regulator power dissipation

Hi!

Imagine a simple DC power supply: transformer secondary, diode bridge, filtering cap, regulator.

Many books calculate power dissipation in the regulator this way:
P = (Vin - Vout) * Imax.

where:
- Vin: the filtered unregulated DC volatage
- Vout: the regulator's output voltage
- Imax: maximum current draw

I think it's correct only when regulating ripple-free DC voltages, but NOT in the case of unregulated DC's that have considerable ripple.

Wouldn't it be more correct to calculate with the mean value of the input voltage (something like rms)?

Let's label bottom and top of ripple with Vrb and Vrt, respectively.

Then I would calculate dissipation this way:
((Vrb + (Vrt-Vrb)/2) - Vout) * Imax

Is this a more proper way or am I missing something?

Thanks,
j

That's correct. But you can simplifiy it to ((Vrt+Vrb)/2) - Vout) * Imax

crutschow said:

That's correct. But you can simplifiy it to ((Vrt+Vrb)/2) - Vout) * Imax

Click to expand...

Thanks for the quick reply!

It's interesting as when calculating with ripple, one may get significantly lower dissipation than the (Vin-Vout)*Imax way, depending on the size of the filter capacitor. In the case of the PSU that I'm designing now it means about 40% difference in the worst case situation.

Of course, by neglecting ripple, one remains on the safe side, maybe this is why I haven't found any references to it.

Thanks,
j

johnsmith1261 said:

Hi!

Imagine a simple DC power supply: transformer secondary, diode bridge, filtering cap, regulator.

Many books calculate power dissipation in the regulator this way:
P = (Vin - Vout) * Imax.

...

Click to expand...

That's perfectly acceptable. The mains ripple voltage is a small percentage of the total DC voltage, and the ripple voltage shape is quite triangular so the measured average is acceptably close to the true value.

You should be de-rating heatsinks etc by a large percentage so a tiny percentage possible error in the reg dissipation calc is meaningless.

Mr RB said:

That's perfectly acceptable. The mains ripple voltage is a small percentage of the total DC voltage, and the ripple voltage shape is quite triangular so the measured average is acceptably close to the true value.

You should be de-rating heatsinks etc by a large percentage so a tiny percentage possible error in the reg dissipation calc is meaningless.

Click to expand...

Hmm... I've double checked my other calculations and indeed there was an error.
The "40%" figure in my previous post is incorrect, it turned out to be under 10%. It seems much more acceptable now

Thanks,
j

Hi there,

Using the average value of the input wave probably isnt too bad really, but
the more exact way would be to use the rms value because you cant calculate
the actual power using the average value, you must use the rms value.
I cant say it will make a huge difference when the ripple is low, but there will be some.

The real question is why someone would want to lower the power dissipation
by relying on the ripple voltage to reduce the total power when designing
the front end more correctly would be the best approach. You have to
realize that the lowest point in the wave will be a critical point where the
ripple might start to push through the regulator if the power line goes to low
line condition, so more capacitance and less input voltage would get the
lowest power dissipation. Of course if you are stuck wth a given transformer
voltage then you have to make due any way you can, but you may actually
be better off with a buck switcher which would offer the lowest power dissipation
given the apparent voltage differential.

MrAl said:

Hi there,

Using the average value of the input wave probably isnt too bad really, but
the more exact way would be to use the rms value because you cant calculate
the actual power using the average value, you must use the rms value.
I cant say it will make a huge difference when the ripple is low, but there will be some.

The real question is why someone would want to lower the power dissipation
by relying on the ripple voltage to reduce the total power when designing
the front end more correctly would be the best approach. You have to
realize that the lowest point in the wave will be a critical point where the
ripple might start to push through the regulator if the power line goes to low
line condition, so more capacitance and less input voltage would get the
lowest power dissipation. Of course if you are stuck wth a given transformer
voltage then you have to make due any way you can, but you may actually
be better off with a buck switcher which would offer the lowest power dissipation
given the apparent voltage differential.

Click to expand...

Yeah, good points. The PSU I'm designing is intended for a synthesizer and for this reason I wouldn't want to make it switching style, but use a linear regulator. Unfortunately the best transformer I could find provides a few volts more voltage than needed for this task and this causes some power to burn in the pass transistor. It's not that bad but not an optimal solution.
But I see your points, having a bigger cap and less voltage would yield the best results. Unfortunately I have to make due with this trafo, as I don't want to have one custom built.

I try to design for wide safety regions, like mains voltages 15% less than nominal (and just in case for 10% over too), take into account component tolerances, keep the pass transistor way below it's maximum allowed junction temperature and so on...

Actually I've written a C# program to verify all of these, so I can feed it the system requirements and component values and tells me if there are any problems. It was necessary as the situation started to get out of hand with so many things to consider and keep in mind

Thanks,
j

Your synth circuit should not use a lot of power. What uses the most current? Maybe there is a low-power IC to replace it.

audioguru said:

Your synth circuit should not use a lot of power. What uses the most current? Maybe there is a low-power IC to replace it.

Click to expand...

The PSU is meant for a large modular synth system with 20+ modules (imagine a wardrobe ). I've read it's advisable to build an at least one amp supply for them. I design it for 1.5 amps for safety and to have enough reserves.

I have never seen an obsolete MOOG synthesizer. They were huge. Now they use a little digital-signal-processor.

johnsmith1261 said:

Yeah, good points. The PSU I'm designing is intended for a synthesizer and for this reason I wouldn't want to make it switching style, but use a linear regulator. Unfortunately the best transformer I could find provides a few volts more voltage than needed for this task and this causes some power to burn in the pass transistor. It's not that bad but not an optimal solution.
But I see your points, having a bigger cap and less voltage would yield the best results. Unfortunately I have to make due with this trafo, as I don't want to have one custom built.

I try to design for wide safety regions, like mains voltages 15% less than nominal (and just in case for 10% over too), take into account component tolerances, keep the pass transistor way below it's maximum allowed junction temperature and so on...

Actually I've written a C# program to verify all of these, so I can feed it the system requirements and component values and tells me if there are any problems. It was necessary as the situation started to get out of hand with so many things to consider and keep in mind

Thanks,
j

Click to expand...

Hi again,


Oh i see, so you are using a program to do some calculations, that's great.
If i knew that i would have recommended using the definition of average
power to calculate the total power dissipation:
P=(1/T)*Integral[0 to T](v(t)*i(t))dt
where
T is one period of the wave
v(t) is the voltage wave
i(t) is the current wave (might be somewhat constant)

You can start with say 1000 samples and calculate the power from that
using both a sawtooth wave where the dips are as low as possible
and using a smoother wave and compare the two if you like.

MrAl said:

Using the average value of the input wave probably isnt too bad really, but
the more exact way would be to use the rms value because you cant calculate
the actual power using the average value, you must use the rms value.
I cant say it will make a huge difference when the ripple is low, but there will be some.

Click to expand...

Not true. RMS voltage would be correct for a resistive load where the current varies with the voltage, but not for a constant load current as is the case here. The regulator input is a constant current as determined by the load current and is independent of the input voltage. Thus the input power is linear function of voltage, or V*I where V is the instantaneous voltage and I is the load current. If V is varying then the average dissipated power is the average value of V times I.

crutschow said:

Not true. RMS voltage would be correct for a resistive load where the current varies with the voltage, but not for a constant load current as is the case here. The regulator input is a constant current as determined by the load current and is independent of the input voltage. Thus the input power is linear function of voltage, or V*I where V is the instantaneous voltage and I is the load current. If V is varying then the average dissipated power is the average value of V times I.

Click to expand...

Hello there,


Yes that is correct Carl.
Here are some numbers using some carefully chosen waveforms:

First with a constant current which would be like the voltage
regulator and a triangular wave with a dc offset:

v(t)=1+t*1
i(t)=1
T=1
vrms=Rms(v(t),t,0,T,100)=1.528
irms=Rms(i(t),t,0,T,100)=1
vavg=Avg(v(t),t,0,T,100)=1.5
iavg=Avg(i(t),t,0,T,100)=1
p(t)=v(t)*i(t)
P_True=(1/T)*Integrate(p(t),t,0,T,1000)=1.5
Pa=vavg*iavg=1.5
Pr=vrms*irms=1.528


Now with a current that is also triangular:

v(t)=1+t*1
i(t)=t
T=1
vrms=Rms(v(t),t,0,T,100)=1.528
irms=Rms(i(t),t,0,T,100)=0.5774
vavg=Avg(v(t),t,0,T,100)=1.5
iavg=Avg(i(t),t,0,T,100)=0.5
p(t)=v(t)*i(t)
P_True=(1/T)*Integrate(p(t),t,0,T,1000)=0.8333
Pa=vavg*iavg=0.75
Pr=vrms*irms=0.882

So in the case of constant current the average voltage works out better
than the rms value for the constant current case, but when the current
varies (for this example) the rms value worked better.

When in doubt, use the integral form


It's still a good idea however to use a larger capacitor and a better matching
input voltage for the regulator because the lowest dip in the ripple has to be
at least as high as a smooth input dc voltage would have to be, so that means
that the power dissipation for a rippling wave input will be higher than when
using a smooth input, providing of course you have control over the input
voltage itself as well as the capacitance.

There is one advantage to having higher ripple voltage from a diode rectifier and capacitor connected to a transformer. It reduces the peak current through the transformer and thus reduces the transformer dissipation. The lower the ripple, the higher the peak current through the transformer due to the shorter time the diodes conduct at the peak of the voltage waveform, and the higher the I²R loss from the transformer winding resistance.

Hi again,


Well, I^2 R loss has two components, the I and also the R. To get a higher
ripple voltage we'd have to lower the capacitance (from a smooth reference
design) and lowering the capacitance would mean the ripple dips would go
lower, and that would mean we would have to increase the number of turns
on the secondary which would increase the resistance of the secondary.

Well, I was assuming a hobbyist application where you have an off-the-shelf transformer and don't have the luxury of adjusting the turns ratio to give an exact voltage. If that case, if you have some extra voltage headroom on the transformer voltage, you could use that to allow a higher ripple voltage.

Otherwise your point is well taken.

Hi there Carl,


Yeah i thought that's what you meant. Some of these problems depend highly
on what we actually have control over and what we really dont.
If i had to recommend something to someone who was that worried about
power dissipation then i would definitely tell them to take a quick look at
some switching regulators. The National Semi "Simple Switcher" family has
a lot to offer especially at the current needed for this application. They are
almost as easy to use as the three terminal linear regulators these days
except for the extra requirement of an inductor and Schottky diode.
Some ripple wouldnt hurt there either.

I've used National Semi's switchers several times for (roughly) 4.2v output
applications, and it's nice to know that i can plug in a 9vdc to 24vdc wall wart
and still get my required output voltage. Another nice thing about the
switcher is that if the input voltage is much above the output voltage, the input
current reduces to a much lower level. For my 4.2v output and 24vdc wall
wart the input current (output current of the 24v wall wart) is only about
200ma for a 1000ma output (about 5 to 1). That's the beauty of true power
conversion
Oh yeah, they dont get hot either for high input-output differentials so the
heat sink can be much much smaller.

Thanks everybody for your replies!
j