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Flyback LED driver has poor quality feedback loop?

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Flyback

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Hello,
We have got a contractor to do a 100W flyback led driver for us, and we believe that he has not done a very good job. The circuit works but we believe that its performance is compromised in the way that he has done it.
The circuit schematic and LTspice simulation are attached.
He has used an external current sense error amplifier with a 0 to 5V supply, and strangly he has used ground as the reference voltage into this error amplifier.
His current sense signal is then going below ground as the current builds up, and he has biased this signal up so that the led driver is in regulation when the inverting and non-inverting inputs of the error amplifier are both at 0V.
The problem is, the opamp’s negative supply is the 0V rail, and when the input to the inverting input of the error amplifier goes below zero, then the opamp does not “know” what this voltage is, because it is supplied from a 0 to 5V supply.
So what I am saying is that the opamp will not respond as strongly as it might to the feedback signal going too high negatively. The feedback circuit in this setup means that the opamp will not be as able to develop sufficient over-drive in order to regulate against higher-than-wanted LED currents.
Do you agree?
 

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Flyback, We should get 1% of your wages for helping. lol
The LT6220 input range is supply to supply. So it should work at 0V but not negative. That is a little scary. Actually it will work to a slightly negative voltage but you probably should not count on it.
There are op-amps that work to -0.6V but will not function at the plus supply. You could change to a different part.
You could move the negative supply pin on the amp to the other side of R6. When operating that will move the supply to -0.3V. (weird thought)
 
Thanks, the fact that you are offering these changes, tells me that you also feel uncomfortable with this feature of using the 0V rail as the reference input to the error amplifier?

I mean, surely the reference voltage cannot be the negative supply rail voltage itself?....the opamp needs to develop an overdrive so that it can respond to the input signal.....and how can it do this if the inut signal cannot possibly go below the reference voltage?.......i am saying that the input signal (the signal on the inverting input of the opamp) cannot go below the reference input, (which is 0V) and it absolutely needs to be able to do that , in order that it can behave like an opamp should in this case. Well, i know that the input signal "can" go below the 0V reference voltage, but the opamp does not "know" what that voltage is, because going below 0V is going below the opamps 0V rail, and the opamp does not know what voltages are below its 0V supply rail.

Surely you agree, that this is at best "shoddy" workmanship?
 
It's true the inverting input has to go slightly negative for a positive output voltage from the op amp but only by a very small voltage equal to the maximum input offset voltage plus the output voltage divided by the minimum open loop gain (should be no more than about a mV total in this case).
If you look at the schematic of the op amp you will see that it has a PNP input stage that operates when the inputs are at or near ground.
These transistor inputs will also conduct when the input is slightly below ground, thus keeping the op amp in the active region, so I see no problem with the circuit as is.
Most Rail-Rail op amps are designed to operate that way.
 
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It's true the inverting input has to go slightly negative for a positive output voltage from the op amp but only by a very small voltage equal to the maximum input offset voltage plus the output voltage divided by the minimum open loop gain (should be no more than about a mV total in this case).

Thanks, -yes, but the opamp inverting input cannot go lower in voltage than the non-inverting input...not even by 1mV.....because the non-inverting input is already at 0V, ie the negative supply rail.

-the opamp needs an "overdrive" voltage, ie, it needs the inverting input to go "below" the non-inverinting input, and it cannot do that......well....it "can", but if it does so, then it has gone below the negative supply rail of the opamp, so that doesnt count, because the opamp does not recognise voltages which are outside its supply rails.

I say that this opamp's dynamic performance is at best, compromised by this kind of connection, do you agree?
 
the opamp inverting input cannot go lower in voltage than the non-inverting input...not even by 1mV.....because the non-inverting input is already at 0V,
The two inputs are independent, so there's no reason the inverting input can't go negative.
 
First you need to notice that R7-R8-V3 combination which brings the negative input above ground, so that with enough current the two inputs will meet at zero.
Second, the feedback loop is all weird and has very high gain for DC and little for AC, which makes me feel that he is using it as a comparator and that is not a thing I would do to keep the current in check, a real comparator would be better.
 
The two inputs are independent, so there's no reason the inverting input can't go negative.
thanks, yes, it can go negative, but what i meant was, when it does go negative, the opamp can't deal properly with that, its dynamic performance will be affected.....the bias'ing of the opamp isnt designed to be properly biased for inputs that go below the negative supply rail...surely?

(BTW its not meant to be working like a comparator, as you can see, it operates fine as a negative feedback loop, but im saying that the dynamic performance of the opamp is limited by the reference voltage being at the negative supply rail.......its performance with the negative feedback loop is not as good, there is less gain and phase margin)
 
Thanks, -yes, but the opamp inverting input cannot go lower in voltage than the non-inverting input...not even by 1mV.....because the non-inverting input is already at 0V, ie the negative supply rail.
.................
Why can't it? So it's at the negative rail. There's nothing absolute about that as far as the inputs are concerned.
As I noted, the PNP input transistors will still work when their base inputs go slightly below ground. If you look at the OP amp schematic, you will see this.
 
I agree with Crutschow. I think the opamp would work as usual even with its inputs at ~ -0.4V. Lower than that and the collector-base junction of the input PNP would probably be forward biased enough to begin to conduct significantly. At that point I guess some weirdness would happen.
If you are concerned about dynamic performance, try applying a large step load in the sim. I suspect the loop will be fast enough to prevent the inverting input going more than a few mV negative, but it would be interesting if you could confirm that.
BTW, I found your sim reached a steady output output current of 3.4A, not the 2.4A shown on the sheet? Also, it runs much faster if you reduce the number of LEDs dramatically.
 
1: Was the circuit designed with a price or budget limit in place.

2: What exactly is the fuss being it does do exactly what it was designed to do?

3: If it's so bad what did your company hire it out rather than do it yourselves in house?
 
I love how Flyback allways complains about the designs some outsource company did for them and how he knows better.
Flyback, if you are so knowledged why don´t you do the designs yourselves and claim all that money wastefully spent on outsourcing?

On a slightly different note, "we believe that he has not done a very good job" seems like that job he made is not up to par with your own liking, would you be so kind and share with us in which actual design specs the product you were given doesn´t meet the requirements your company specified, or the general practice?
Have you determined the reliability will not be what you asked for? If so, what exactly are the problems? So far we got to the point that the opamp is well in its oprational limits all the time except for brief periods when the circuit suddenly gets overloaded, i.e. an LED goes short. If my mental calcultaions are correct you would need roughly 6A flowing through the diodes to get you out of the Vdd-0.3V input operating range typical for most opamps that include ground in their input range.
 
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Those who can do. Those who can't pick apart the work of those who can for not being perfect.
 
The voltage at the inverting input of the op-amp won't be allowed to go more than a few millivolts negative, due to the connection to the output pin of the op-amp.

As soon as the voltage at the inverting pin goes even slightly below that of the non-inverting pin, the output pin starts to go positive. That voltage is coupled back to the inverting pin.

As long as the circuit is designed to allow the op-amp the necessary freedom of operation, you have a voltage divider whose midpoint is held at a voltage equal to that of the non-inverting pin. What the output pin does in order to keep the summing junction at that point, is what makes any negative feedback op-amp circuit useful. In this case, the output pin drives the opto-coupler, whose output modulates the pulse width of the PWM controller.
 
the only way this opamp can ever "know" that the inv pin has gone below the non-inv pin is because of noise on the non-inv pin...that is not a good way to operate an opamp you surely agree?

If the inv input goes below 0v, then the opamp doesnt "know" that...the opamp's negative supply is the 0v rail......so anything below that on its inputs and the opamp surely doesnt "know" what that is...the dynamics are affected and the normal transfer function for the opamp is no longer accurate..thereby we can no longer calculate a reliable gain / phase margin figures for the smps...surely?
 
I disagree. That is a perfectly good way to run an op-amp.

It would be noise if it were random and unpredictable. What you have here is a string of regular pulses, each one in time with the flyback period and whose amplitude corelates with the load current. And, as I said, the voltage at the inverting pin is not allowed to go negative because it gets pulled back to the reference by the positive current coming from the ouput pin of the op-amp.

You need to think of that node as if it were a zero impedance current summing junction (which it is,) where all of the currents in and out of the node sums to zero.

You have made an LT Spice model of the circuit. I have run it, and everything seems to work fine. The voltage at the inverting pin never goes negative, and the output voltage of the op-amp doesn't come close to it's +5V rail. Is it and accurate model? Are there real problems that you see that are not reflected in that model?
 
i agree it appears to work on a general scale...but how about the transfer function of the opamp...is it accurate when operated like this...and if not, then that means the feedback loop calculation for the smps has gone awry....bad news.......

I can "improve" this circuit by adding a high side current monitor ic, and feeding that into an error amplifier along with a reference voltage....and doing it like that....at least its standard way...and i am confident the transfer fucntion is what it says on the tim so to speak.
 
Can you show this transfer function that has supposedly gone awry?

You can certainly change it, but it would be an improvement only in your own mind.

And, if you cannot get your head around how this circuit works, then that may be what you need to do to give yourself some level of confidence in the product.
 
the only way this opamp can ever "know" that the inv pin has gone below the non-inv pin is because of noise on the non-inv pin..
How do you work that out? :confused: The opamp simply amplifies the difference between the two inputs. Since its open-loop gain is extremely high that difference can be extremely small.
 
The problem is, the opamp’s negative supply is the 0V rail, and when the input to the inverting input of the error amplifier goes below zero, then the opamp does not “know” what this voltage is, because it is supplied from a 0 to 5V supply.
I think the other members have already covered this, but they've stated that the input range actually exceeds the guaranteed input range specified in the datasheet. Because the voltage on the negative input will generally only be at most a few millivolts outside this range, it's expected that the opamp will have no trouble with it. If you are still concerned, you can move the negative supply of the opamp to the other side of the current sense resistor (also mentioned above by someone).
So what I am saying is that the opamp will not respond as strongly as it might to the feedback signal going too high negatively. The feedback circuit in this setup means that the opamp will not be as able to develop sufficient over-drive in order to regulate against higher-than-wanted LED currents.
Do you agree?
I don't agree. Higher-than-wanted LED currents will cause the negative input to be pulled lower, and the opamp output to increase. Considering that the output would be sitting 0V for low currents, and increasing to some higher voltage when the current exceeds the set-point, there is no need for "overdriving" anything here. The rate at which the output changes is limited a lot by the large capacitance in its immediate feedback loop. If you remove the capacitor, you will have the opamp acting like a comparator, and you will have large output current ripple. If you reduce the capacitance to something more suitable, you will have a better response time and less overshoot of the current.

Someone above mentioned that the current is 3.4A instead of 2.4A: that is because of the large capacitor just mentioned causing slow response and large overshoot of the output current. The current will reach its expected steady-state value of 2.55A, it just takes a while to get there. Reduce the cap to 4.7nF and the overshoot is not as evident.
 
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