Hi All

Attached is a diagram of a full wave power supply with a centre tap and 2 diodes.

The question asks which waveform is the output waveform measured at point A with respect to point B.

The answer is (b) as with the diodes connected as shown, point A will be negative with respect to point B.

Hmmmm. Try as I might I can't work out why. The way I see it is that electron flow (negative) will start at the centre tap and travel to point B first, then through the resistor to point A then back home through the top diode.

So shouldn't A be positive with respect to point B?

Really confused here...

 

The way I see it is that electron flow (negative) will start at the centre tap and travel to point B first, then through the resistor to point A then back home through the top diode.

Electron flow is in a direction opposite the arrows on the diode symbol. Confusing, isn't it! So electrons go from point A to point B. Conventional current (defined for positive charge flow) would be from point B to point A; hence B is more positive than A.

 

The diode's arrow points to the direction positive charges would flow. However, because electrons are negatively charged they flow in the opposite direction and thus the correct answer is b.

 

Thank you both for your responses, I am hoping you will keep in touch with this thread as I am still not clear on exactly how the current flows in these types of circuits (half wave, full wave and Bridge Type power supplies) and have a few more questions in this area that I would like to ask along the way.

Refering back to the original attachment here is how I think the current flows:

If we start with the top of the secondary coil being negative the current will flow through the top diode then through A, then RL, then B then back to the centre tap.

Then (if I am correct so far?) the wave reverses so the bottom of the secondary coil now becomes negative and the current will flow from the bottom of the coil, turn left then right, then though A, then RL, then B, then back to the centre tap.

This means the polarity of the secondary coil will alternate with the AC current (I think?).

So no matter whether the current is going "forwards" or "backwards" it will always travel through A first.

Is this description of current flow correct?

I need to know as I have some more complex power wave circuits where current flow confuses me so I thought I would try and understand this one first.

If my description of the current flow is not correct could you please describe the current flow and include both directions. Thanks!

 

I'll try to explain it as electron flow.

If the top of the secondary is negative then the current (electron) will flow through the top diode it is blocked by the bottom diode so it has to go through the load from the point A to point B exit point B and flow to the tap.

If the secondary winding is positive on top, then ELECTRON flow will flow from the negative side of winding (secondary) through the bottom diode it is blocked at the top diode so current must flow again through the load enterung point A and exiting point B back to the tap.

Electron current flows from negative to positive, so in both alerations the current is entering the load at point A and exiting at point B making point B more positive then A.

 

Thanks Hobbyist, good explaination.

Two more questions:

Q1.
Could I just ask if AC swaps polarity when the electron flow changes direction, how does it make sense to have a positive and negative wire at an AC power socket

Q2.
Attached is an unfiltered bridge-type power supply. My text says if you reverse the direction of all four diodes, you will "reverse the output polarity" meaning the "output voltage will be a series of negative pulses".

This confuses me no end If the electron flow is alternating how can you get negative or positive pulses? Shouldn't pulses always be negative regardless of the direction of the current because you always follow electron flow - which is negative??

Or does it mean that because "ground" in the diagram is negative a voltmeter would show a negative voltage when the diodes are reversed?

Or does it mean you have not reversed the voltmeter terminals or what

Any feedback appreciated...

 

Hello,

Take a look in this PDF : http://technology.niagarac.on.ca/courses/etec1120/Files/Unit12.pdf

Greetings,
Bertus

 

If you understand it when the output is positive pulses then you should understand it when the pulses are negative.

If the pulses are positive or negative depends where you place the probes of the voltmeter.

Reversing the diodes has the same results as reversing the voltmeter's probes without reversing the diodes.

 

Hello bertus, thanks! That reference is really good! It cleared it up for me in the half wave description

Thanks again!

 

Thanks mik3, makes it even clearer.

 

Hi again all,

Can't thank you enough for clearing things up.

I have another question.

The attached circuit is a half wave "filtered" power supply. I am asked to calculate the maximum output voltage when the supply is "unloaded".

My attempt:

As is an "unloaded" supply the max output voltage is the peak voltage:

So: Peak = rms x 1.414 = 14v x 1.414 = 19.8 V.

I think the trick here is NOT to factor in:

Eout = 0.318(Ein(peak))

But instead just calculate the peak because the circuit is unloaded.

So the answer is 19.8V. Is my thinking correct?

 

If the output is taken between points A and B then the maximum voltage is 19.8V peak as you calculated.

If the output voltage is taken between A and C or B and C then the maximum voltage is 19.8/2 V.

 

Thanks mik3.

Yet another related question. If the "filter" (a capacitor) is removed from this circuit so its now an unfiltered half wave what would be the measured output voltage?

My attempt:

I am guessing because there is no capacitor in the circuit the "unloaded" half wave power supply the voltage through the rectifier (diode) would not reach peak voltage because there is no capacitor to hold the peak charge so this is the formula to use:

Eout = 0.318(Ein(peak)) = 0.318 x 19.8 = 6.3 V.

I can only guess the voltage drop is due to the diode in the "unloaded" rectifier? If so, I am not sure why its the peak reduced by 0.318

 

The peak voltage will be the same.

It is the average voltage that changes.

 

So is 6.3V the output voltage or not?

 

So is 6.3V the output voltage or not?

If you are looking for the DC voltage, yes.

If you are looking for the peak, no. It is the same as with the filter.