Output of a half wave rectifier

[esha]

In a half wave rectifier only a single diode is present. One end of the secondary wire of the transistor is connected to the p side of diode while the other to the load resistor. The n side is connected to the load resistor. When the diode is reverse biased no current passes through it. But current does pass through the other wire of the secondary which eventually meets the load transistor. So wouldn't the output in this case be an ac current?

 

[davenn]

In a half wave rectifier only a single diode is present. One end of the secondary wire of the transistor is connected to the p side of diode while the other to the load resistor. The n side is connected to the load resistor. When the diode is reverse biased no current passes through it. But current does pass through the other wire of the secondary which eventually meets the load transistor. So wouldn't the output in this case be an ac current?

hi there

just a few wrong words in there

"One end of the secondary wire of the transistor is connected..." should be One end of the secondary wire of the transformer is connected...

"But current does pass through the other wire of the secondary which eventually meets the load transistor."

should be ... But current does pass through the other wire of the secondary which eventually meets the load resistor.

well it's a pulsed AC ... have you looked online to see what the waveform looks like ?

the top is the full AC voltage on the secondary side of the transformer
the lower is the half wave rectified voltage across the load resistor after the diode

half-wave rectification in general uses larger value smoothing capacitors to help keep the voltage from sagging (dropping) so much between each pulse.

half-wave rectification is also generally only used where the equipment isn't sensitive to lots of AC ripple voltage
Dave

 

[esha]

thanks for correcting me... m new at this so i mixed up all the terms... i know that the graph of the half wave rectifier looks that way... but my point is the diode less wire of the secondary will conduct current no matter what happens... so why didnt the graph consider thqt

 

[davenn]

but my point is the diode less wire of the secondary will conduct current no matter what happens... so why didnt the graph consider thqt

no it doesn't, there is only current flowing in one direction, and that is the 1/2 cycle when the diode is conducting
NO current is flowing during the other 1/2 cycle

 

[phinds]

thanks for correcting me... m new at this so i mixed up all the terms... i know that the graph of the half wave rectifier looks that way... but my point is the diode less wire of the secondary will conduct current no matter what happens... so why didnt the graph consider thqt

And just where is that current going to come from?

 

[esha]

No current is flowing through the DIODE. But current should flow from the other wire

 

[esha]

from the secondary transformer due to mutual induction

 

[davenn]

No current is flowing through the DIODE. But current should flow from the other wire

no ... why would you say that ? there is a hole in the circuit and for current to flow, there needs to be a complete circuit

from the secondary transformer due to mutual induction

no

Dave

 

[esha]

why is there a hole in the circuit

 

[davenn]

why is there a hole in the circuit

you have already answered that yourself

No current is flowing through the DIODE.

because the diode is reversed biased and call it a hole in the circuit or maybe a closed door
whichever description, no current is flowing through it in the other 1/2 cycle, therefore no current flows anywhere in the circuit during that 1/2 cycle

 

[esha]

ohh... i got it... i was being so stupid

 

[davenn]

i was being so stupid

no, never say that of yourself ... never put yourself down like that
it was just a small lack of understanding

Dave

 

[esha]

but when the circuit is complete the diode wire has a pulsating current while the other wire has an ac current. Wouldnt that make things messy at the load resistor

 

[davenn]

but when the circuit is complete the diode wire has a pulsating current while the other wire has an ac current. Wouldnt that make things messy at the load resistor

when the diode conducts ( and it conducts in ONE DIRECTION only), there is current flowing in one direction only. so the load resistor sees just that pulsed voltage across it

 

[esha]

i mean the other wire... during forward bias circuit is complete. Current naturally flows through the other wire

 

[cnh1995]

i mean the other wire... during forward bias circuit is complete. Current naturally flows through the other wire

What "other" wire? Do you mean transformer primary?

 

[esha]

no the other wire coming from the secondary and directly meeting the load transistor

 

[cnh1995]

no the other wire coming from the secondary and directly meeting the load transistor

There's no transistor. I assume you mean transformer again.
This is the circuit we're talking about.

So what is your exact question here?

 

[esha]

lets consider the wire connected to diode A and the wire connected to the load resistor B. When forward bias occurs then current flows through A and B. A due to the diode has an output current in the form of pulses. But B has no diode therefore the current would be ac through it. Both of them will pass through the load transistor. I am saying that the output would be quite messy then

 

[davenn]

lets consider the wire connected to diode A and the wire connected to the load resistor B. When forward bias occurs then current flows through A and B. A due to the diode has an output current in the form of pulses. But B has no diode therefore the current would be ac through it. Both of them will pass through the load transistor. I am saying that the output would be quite messy then

no, you are still not quite understanding

look at cnh1995's circuit ...

and what I said in my previous post

during the positive 1/2 cycle, when the diode conducts, it ONLY CONDUCTS in one direction ( we will use conventional current flow), in the above circuit that is out of the top of the transformer secondary, where the red + is through the diode, through the load resistor through the bottom wire and back into the transformer

during the other 1/2 cycle (the negative 1/2) there is no conduction, the diode door is shut
so current cannot flow into the bottom wire through the resistor and up to the diode.
there has to be a complete circuit path for the current to flow

so again repeating what I said earlier, the load resistor ONLY sees current flowing through it in one direction
and it see only pulses of voltage as the diode conducts for each positive 1/2 cycleDave

 

[davenn]

But B has no diode therefore the current would be ac through it. Both of them will pass through the load transistor.

during the other 1/2 cycle (the negative 1/2) there is no conduction, the diode door is shut
so current cannot flow into the bottom wire through the resistor and up to the diode.
there has to be a complete circuit path for the current to flow

OK, let's look at this another way using something that works in a similar way

lets take the top wire, diode, resistor and the bottom wire as a pipe full of marbles ( I assume you know what marbles are )

the diode is place at a point in the pipe and you can think of the diode as a one way valve ( only let's the marbles flow in the pipe in one direction)

now the pipe is full of marbles, the only way to push another marble into the pipe is if one leaves the other end of the pipe ... OK ?
now if one marble is pushed in the top end of the pipe, the one way valve ( diode) let's one marble pass through it and it pushes all the other marbles in front of it and they all move one position along the pipe and one marble drops out the other end ( there is the start of a flow = current)

BUT is we try and push a marble in the bottom end of the pipe, it won't go in, because the one way valve will not let a marble pass
and therefore ALL the marble in the pipe stay where they are and the one you are trying to push in WILL NOT go in ... so there is no flow of marbles = no currentdoes that help ?

 

[esha]

yea it does... U r really a persistent teacher... I do wish i had one like u in real life

 

[davenn]

yea it does...

awesome

So if we go back to the marbles in the pipe being the electrons in the wire.
It means for every electron that enters one end of the wire, an electron must leave the other end of the wire to make room for it
If the diode stops the movement of electrons in one direction, then no electrons can move in the wire in that direction
and therefore no electron(s) can enter the wire

 

[esha]

thanks a lot

 

[sophiecentaur]

but when the circuit is complete the diode wire has a pulsating current while the other wire has an ac current. Wouldnt that make things messy at the load resistor

I think this is your basic problem. In a single loop of circuit, there is the same current at every point. If there's a 'switch' (i.e. diode) which is turned off (reverse biased) in one leg then the zero current applies all round the loop. In your simple circuit, the current waveform is identical all the way round.
This only applies to circuits which are much smaller than the wavelength of any electrical disturbance. RF circuitry can't make the same assumptions because lengths of wire also behave as Capacitors and Inductors at high frequency but. one thing at a time.

 

[esha]

ohh.. ok