More Help Needed with LM5045

Houari Cobas

Other Parts Discussed in Thread: LM5045

Hello, thank you for take a look to this help request.

I am facing a problem and would like to have some advice on it. I am designing an SMPS power supplay with LM5045 like you saw in the previous help request (full bridge PWM controlled with current doubler output topology). The power supply have 5 outputs, +/- 170 (6.8A max), +/- 12V (1 A), and 12V (4A).

I am facing a problem with the +/-170V output. I need this output to power an Class AB amplifier which will have at his output an Sinusoidal wave form with 40 Hz frequency. So this output need to maintain fix the +/- 170 V (with Vomax=171V and Vomin=169V) but the ouput current will vary from 50 mA to 6,8 A following the load current needs. Both outputs will work 180° out of fase.

In the output stage design, i must calculate the LC filter.

For the inductance calculation i have the following equation:

Lmin= ((N2/N1)*Vdmax-Vomin)/(Irmax/ton)

where

N2 -> secundary turns (9)

N1 -> primary turns (17)

Vdmax-> maximum primary voltage (400 V)

Vomin-> minimum output voltage (169 V)

Irmax-> The ripple for the maximum average inductor current

ton -> conduction time

ton= (Vomin*N1*Ts)/(2*Vdmax*N2)

the operation frequency selected is 300k Hz, so the ton = 1.32 us.

I was taking a look to some books and i saw that in this kind of topology the current flow trough the inductor with a triangular waveform centered at the load current in a continuous current mode. That triangle waveform would be the current ripple which must be filtered by the inductor.

The first problem is that like i said before, in my application the load current will vary from 50 mA to 6,8A, the load current will not be fixed. How can i handdle this problem? It is really a problem, the power supply will can work with that current variation? What i am looking is that if i fix a ripple for the maximum current, then at the minimum load current, minus the mentioned ripple, the inductor current will fall to 0.

Like i can not have current ripple higher than 50 mA (minimum output current), i select an current ripple of 2,5 mA.

When i do the calculations, Lmin=22.58 mH.

For the capacitor calculation i have this equation:

Cmax = (Irmax*ton)/Vr

where

Vr-> Ripple voltage for the minimum output voltage (in my application the output is fixed to 170 and the maximum ripple i can have is 2V)

That make Cmax = 1.65 nF.

I think this values have something wrong, can you give an idea about what is wrong here?

Another think is the following, when i going to calculate the Rslope with the equation (i will do like i was just one output with 170 V (6,8 A max)):

Rslope=(Vout*Rcs)/(Lfilter*Fosc*Islope*Ntr)

where

Vout -> output voltage

Rcs-> sense resistor (0.1 ohm)

Lfilter -> inductor (22.58 mH)

Fosc-> oscilation frequency (300 kHz)

Islope -> down slope of the filter inductor (i calculate it like ratio between the inductor ripple current and the toff time, i do not know if it is correct, Islope=(2.5 mA)/(2.01us)=1244A/s

Ntr -> turn ratio with respect to the secondary (9/17=0,53)

Taking into account all of these values Rslope = 3.76 uohm which i think is not correct.

So please, could you tell me what is wrong here. What i am not looking correctly?

thank you very much

Houari

over 13 years ago

0 Terry Allinder over 13 years ago

TI__Expert 3545 points

Houari,

First I would not calculate the output inductor the way you have done it.

1-L=Vs-Vout/di X ton. Vs is the transformer secondary voltage, if Vin 400Vdc, VS is 211Vpk.

2-di should be based on the maximum output current, 6.8A.

3-I would normally start out with di 10 or 20% of the maximum load current, so let pick 10%, or 0.68A, and I will just a 50% duty cycle, or ton 1.67us

4-L=((211-170)/ 0.68)X1.67us= 100uH, not 22mH

Terry

0 Houari Cobas over 13 years ago in reply to Terry Allinder

Prodigy 180 points

Hello Terry, thank you very much for your help which is being too valuable.

I have a doubt about the functioning of this circuit. I readed that it wortk in continuous current mode. So the inductor current should not be 0. But in the application i want to use the circuit the each output current can go down almost to 10 mA. For example, my outputs with +/- 170 V will work in contrafase. I will use the that output as an Class AB amplifier power supply. the amplifier output current will be sinusoidal with a peak amplitude of 6.8A. Like it is sinusoidal it will pass trough 0. In the positive semicilce the 170V power supply output will be working and will supply the current, in this period the -170V current will be the class AB polarization current, almost 10mA. In the negative semicile will happen the oposite. So my question is, the power supply will be regulating his ouput even when his output current fall down to 10 mA?

Are you agree with the capacitor equation?

How do i ensure that it will be regulating all the time?

Like mentioned before i will have five outputs, +/- 12V, +/-170V and 12V aux. I pretend to use a cross regulation techinque. Do you think it will work in such scenario?

Another thing it is possible with a current doubler topology to generate a negative voltage? I was trying to see several articles, but i just see positive ouput voltage. I ask you that because i would like to use the sinchronous rectifier option, but when i try to implement some circuit like that i see that mosfet´s gates voltage will not referrenced to ground.

Once more time thank you very much.

0 Terry Allinder over 13 years ago in reply to Houari Cobas

TI__Expert 3545 points

Houari,

Design the output inductor the way I described it. You can still use current mode control at light loads. You need to design the control loop to handle light loads. You may need to use a Network Analysis to measure the gain and phase. With all the loads, that would be the best way to be sure your control loop is stable.

I'm not sure how well the other output will regulate using cross over regulation, there are a lot of factor that contribute to the error. One of the biggest is the coupling in the transformer. If you think your MOSFETs Gate voltage are not referenced to ground, TI make a lot of high side drivers.

Terry

0 Houari Cobas over 13 years ago in reply to Terry Allinder

Prodigy 180 points

Hello Terry, i would like to know if you have some sugestion to me in order to overcome the problem concerning the severals outputs i have and the output voltage regulation.

thanks

0 Terry Allinder over 13 years ago in reply to Houari Cobas

TI__Expert 3545 points

Houari,

Your trying to regulate +/- 170V I believe to +/-1V that will be difficult. You will need to sense across the +/-170 and feed that back. Where are you located? Have you tried to contact a local TI FAE, if you can they maybe able to sit down one-on-one a help you.

Terry