of 2)

Now, most meters when measuring AC voltages actually measure the 'average' value so you need to use the above equations if you need the rms or peak voltage.

The mathematical relationship between the average and rms values is 1.11, arrived at by dividing 0.707 by 0.636 and is called 'The Form Factor' of the wave.

It hadn't occurred to me that the value 1.57 is half of pi, but since we are dealing with sine waves here it is probably no coincidence.

Regards

[pwdrive]

That's why I was confused, it did not fit my understanding either, Vout = Vin x .707 (less any forward drop)

Lawrence.

[Joe]

I used to know at least some of this stuff - 0.7071 and 1.414 are certainly ringing distant bells. Isn't the former the cosine of 45°?

I may have learnt it in the past; I'm not claiming to have ever understood it though.

- Joe

[Yorkie]

I am not worthy!

[Robert Darwent]

Of course! 50Hz AC is 3.14 x RMS Voltage yes? So half wave (zero voltage on the negative going half cycle) is half of 3.14 = 1.57.

After giving this a little more thought, you were almost correct David. If you'd put 3.14 x AVERAGE Voltage, you would have been spot on!

A sine wave varies between;

+/- 1 over 0.707 x its RMS value (with 1 over 0.707 = square root of 2)

or

+/- 1 over 0.636 x its AVERAGE value (with 1 over 0.636 = half of pi)

In my reply to Joe earlier, what I was telling him to do by multiplying his meter reading by 1.57 (1.5 to make it easier), is converting the AVERAGE value measured by his meter to the PEAK value;

1.11 (0.707 / 0.636 the form factor of a sine wave) x 1.414 (square root of 2) = 1.57 (half of pi)

Phew, talk about alternative ways of saying the same thing!

Hope this makes things a little clearer, but if it doesn't don't worry I've been confusing myself just as much before I got my head around it!

Regards

[pwdrive]

Yes you are worthy David, it's just a slip of the keyboard. Thanks Robert for the explanation.

So to recap, pk = rms x 1.414, rms = pk x .707, pk to pk = rms x 1.414 x 2

For diode dropper calc, Vout = Vin x .707 then use this product to calculate the value of any additional series resistance required to drop the voltage down to the required total heater supply voltage at the required heater current (have neglected any forward voltage drop by the diode)
Is this correct?

Lawrence.

[Robert Darwent]

So to recap, pk = rms x 1.414, rms = pk x .707, pk to pk = rms x 1.414 x 2

For diode dropper calc, Vout = Vin x .707 then use this product to calculate the value of any additional series resistance required to drop the voltage down to the required total heater supply voltage at the required heater current (have neglected any forward voltage drop by the diode)
Is this correct?

Yes Lawrence, I believe that is all correct.

Regards

[pwdrive]

So to recap, pk = rms x 1.414, rms = pk x .707, pk to pk = rms x 1.414 x 2

For diode dropper calc, Vout = Vin x .707 then use this product to calculate the value of any additional series resistance required to drop the voltage down to the required total heater supply voltage at the required heater current (have neglected any forward voltage drop by the diode)
Is this correct?

Yes Lawrence, I believe that is all correct.

Regards

Thanks Robert, diode or capacitor droppers can be a bit of a trap for the unweary, so I guess this thread has been good for the advancement of the black art of radio repair/restoration.

Lawrence.

[Robert Darwent]

For diode dropper calc, Vout = Vin x .707 then use this product to calculate the value of any additional series resistance required

Actually, Lawrence that is not quite right;

Dropper Resistance = (Vs - (Vh x 1.414)) divided by (Ih x 1.414)

where;

Vs = supply voltage
Vh = heater voltage
Ih = heater current

Regards

[pwdrive]

For diode dropper calc, Vout = Vin x .707 then use this product to calculate the value of any additional series resistance required

Actually, Lawrence that is not quite right;

Dropper Resistance = (Vs - (Vh x 1.414)) divided by (Ih x 1.414)

where;

Vs = supply voltage
Vh = heater voltage
Ih = heater current

Regards

Hi Robert, I always thought that the diode delivers half power therefore the voltage
Vout must be .707 of Vin, that was the basis of my last post, ie: 240 volts in 170 volts out, so R = 170 minus Vh divided by Ih (at 240v mains) dissipation being the usual I sqrd R, I think it basically amounts to the same thing.

Lawrence.