Re: Opto-couplers in smps - part 3 (final chapter)

On 7 Feb, 09:51, Terry Given <my_n...@xxxxxxxx> wrote:
Steve wrote:
On 6 Feb, 21:58, Terry Given <my_n...@xxxxxxxx> wrote:

Steve wrote:

On 6 Feb, 11:16, Terry Given <my_n...@xxxxxxxx> wrote:

Steve wrote:

On 5 Feb, 16:35, "Paul Mathews" <o...@xxxxxxxxxxx> wrote:

On Feb 5, 2:48 am, "Steve" <stefan.toftev...@xxxxxxxxxxxx> wrote:

On 5 Feb, 08:58, Terry Given <my_n...@xxxxxxxx> wrote:

Eeyore wrote:

Steve wrote:

Hi once again,

I've find an optocoupler called TCET 1102 G.
According to its data***, this coupler has an Fc of 110kHz.
Data***:http://www.dzsc.com/pdf_down.asp?id=642422

110kHz with a diode current of 10mA and a load resistance of 100 ohms. I suspect
you're unlikely to be using it like that.

Will this mean I may use it in my smps loop which has a unity gain
frequency of 3kHz.

It won't have an Fc of 110kHz most likely under the conditions you're likely to
be using it I reckon.

Graham

wot he said. Its trivial to measure the -3dB frequency response of the
opto in your circuit; do so. That way you'll KNOW the phase shift the
opto introduces.

Conversely its astonishingly hard to measure the total loop response
unless you have a fancy wodgy. thats wy step responses are often used
(and why said wodgies exist).

you can also do this: run the smps from an isolated supply, and
completely bypass the opto. Do a step response, check its fine, then put
the opto back in and do it again. you'll pretty soon see if the opto
adds too much phase shift.

Cheers
Terry- Dölj citerad text -

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Though it seems like a good opto for this purpose?- Hide quoted text -

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Yes, although 100 kHz is about the usual fc for most optos operating
at around 1 mA LED bias and loaded on the phototransistor side such
that saturation is avoided. If you're connecting the LED anode to your
p.s. output without intermediate filtering, note that the frequency
characteristics of the circuit driving the cathode will not
necessarily dominate. For example, the classic TL431 integrator
driving the cathode can be understood to provide something like an AC
ground at the cathode at high frequencies, with the optocoupler then
coupling h.f. ripple on the LED anode directly to the output
phototransistor. The resulting midband rise in feedback has probably
saved more than a few designs and thrashed as many others.
Paul Mathews- Dölj citerad text -

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Ok
First I thought of using an op-amp with compensated feedback for
driving the led, but
since yesterday I'll try the TL431 instead.

it makes no difference if the LED is fed from the supply you are trying
to regulate.

The opto-coupler led anode will be coming right from one of the supply
lines, through dropping resistor / zenerdiode combination.
Across the zener there's a filter cap. This will bring noise down a
bit.
To the "input" of the TL431 there are a pole-zero network and across
the TL431 there's another pole-zero network and
one slope. There might come through some h.f ripple through the
compensation components across the 431.
This would surely also increase midband feedback / lowering the gain /
cause unwanted phase shift... Oh dear.

what you have described should avoid the issue Paul mentioned.

When the opto LED circuit runs off the supply to be regulated, the
overall circuit gain is always > 1 - draw a block diagram, its pretty
clear why. Even when the TL431 is a dead short, there is always the
direct path from supply thru led and series R.

with the zener pre-regulator and associated cap feeding the opto LED,
you effectively remove this path.

there is nothing "wrong" with either approach, you just need to
understand your loop before attempting to close it. As others mentioned,
most SMPS do this.

Cheers
Terry- Dölj citerad text -

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Hello Terry.

I get your point of the gain > 1 since the power to the led comes from
the same point as the
regulation does. If led power drops, the power will increase (since
led only will limit the pulsewidth)

just examine the maths, rather than hand-wavey arguments ;)

But I'm little confused of the difference in using a zener pre-reg.
I see it's a parallel path across the led, but........ hmmm

Most greatful for any additional comments,

Thanks!
Regards
Stefan

When you write an expression for LED current, it looks something like this:

Iled = (Vout - Vled - Vk)/Rled

Vk = (Vout - Vref)*H(s)

Vk = kathode voltage of TL431, ie the error amp output voltage (I've
carefully ignored the voltage divider here). H(s) = xfer function of
TL431 cct.

when you look at the small-signal response (perturb and toss out DC &
high-order perturbation terms - IOW partial derivatives wrt Vout) Vled
disappears, as does Vref. You actually end up with:

Iled(s) = (Vout(s) - Vk(s))/Rled
^^^^^
this is the key bit here

H(s) = -Zf(s)/Rtop

Zf(s) = impedance of TL431 feedback network

Rtop = top resistor in voltage divider; the bottom resistor always has a
constant voltage = Vref across it, due to the magic of -ve feedback, so
it falls out of the equation. if you thevenise the voltage divider and
do the sums, you will see this happen.

Vk(s) = Vout(s)*H(s)

so you get:

Iled(s) = [Vout(s) - Vout(s)*H(s)]/Rled

Iled(s) = Vout(s)[1-H(s)]/Rled

now you see where the 1 comes from. Even when H(s) goes to zero, the 1
doesnt.

If you use a pre-regulator, it is:

Iled = (Vprereg - Vled - Vk)/Rled

and, by definition (the regulator bit) Vprereg has no dependance on
Vout, so its partial derivative wrt Vout = zero - it looks like an AC
short circuit.

you then end up with:

Iled(s) = -Vout(s)*H(s)/Rled

which will go to zero whenever H(s) goes to zero.

HTH

Cheers
Terry- Dölj citerad text -

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Ok, but I'm forced to use the pre-reg since the tl431 cannot withstand
this high secondary voltage.
I'm regulating somewhere around 100V.
But when you say that gain will always be > 1 , in which frequency
does this apply?

read harder. it sounds like English is not your first language.

if you use a pre-regulator (you do) then that doesnt apply.

I try to have a closed loop response unity gain frequency of 3kHz,
since Fswitch is only 30kHz due to the slow IGBT devices used as main
switches.

just analyse the transfer function of your entire secondary circuit,
using much the same technique I illustrated above. Dont forget to
include the frequency response of the TL431 itself.....

or, post an ASCII drawing, and we'll help you do it.

Cheers
Terry- Dölj citerad text -

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No, English is not my first language, you're right, but..........
no, never mind.

I'll read harder, and try to understand.

Thanks a lot for your assistance, I really appreciate it.

Regards
Stefan


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