Conduction Angle For Half-Wave vs Full-Wave Rectifier

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
On a quiz my professor wanted to know the percent of the input voltage cycle that the diode conducts for the following circuit, where the graph represents output voltage versus input voltage. For this case, the correct procedure was to use the conduction angle formula and divide the result by 2 * pie. What if the circuit was a full-wave rectifier? Would you simply divide by pie instead?:

Conduction angle: SquareRoot( (2 * RippleVoltage) / (Max Output Voltage) )

example1.jpg
 

wayneh

Joined Sep 9, 2010
17,153
Sounds good to me. It's just semantics, but I'd still say you divide by 2π because that's one complete cycle of the input. Of course you get two output cycles.
 

WBahn

Joined Mar 31, 2012
26,398
Do you understand where that formula for conduction angle comes from? If so, then you should be able to reason out whether your proposed solution is correct or, even better, follow the same line of reasoning to come up with a corresponding formula for a full-wave rectifier?
 

WBahn

Joined Mar 31, 2012
26,398
Sounds good to me. It's just semantics, but I'd still say you divide by 2π because that's one complete cycle of the input. Of course you get two output cycles.
It doesn't matter whether you call a cycle the period of the input or the period of the output, the diode is conducting for twice the duty cycle that it was before. But that's with the same ripple voltage. If the voltage droops at the same rate, then the ripple voltage is only going to be about half of what it was. So if the question is what the conduction angle would be for that same circuit except with a full-wave rectifier, then that would need to be taken into account -- and that's a nonlinear effect (but one that is pretty easy to see the estimated effect on the answer).
 

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
Ok, so I looked in a different textbook and received a clearer explanation. The equation gives the amount of degrees over which the diode conducts. So, you just see how many times the diode conducts over the time period in question and multiply it by the equation result.
 
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WBahn

Joined Mar 31, 2012
26,398
Ok, so I looked in a different textbook and received a clearer explanation. The equation gives the amount of degrees over which the diode conducts. So, you just see how many times the diode conducts over the time period in question and multiply it by the equation result.
So, based on your current understanding, it they were to take that same circuit and make it full-wave rectified, but kept the same R and C, would you expect the conduction angle to double?
 

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
During a single cycle of the input voltage it would conduct over a region of the cycle equal to the conduction angle, twice, once on the positive half of the sin-wave, once on the negative half of the sin-wave. So either 2 x formula / 2pi or formula/pi, for percent of cycle conducting.
 

WBahn

Joined Mar 31, 2012
26,398
During a single cycle of the input voltage it would conduct over a region of the cycle equal to the conduction angle, twice, once on the positive half of the sin-wave, once on the negative half of the sin-wave. So either 2 x formula / 2pi or formula/pi, for percent of cycle conducting.
So you're saying that the ripple voltage will be the same in either case?
 

WBahn

Joined Mar 31, 2012
26,398
The formula for the conduction angle stays the same, but the ripple voltage is half as much for the full-wave case.
I agree that the ripple voltage is half as much. But why would the conduction angle formula stay the same?

Take some time and make sure you understand how to DERIVE the conduction angle formula for the half-wave rectified case. Then derive the formula for the full-wave rectified case. Then plug half the ripple voltage into the full-wave formula.

What is the ratio CAfull / CAhalf when the R and C remain the same?
 

MrAl

Joined Jun 17, 2014
8,479
Hello,

I've seen arguments like this when calculating the Fourier components of a half wave and full wave rectified sine.
For the half wave nobody seems to have a problem but for the full wave there's always a debate about what the fundamental should be.

For example, for a 50 Hz system and full wave rectifier, the output has lowest frequency of 100Hz so that is sometimes referred to as the fundamental. However, i always argue that the harmonics should always be in relation to the original fundamental so that we can compare it to other systems of the same type, unless of course there is some very specific reason for doing otherwise.

In this case if we say have a conduction angle of 1 degree with half wave rectification it's 1 degree, but if we have one diode conducting for 1 degree in a full wave rectified system where that 1 degree is over one cycle of the original frequency of 50Hz, then we might want to call it 2 degrees because we end up having two diodes conducting for 1 degree over the 50Hz cycle, and so the total conduction is 2 degrees.

Perhaps it is better to specify what the conduction period refers itself to, either a half cycle of the original or the full cycle of the original. After all, a half cycle of a full wave rectified sine is a full cycle at twice the original frequency even though it is not a sine anymore.
 

Thread Starter

WilkinsMicawber

Joined Jun 5, 2017
29
Alright, so after spending a bunch of time thinking about it (way more than should be necessary), you just do VmSin(theta) = Vm - Vr, then subtract the answer from 90 degrees to get the magnitude of the angle.

I'm not sure why the shorthand formula was introduced in the first place. No need to memorize it, doesn't save any time.

As far as I can tell, there is no difference from the full wave rectifier, as the same factors are at play, only the ripply voltage is half as much because the capacitor discharges for half the time as the half-wave before recharging.
 

WBahn

Joined Mar 31, 2012
26,398
Alright, so after spending a bunch of time thinking about it (way more than should be necessary), you just do VmSin(theta) = Vm - Vr, then subtract the answer from 90 degrees to get the magnitude of the angle.
Yes, although that reduces to Vm·cos(θ) = Vm - Vr

I'm not sure why the shorthand formula was introduced in the first place. No need to memorize it, doesn't save any time.
There's a number of reasons. One, a lot of people that need to work with this stuff have very poor math skills. There are books out there with nothing but page after page of formulas because of the number of people working in the field that have zero algebra skills and so they need a formula for everything that has every single thing needed exactly as they need to plug it in. Second, people are much more likely to make mistakes when trig function are involved, if for no other reason than many people can't keep the distinction between radians and degrees clear. In addition, algebraic equations are much more likely to work better when working on larger problems than trig equations are.

As far as I can tell, there is no difference from the full wave rectifier, as the same factors are at play, only the ripply voltage is half as much because the capacitor discharges for half the time as the half-wave before recharging.
Consider that there are two factors at play. First, the total charge that needs to be replenished during each period is the same for both circuits, namely half of the charge that was consumed by the load. With two conduction periods per cycle, each cycle only has to replenish half the charge of the single conduction period, so if the current is the same throughout the conduction period, the total conduction time would need to be the same, just split into two parts, so each conduction period would be half the length but the total fraction of the period that the diodes are conducting would be the same. However, because the voltage only droops half as much, the voltage difference between the supply and the capacitor starts out smaller and this means that the things aren't as simple as they might seem at first.

If you use the formula you started with, you would discover that it predicts that CAfull/CAhalf = 1/√2. But this has to be taken with a grain of salt because there are three approximations that go into that formula and two of them have at least the reasonable potential of impacting this result noticeably.
 

MrAl

Joined Jun 17, 2014
8,479
Alright, so after spending a bunch of time thinking about it (way more than should be necessary), you just do VmSin(theta) = Vm - Vr, then subtract the answer from 90 degrees to get the magnitude of the angle.

I'm not sure why the shorthand formula was introduced in the first place. No need to memorize it, doesn't save any time.

As far as I can tell, there is no difference from the full wave rectifier, as the same factors are at play, only the ripply voltage is half as much because the capacitor discharges for half the time as the half-wave before recharging.
Hi,

If you want to figure this out right, you can approximate the conduction part as the top of a sine wave, then calculate when the discharge wave breaks from the sine on the right of a peak, then calculate using the capacitor/resistor discharge curve when it meets up with the next rising sine edge, and that gives you the info you need to calculate the angle to some approximation that isnt too bad although still a bit theoretical.
You can compare results with that formula you were given.

The ripple voltage does approximately halve as we go from the half wave to the full wave version, but it wont be exactly half and you will see that by doing the above. The reason it wont be exactly half is because the exponential discharge is not linear, so that when you double the time you dont halve the voltage, at least not exactly anyway. It may be an allowed approximation however.
 

wayneh

Joined Sep 9, 2010
17,153
It doesn't matter whether you call a cycle the period of the input or the period of the output, the diode is conducting for twice the duty cycle that it was before. But that's with the same ripple voltage. If the voltage droops at the same rate, then the ripple voltage is only going to be about half of what it was. So if the question is what the conduction angle would be for that same circuit except with a full-wave rectifier, then that would need to be taken into account -- and that's a nonlinear effect (but one that is pretty easy to see the estimated effect on the answer).
I completely overlooked that the ripple would be less less and therefore the conduction angle drops. So, it’s not so simple. An interesting practical implication is specifying the rectifier diodes.
 

MrAl

Joined Jun 17, 2014
8,479
I completely overlooked that the ripple would be less less and therefore the conduction angle drops. So, it’s not so simple. An interesting practical implication is specifying the rectifier diodes.
Hi,

Yes, but here i believe we were talking ideal theory where the diode has little or no voltage drop.
 

wayneh

Joined Sep 9, 2010
17,153
Yes, but here i believe we were talking ideal theory where the diode has little or no voltage drop.
I was just referring to the current and heat. The average heat is not so much affected by the nuances of conduction angle, so I think I overstated the impact on design. I mean, if you know the average load current and the topology (half or full rectifier), you know most of all you need to know to specify the right diodes.
 

MrAl

Joined Jun 17, 2014
8,479
Hello again,

Here is an exact solution equation for the ideal half wave rectifier:
w*C*log(K^2/(w^2*C^2*R^2)+K^2)*R-2*atan(1/(w*C*R))+2*asin(K)+3*pi=0

where K is 1 minus the percent ripple, where the ripple is measured peak to peak.
C is capacitance value, R is load resistance. w=2*pi*f.
"log()" is the natural log usually written as "ln()".
So for 10 percent ripple, K=0.9 .

This is not an approximation but an exact equation which when solved for whatever variable is needed results in a theoretically exact value for that variable. Normally we would solve for either K or C. This does of course assume ideal components especially the diode which is considered to have zero resistance when conducting, and that is often acceptable when talking about rectifier circuits especially in an academic setting.

The exact solutions are possible for the ideal cases (full wave or half wave) because there are certain simplifications that do not appear in the non ideal cases when we have to move to much more complicated expressions.
 
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wayneh

Joined Sep 9, 2010
17,153
Here is an exact solution equation for the ideal half wave rectifier:
w*C*log(K^2/(w^2*C^2*R^2)+K^2)*R-2*atan(1/(w*C*R))+2*asin(K)+3*pi=0
I'm highly skeptical there is an exact solution. Can you document the source of this? A numerical solution is a piece of cake by iteration with a modern spreadsheet but as I said, I doubt there is an analytical solution. I spent quite a long time looking for one with no luck.
 
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