Forum members never cease to amaze me

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ohmhead

ohmhead

Senior Member
Location
ORLANDO FLA
Well Gar , lets see we have a test circuit= 23.16 volts ac supply to a diode and resistor in a series circuit R= 13.79k & D= 1N4005 on input to cathode wve got 23.16 volts ac and on the anode side wve got 10.16 dc volts output .By Kirchhoffs voltage law the voltage drop across the load plus the vd across the diode must add up to equal the source voltage . The peak voltage across the load is the peak voltage of the source voltage then by ohms law the peak load current and diode current of same circuit . We can show that the average value of a half of a sine wave over a full ac cycle is the peak value divided by pie 3.1416 . You must change rms voltage to peak voltage 23.16 x 1.41 = 32.65 v next 32.65 peak volts divided by pie 3.1416 = 10.39 volts dc output of diode on anode . The load voltage waveform with applied voltage waveform shows that there is one pulse of load voltage for each full ac cycle of applied voltage the fundamental frequency in the output is the frequency of the supply .When a diode is used in a half wave rectifier with a resistor load ratings are important like peak diode current is the peak load current . peak inverse voltage is the peak voltage of the ac source . diode current is the current in the load . Yes we know how a meter works and thats why the voltage is lower on that diode output effective = rms meter see rms only . Take care Gar
 
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chris kennedy

chris kennedy

Senior Member
Location
Miami Fla.
Occupation
60 yr old tool twisting electrician
ohmhead said:
Well Gar , lets see we have a test circuit= 23.16 volts ac supply to a diode and resistor in a series circuit R= 13.79k & D= 1N4005 on input to cathode wve got 23.16 volts ac and on the anode side wve got 10.16 dc volts output .By Kirchhoffs voltage law the voltage drop across the load plus the vd across the diode must add up to equal the source voltage . The peak voltage across the load is the peak voltage of the source voltage then by ohms law the peak load current and diode current of same circuit . we can show that the average value of a half of a sine wave over a full ac cycle is the peak value divided by pie 3.1416 . You must change rms voltage to peak voltage 23.16 x 1.41 = 32.65 v next 32.65 peak volts divided by pie 3.1416 = 10.39 volts dc output of diode on anode . The load voltage waveform with applied voltage waveform shows that there is one pulse of load voltage for each full ac cycle of applied voltage the fundamental frequency in the output is the frequency of the supply .When a diode is used in a half wave rectifier with a resistor load ratings are important like peak diode current is the peak load current . peak inverse voltage is the peak voltage of the ac source . diode current is the current in the load . Yes we know how a meter works and thats why the voltage is lower on that diode output effective = rms meter see rms only . Take care Gar
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I'll just get a new meter.
 
ohmhead

ohmhead

Senior Member
Location
ORLANDO FLA
roger said:
Chris, just stay with Gar, he'll get you through the weeds.

Roger
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Well Roger Gar was correct and what i found was words to prove what he said was correct were here to learn but were also able to show why in simple terms how that problem or question was found . I think Gar is the only one who can understand this . yes i do Gar its hard for a electrician to look at dc and ac at the same point in time testing take care
 
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