Power losses for half/full-diode-bridge ??

Hello,
I have some concern regarding an equation for calculating power losses for half - and full diode-bridge.
Can someone please check if this equation is correct or not? See the attached image.

"

Fig. 1. Typical diode full-wave bridge.



Full-wave diode bridges are found in many electronic systems (Fig. 1). At typical AC power line voltages, the drop across the diodes has little impact on the rectified output voltage and diode power dissipation. However, diodes in low voltage, high power rectification applications dissipate significant power and the inherent diode drops take a significant bite out of the operating voltage. For example, power dissipation of the diode bridge in Fig. 1 is:



Equation: P = 2 × 0.6 V × IL (1)

Where:

IL = Line current in amperes

P = Power dissipation in watts

0.6 V = Typical voltage drop for one conducting rectifier diode"


What I understand from this equation, this gives total power losses for a whole bridge, by multiplying with 2.
If I had multiplying with 1, I would obtain half of the power losses (which would be for a half-diode bridge).

I really need a confirmation of this soon...

Ref:
https://www.powerelectronics.com/pm...er-cuts-power-dissipation-improves-efficiency

That power equation is an idealistic one which ignores the fact that the diode voltage drop increases as the load current increases.
What is "a half-diode bridge"?

alec_t said:

That power equation is an idealistic one which ignores the fact that the diode voltage drop increases as the load current increases.
What is "a half-diode bridge"?

Click to expand...

Yes, but why multiplying with '2'?

For my case I have this "Totem-pole" that I have talked about earlier. It looks liks this, see the attached image. Where there are two diodes and two mosfets, so thats why I called it "half diode-bridge".

mading2018 said:

Hello,
I have some concern regarding an equation for calculating power losses for half - and full diode-bridge.
Can someone please check if this equation is correct or not? See the attached image.

"

Fig. 1. Typical diode full-wave bridge.



Full-wave diode bridges are found in many electronic systems (Fig. 1). At typical AC power line voltages, the drop across the diodes has little impact on the rectified output voltage and diode power dissipation. However, diodes in low voltage, high power rectification applications dissipate significant power and the inherent diode drops take a significant bite out of the operating voltage. For example, power dissipation of the diode bridge in Fig. 1 is:



Equation: P = 2 × 0.6 V × IL (1)

Where:

IL = Line current in amperes

P = Power dissipation in watts

0.6 V = Typical voltage drop for one conducting rectifier diode"


What I understand from this equation, this gives total power losses for a whole bridge, by multiplying with 2.
If I had multiplying with 1, I would obtain half of the power losses (which would be for a half-diode bridge).

I really need a confirmation of this soon...

Ref:
https://www.powerelectronics.com/pm...er-cuts-power-dissipation-improves-efficiency

Click to expand...

Hello there,

That equation would be "somewhat" true if there was a resistive load only or perhaps an inductor input filter. With no inductor and an output filter cap the diode waveshape is very peaked and narrow. The capacitor makes it very hard to calculate the voltage drop of the diode(s) without doing a complete analysis and using something like a spice model of the diode. In fact, the best bet if you are not heavy into the math is to use a simulator and measure the power in one diode and then go from there.

With just a resistive load, the current peak follows the sine shape and so we might use an approximation like we see in that equation, but when we add a capacitor what happens is the diode does not conduct at all for some parts of the cycle, then suddenly the current jumps up creating as current surge that is hard to calculate without a good spice model. So that equation is a very very vague approximation at best.

We could easily do a simulation and compare results. If we did get lucky the equation would match the simulation results, but it would be pure luck as there is no physical theory that i know of that says that should be true.

mading2018 said:

why multiplying with '2'?

Click to expand...

Because in either half cycle the current has to pass through two diodes of the bridge.

It is true this version of the PFC saves 0.6 volts out of 220 volts. My last PFC pulls 1A from the power line. That is 600mW savings. I think it might cost 600mW to drive the top MOSFET like this. The theory is that because current flows through less parts there is less power loss.

Reality is that the number of parts will increase greatly. More parts less reliability. More parts more cost. There is no IC to do this job. More parts more power loss. More Gates to drive = more power loss. I think the total power saving will be 1/2 what you think.

I think for small power supplies this is no help. Probably at 200 watts there is no savings in power.
If you are to build a 3000 watt power supply this PFC will save power. If the IC exists then yes make a supply like this.

The "market" has said clearly that this type if PFC is not worth investing in. TI, Linear, ST and all the PWM companies have not made parts.

alec_t said:

Because in either half cycle the current has to pass through two diodes of the bridge.

Click to expand...

Is it the same concept for a full-bridge that have MOSFETS instead for diodes?

If I calculate the power losses for one MOSFET, and then I take that multiplying by 2, I will get the power losses for the whole bridge? Is should be correct since they are also operating as pair
as the case with diodes.

ronsimpson said:

The "market" has said clearly that this type if PFC is not worth investing in. TI, Linear, ST and all the PWM companies have not made parts.

Click to expand...

Can you please show me that source? I haven't found information about that.
I asking cause we need to selected a PFC for low-power application (around 900W) that we should have...boost or totem? We are considering Totem cause it seems it have a higher efficiency, and thereby lower losses compared to boost.. For our case, it would be okay to select Totem since we have 900 W which is more than 200 W.

mading2018 said:

Can you please have that source?

Click to expand...

The source is Ron Simpson.
I have looked at 45 different companies that make PWN ICs and none have Totem PFC.
I have looked at 20 different companies that make PFC ICs and none have Totem PFC.

Here are some application notes. Several time they use large DSP micro computers as the PFC IC. This is a very large project. Very complicated. Good luck.

https://www.fujitsu.com/downloads/MICRO/fme/transphorm/TDPS2800E2C1_AppNote_rev1.1.pdf
https://unitedsic.com/wp-content/uploads/2016/11/Totem-Pole-PFC-AppNotes.pdf
**broken link removed**
https://www.ti.com/lit/an/slyt718/slyt718.pdf
https://training.ti.com/sites/defau...g a 99% Efficient Totem Pole PFC with GaN.pdf

mading2018 said:

If I calculate the power losses for one MOSFET, and then I take that multiplying by 2, I will get the power losses for the whole bridge?

Click to expand...

In theory, yes. Reality has a nasty habit of disagreeing with theory, though .

alec_t said:

In theory, yes. Reality has a nasty habit of disagreeing with theory, though .

Click to expand...

Hmm..okay But if only look at the conduction losses, which is this equation for power losses for one MOSFET:

P = rdson * Iavg^2.

I guess I should take that multiply with 4? Cause I have four mosfets in the bridge. The power losses that I need can be estimated for my case.

You can't use the average current through a switching mosfet to calculate it's power dissipation, unless the duty cycle is 100%. Otherwise, you need to calculate the power dissipated when it's on, then factor the duty cycle.

Let's take a 1 ohm mosfet in a system with a 5 amp average current. IF it were on 100% of the time, then
(5^2) = 25 Watts instantaneous power.
25 * 100% duty cycle = 25 Watts average power.

But, mosfets in switching power supplies are almost never on 100% of the time. Duty cycles can literally be anywhere from 1% to 99%

So let's look at the same system at 50% duty cycle. 5 Amps average is 10 Amps at 50% duty cycle
(10^2) = 100 Watts instantaneous power.
100 * 50% duty cycle = 50 Watts average power.

Now let's look at 25% duty cycle. 5 Amps average is 20 Amps at 25% duty cycle
(20^2) = 400 Watts instantaneous power.
400 * 25% duty cycle = 100 Watts average power.

So you can see, the actual duty cycle can have a big impact on the I^R power dissipation calculations.

What complicates the matter is that, as the supply regulates, the actual duty cycle varies, so you need to choose a worst case condition to base your calculation on if you're engineering for reliability.

..... or best case if you're writing an internet white paper. lol

NOTE: This is not limited to mosfets, but is also true of any resistive component that pulse currents flow through. Including transformer and inductor windings.

ChrisP58 said:

So let's look at the same system at 50% duty cycle. 5 Amps average is 10 Amps at 50% duty cycle
(10^2) = 100 Watts instantaneous power.
100 * 50% duty cycle = 50 Watts average power.

Click to expand...

Aha okay, I see. so if my duty cycle is 0.5 which 50 %.
However, I talked to a teacher and he suggested that I need to have the "switch current (RMS)".

So if my input current and voltage is a DC . Input current = 2.38 A DC and input voltage = 400 V DC.
How do I obtain the switch current (RMS) "Irms" from a MOSFET? according to the equation below:

Pcond = rdson * Irms^2 * D.

mading2018 said:

Aha okay, I see. so if my duty cycle is 0.5 which 50 %.
However, I talked to a teacher and he suggested that I need to have the "switch current (RMS)".

So if my input current and voltage is a DC . Input current = 2.38 A DC and input voltage = 400 V DC.
How do I obtain the switch current (RMS) "Irms" from a MOSFET? according to the equation below:

Pcond = rdson * Irms^2 * D.

Click to expand...

Basically, RMS is a way of finding the DC equivalent value of a complex waveform. What do you think the RMS (DC equivalent) value of a DC value is?

So your duty cycle is always 0.5 (50%)?
This is a battery charger, isn't it? Will it stay at the same duty cycle as the battery goes from fully discharged, to fully charged, and on to float mode? I highly doubt that.

I think Idc =Idc(rms)=2.38 A.

But in that case I need to choose a worst case scenario as you said (I missed that..) but what could that be?
What would you choose for the duty cycle to be for a worst case scenario?

--edit--
The charger is suppose to operate as low-power charger, so I guess the current and duty cycle change very slowly as well? But since I do not consider the battery in my project, I mean we are not actually gonna charge anything, a duty cycle of 50 % is okay to assume?

mading2018 said:

Is it the same concept for a full-bridge that have MOSFETS instead for diodes?

If I calculate the power losses for one MOSFET, and then I take that multiplying by 2, I will get the power losses for the whole bridge? Is should be correct since they are also operating as pair
as the case with diodes.

Click to expand...

Hello again,

I think you might also have to consider other modes of operation for this kind of circuit, as well as the two diode two mosfet version. That's because it does not necessarily have to function as ONLY a rectifier. It also has a BOOST mode of operation which can change things quite a bit.
For full wave rectification we would probably just run this as a synchronous bridge rectifier, which means the mosfets are turned on in a way that mimics the way the diodes would turn on in a full bridge, but in boost mode we could for example turn two (or one) lower mosfet 'on' and that would charge the inductor in a way that a regular boost circuit works. So it would be a boost / rectifier circuit. That is the way i think it is meant to be used anyway. For this you'd have to consider calculating losses in a different way.
You could look into this more.

mading2018 said:

I think Idc =Idc(rms)=2.38 A.

But in that case I need to choose a worst case scenario as you said (I missed that..) but what could that be?
What would you choose for the duty cycle to be for a worst case scenario?

--edit--
The charger is suppose to operate as low-power charger, so I guess the current and duty cycle change very slowly as well? But since I do not consider the battery in my project, I mean we are not actually gonna charge anything, a duty cycle of 50 % is okay to assume?

Click to expand...

I'm confused.

If you're not going to build it. Or even to model it doing any simulated work, then what value is there in knowing what the losses in the components are?

ChrisP58 said:

I'm confused.

If you're not going to build it. Or even to model it doing any simulated work, then what value is there in knowing what the losses in the components are?

Click to expand...

I simulate and calculate it. I talked to a teacher that said the duty cycle could be 50 %. What would you set the duty cycle to for a worst case scenario？