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https://res.cloudinary.com/engineering-com/image/upload/v1574757210/tips/ThyristorControlledRectifier_uxfgok.pdf Attached is the circuit diagram and It is a Fluke 77 series multimeter.

Although the question was put in that way, later in the discussion I pointed out that the main interest was the AC and DC voltages that I was measuring across the coil. But Waross pointed out that it could be due to metering errors.

Not yet, it is not speculation it is based on well established formulas.

The reason why I calculated the Vrms (Vp / √2) of the rectifier output rather than the Vav (2Vm /π) is to give the DC equivalent for the same power as the rectifier output is a complex waveform.

Why do you need the coil to have enough inductance please, is it because of its filtering effect.

So what is the correlation between AC and DC voltages that I measured across the coil in your opinion.

DC current was measured in series with the coil and the AC and DC voltages across the coil. In fact, Also, I have measured the AC current and it was 2.54 A. Both were measured with a multimeter changing modes between AC and DC. I was only interested in DC current to the coil but also measured AC...

Hi itsmoked, My sketchy maths check, The Vrms of a full bridge rectifier output is Voltage peak / √2 and the peak voltage of an Vrms = Vrms * √2. Therefore, 110V Ac rms input to a full bridge rectifier will yield 110V rms at the output, 110 * √2 / √2. Hi IRstuff, The reason why raised the...

I have a full wave rectifier thyristor control circuit on a 10 Ohms resistor coil which produced the following results with 240V AC rms. DC Current through coil (A) DC Voltage across Coil (V) AC Voltage across coil (V rms) 1.15 10.8 92.0 2.43 22.5...

if you measure the outputs A and B of a pulse width modulation circuit at zero duty cycle, it will measure equal voltage at A and B outputs with the same duty cycle with respect to ground resulting in 0V. What will the output will look like please if measured between A and B outputs please.

Having re-considered the options suggested, i will use a diode and a resistor in series with the heating element. Heater 110V 40 W I = 40 / 110 = 0.36 A Heating element resistance = 110 / 0.36 = 305 R Diode will reduce the 240 rms to 240 * 1.414 / 2 = 170 V rms A 300 R resistor in series with...

Calculations rule out using a diode so I will go ahead and use a resistor in series with the heater coil as planned originally and waste some energy in the process.

Hi Itsmoked, The way I look at it was as follows, Vdc across the heater coil is 240V rms * 0.45 = 108 through a diode (half wave rectification). Current across the heater coil is V/R = 108/302 = 0.36A and power = I^2 * R = 38.6 W. Therefore, 110V AC directly or 240V AC through a diode will...

Hi Keith and Waross For a half wave rectifier Vdc = 0.45Vrms = 240 * 0.45 = 108 V DC Effectively, the voltage across the heater will be the same as applying 110V AC so the power will be almost the same (110V 40W) whether you apply 110V AC directly or 240V AC through a diode.

Considering all the options suggested, I have decided to use a diode in series with the heater.

Thank you for your replies but I would like to use an appropriate power rated resistor, any comments on use of resistor please.

I have a 7.5 kW motor with 110V, 40W anti condensation heater. But i have only 240 V supply for the heater, can I put a resistor in series with the heater to reduce the voltage please.