# What is the DC Gain of this Flyback power stage transfer function?

Discussion in 'Mathematics and Physics' started by Flyback, Jan 13, 2014.

1. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
Hello,
The following equation (2A-22) , on page 232 of Basso's book (switch mode power supplies) is the small signal power stage transfer function for a buckboost converter, in current mode, CCM.
Please advise on what is the DC gain of this transfer function?
I am actually using it for a flyback (flyback is derivative of the buckboost) and my Ns/Np for the flyback is 0.837.
Pages 225, 226 explain what the parameters are
Pages 234, 235, explain how to convert equation 2A-22 for a flyback
Page 233 also shown as it shows the poles and zeros

File size:
720.7 KB
Views:
252
File size:
437.6 KB
Views:
118
File size:
129.4 KB
Views:
137
File size:
161 KB
Views:
118
File size:
149.6 KB
Views:
138
File size:
164.6 KB
Views:
147
File size:
609 KB
Views:
250
2. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ
Hi,

For that transfer equation itself it looks like the DC gain is:
Vin*R/(Ri*(Vin-2*Vout))

which is just the first part of that thing.

3. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
I agree....and you get that by putting in w=0.

The thing is, i am using a flyback converter...which is the transformer isolated version of the buckboost (as the page 234 says)
So I believe I need to adjust your above equation as follows.....(Some of these adjustments taken from p234, 235)

R becomes R/N^2 where N=Ns/Np
Vout becomes Vout * Np/Ns

Do you agree with this?

Its just that the book doesn't say how "vout" is to be adjusted when converting from buckboost to flyback

Joined:
Jan 12, 1997
Messages:
-
Likes:
0

5. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ

Hi,

Actually it is the limit as w approaches zero. But it would be better if i saw both schematics side by side before answering this. Did you post both yet in this thread?

6. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
Thanks, here is the flyback, there isn't a buckboost, the buckboost equations are being used to derive the flyback equations

File size:
18.2 KB
Views:
427
7. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ
Hi again,

So what changes in your flyback to make it into the buck boost you are talking about?
I dont feel comfortable answering a question if i can not see the schematic as i am sure you can understand.

8. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
a flyback is a transformer isolated form of a buckboost.
The transformer is the salient point....I am not sure whether I have to put "vout" in the calculation, or "vout" as seen from the primary, ie vout*np/ns..(I am transferring everything over to the primary)

9. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ
Hi,

Ok then let me see if i can find a reference circuit to look at.

It sounds like you are correct because it sounds like you are including the gain and the effect of the inductor being moved, but i like to see a circuit so i can see exactly what is happening and what has changed. I am not as experienced with the buck boost as i am with the other topologies because i never needed that for anything other than simple LED drivers, and i did not have to go into detailed analysis. The point which i would like to clarify and maybe you know this one, is what is the effect of the pulse width on the output. For example, what if we vary the pulse width by 10 percent does that mean the output goes up by 10 percent? An even more pressing point is, are you looking at the buck boost as being in the boost mode only? That will simplify things also.

10. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
Pulse width does occur in the transfer equation 2A-22 on page 232 of the top post (D')
D' = 1-D, where D is the duty cycle which signifies the pulse width.

11. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ
Hello again,

I found a reference and it appears that the only difference is in the turns ratio. Thus if the turns ratio is 5 then the output will be 5 times greater. The inductance is always in series with the switch. There may also be a change in sign so instead of 5 we might have to use -5.

I was wondering why you wanted to change "R" in your original calculation.

What we could do to test this is to simulate the simplest buck boost and the simplest flyback and see if we get the correct results. Do you feel up to this task?

12. ### FlybackWell-Known Member

Joined:
Jan 5, 2007
Messages:
2,009
Likes:
34
thanks, I changed R because of advice on Pg 235, Basso.
The simulation you suggest is a great idea....simulate both,..give each a transient load and see if they respond the same.
I will be on to this soon, I hope sooner than later.