Spice simulation

[Rafael L]

The loop gain probe (or Middlebrook probe) is the combination of current source and voltage source that you have placed inside the feedback loop. You have successfully done this in your attached ex_mid.asc. You computed the formula:

-1/(1-1/(2*(I(Vi)@1*V(x)@2-V(x)@1*I(Vi)@2)+V(x)@1+I(Vi)@2))

What does this represent? It represents A_OLB. But it is very accurate, because it takes into account all impedance interactions. For example, there is an impedance interaction between the feedback loop and the impedance at the inverting input of the op-amp. So we have this result:

A_OLB = loop gain = -1/(1-1/(2*(I(Vi)@1*V(x)@2-V(x)@1*I(Vi)@2)+V(x)@1+I(Vi)@2))

It is this expression we must analyze to determine stability. Let's do that. Let's find the phase margin using your attached ex_mid.asc. We run ex_mid.asc and plot the Middlebrook expression:

-1/(1-1/(2*(I(Vi)@1*V(x)@2-V(x)@1*I(Vi)@2)+V(x)@1+I(Vi)@2))

which is the same thing as A_OLB. It's about 42.5 dB at low frequencies. Next, we find the frequency at which the magnitude of this expression is 0 dB. It is 716 kHz. Since this is the loop gain, and its value is 0 dB (= voltage ratio of 1) at this frequency, we call this the unity loop gain frequency. So we can say the unity loop gain frequency is 716 kHz. Now we look at the phase at this frequency. It is -89.75 degrees. A measure of the stability of a feedback circuit is its phase margin. To compute it, we see how much the phase of A_OLB at the unity loop gain frequency differs from -180 degrees. Here is the formula:

phase margin = -89.75 - (-180) = 90.25 degrees.

This is very stable. A good rule of thumb is that phase margins much less than 80 degrees will start to give some overshoot and/or ringing on a small-signal square wave when doing a transient simulation.

So, to sum up:
1) Look at A_OLB = loop gain = -1/(1-1/(2*(I(Vi)@1*V(x)@2-V(x)@1*I(Vi)@2)+V(x)@1+I(Vi)@2)) to analyze stability.

2) Find the frequency for which A_OLB is 0 dB. This is the unity loop gain frequency.

3) Find the phase shift of A_OLB (call it phi) at the unity loop gain frequency.

4) Compute the phase margin from phase_margin = phi - (-180)

I hope this makes sense.

Andy , thank you very much for detailed explanation.
Yes, I understood, appear more questions, would be correct say that: AOL would only gain in voltage (dBu), Middlebrook would product of voltage and current (dB).
My question is to determine the quantity of negative feedback, I think that the correct to determine the quantity of negative feedback would be only the voltage output and not by power (V-out x I-RL).
Or would the value obtained in the simulation Middlebrook minus
Beta (R2/R1+1).......Not it has a voltage reference here!
I wonder which of the two right?

Thank you again

 

[unclejed613]

what middlebrook tells you is the loop gain, which is sort of an oxymoron, because it's actually attenuation. it's the difference betweem open loop gain and closed loop gain. if the open loop gain is 100db, and the closed loop gain is 10db, then the loop gain is 90db. this is useful to know, because you can use it to find the open loop gain without a lot of tweaking of dc offset voltages (required to keep am op amp running open loop from sticking to a rail). so if your closed loop gain, and the middlebrook circuit shows a loop gain of 90db, then the open loop gain is 100db


.............................................................................................................

maybe i should think about keeping a plain vanilla copy of LTSpice just for sharing purposes. up to this point i think most of my posts were either .jpg pix of a simulated circuit or using the default transistors. i like having 3rd party models in my standard.bjt file because it saves time, but i don't want to give someone fits if i upload a model and they have to spend a week getting the models for the devices. i know that 2N3904 and 2N3906 models will do just about anything, including providing hundreds of amps emitter current.... the only problem is the real devices have real limitations..... in other words, before stating that a circuit works, check your device currents and voltages and make sure you are within the real device specs. sims can help you reduce your burnt parts count in a prototype, but only if you pay attention to the actual device specs vs what you see in a sim.

 

[Rafael L]

In this case value middlebrook loop gain, is the amount of negative feedback ?

In the circuit of my post here the loop gain middlebrook is 42dB, the gain in closed loop is (10K/1K+1=11)
in dB (20 log 11=20.8 dB)
Is correct to affirm that:
Open loop:42dB+20.8dB= 62.8dB( loop gain considering the impedance circuit )
Negative feedback: 42dB

I am sure?

Thanks

 

[Rafael L]

In my post above, I want to know the real value of my circuit.
I know that for gain open loop in voltage is 84dB (AOL), gain closed-loop 20.8dB (B, Beta ) or AOLB if I measure VOUT, negative feedback is aprox~64dB.
OBS: My circuit in link the post above.
I found that middlebrook I would be the open loop gain(AOL, considering the impedance).
Andy wrote what! probe middlebrook is AOLB, but the measured value is greater that gain closed-loop, in this case can only be:difference between open-loop and closed-loop or gain open loop (attenuated by impedances).
This is what I want to know...

thanks

 

[unclejed613]

it also depends what frequency you measure at if you measure the loop gain at 10HZ you will get a much higher result than if you measure it at 1khz. this is because above a certain frequency (called the dominant pole frequency) the Aol begins rolling off at -6db/octave. a lot of op amps (and the LTSpice generic model op amp) have a dominant pole frequency of about 10hz. you can change this in the LTSpice generic op amp model by changing the GBW parameter. you can also change the Aol.

yes the loop gain is the negative feedback, which in most cases is derived from an attenuator. the gain portion of it is inside the amplifier, the attenuator is external.

 

[andy_c]

In this case value middlebrook loop gain, is the amount of negative feedback ?

It is close, but not exactly the same thing. What's normally called "the amount of feedback" is the ratio of the open-loop gain to the closed-loop gain. The open-loop gain is of course A_OL. The closed-loop gain is A_OL/(1 + A_OLB). When you divide these two, you get:

amount_of_FB = (1 + A_OLB)

So you could say "the amount of feedback is equal to the loop gain plus one". If the loop gain is large, it's nearly the same thing as the amount of feedback.

 

[stinius]

Hi Andy

Maybe its nitpicking, but the formula for closed loop gain should be:

AOL/(1 - AOL B)

When you have negative feedback (-B), you will get AOL/(1 + AOL B)

 

[andy_c]

Hi Andy

Maybe its nitpicking, but the formula for closed loop gain should be:

AOL/(1 - AOL B)

When you have negative feedback (-B), you will get AOL/(1 + AOL B)

This goes back to my earlier conversation with Rafael, where I showed the diagram for the feedback system here For that case, you will see a plus sign before the A_OLB.

 

[Rafael L]

Hi

yes the loop gain is the negative feedback, which in most cases is derived from an attenuator. the gain portion of it is inside the amplifier, the attenuator is external.

unclejed , yes I got the same conclusion

The closed-loop gain is AOL/(1 + AOLB)

I believe "AOLB" is the value probe middlebrook, really, get my closed loop. (if AOL x B denominator be greater than numerator which results in value <1)
Remember that, to use the formula must convert dB to volts (anti-log, I do not know the translation this Mathematics function in English), just select the option in LTspice "linear" for scale in volts, in program is automatically.

So you could say "the amount of feedback is equal to the loop gain plus one". If the loop gain is large, it's nearly the same thing as the amount of feedback.

Andy I did simulations with different value the ratio, and really see that the value middlebrook, is the amount of negative feedback,
I just add value middlebrook with value closed loop, the result is AOL.
Strangely I had a high value using the formula V(out)/(V(a)-V(b)), to the values of between Beta 5 to 15, this circuit here.
another way of measuring AOL , is put Q2 in earth and disconnect the link feedback, I am with a value of AC=1 then the biggest difference that I have between 1 and 0 in circuit , is 1V.
this would be the fixed gain AOL.

Thanks for answer my questions.

 

[klewis]

Re: Re: Re: Re: Fairchild KSC3503/2SC3503 Model

Fairchild has a revised spice models for the SC3503 from 03.11 2008.
I have tried to get them, but it seems like Fairchild has some problems sending them.
You have to order the models by email, it usually takes seconds to receive them, but I ordered the yesterday and still no response.
Anybody else has the same problem?

Cheers
Stinius

I had the same problem around the same time frame, and sent Fairchild a note to have their web folks look into it. I don't know if it was me, but, I'm now able to get models without a problem. If you haven't tried in the last week or two, give it a try now.

Ken

 

[Rafael L]

Oscillation...

Hi

I added more transistors output, is showing a peak in the dc closed-loop this indicates a local oscillation?

The original circuit has the resistors of 10 ohms, the base of the transistor output.

Thanks

 

[unclejed613]

yes, that peak shows instability. and i suspect your input signal was 0.5V or less? and that the gain was still greater than 1 when the phase line crossed the 180 degree mark.

 

[Rafael L]

yes, that peak shows instability. and i suspect your input signal was 0.5V or less? and that the gain was still greater than 1 when the phase line crossed the 180 degree mark.

I have noticed, that exceed the response in frequency (>800KHz in closed loop)even with good phase margin, always have this peak.
I have to limit the response placing a capacitor short circuit ratio

To simulate the correct, I would have to include a mathematical model thermal output stage?

http://peufeu.free.fr/audio/articles/Thermal modeling of power systems.pdf

 

[unclejed613]

just because you have instability, doesn't mean it will always show up as oscillation under normal operation of the amp into a resistive load, however 2 different operating conditions might cause oscillation, running the amp into a moderately capacitive (around .01 or ,02uf across the load resistor) load, or running the amp into clipping, with the worst case being clipping into a capacitive load. loads tend to get capacitive for two reasons, long speaker cables and crossover networks in the speakers. the first, you have some control over, the second you have less control over. as far as clipping goes, the only way to avoid it completely is by using a dynamic limiter that lowers the gain of the amp before it clips. this is a form of compression, and works well for average listening, but sounds artificial if you try to push the amp. the other way to keep the amp out of clipping would be to use "soft clipping" which clips the input signal before the amp outputs reach saturation. both of these methods are in use by various manufacturers. Peavey and Fender use the first method because some of their solid state amp designs tend to misbehave when they saturate. NAD used (maybe even continues to use) the second method, but it's a user settable option. NAD's amps are usually pretty stable so they dont require the use of soft clipping, but they include it because some listeners prefer the smooth tube-like clipping to transistor saturation clipping.

 

[unclejed613]

your simulation data shows another indication of instability that you never hear much about. the slope of the gain rolloff is -48db/decade. -20db/decade (IIRC) is normal for a stable amplifier. the steeper the slope, the more unstable the amp is.

 

[Rafael L]

your simulation data shows another indication of instability that you never hear much about. the slope of the gain rolloff is -48db/decade. -20db/decade (IIRC) is normal for a stable amplifier. the steeper the slope, the more unstable the amp is.

This brute rolloff (48dB or even more), is due to peak in closed-loop, if I limit response in frequency in closed-loop, the curve rolloff improves

 

[simulation11]

Very good discussion. Thank you for your posts.

simulation rachat credit​

 

[unclejed613]

This brute rolloff (48dB or even more), is due to peak in closed-loop, if I limit response in frequency in closed-loop, the curve rolloff improves

actually it's there for the same reason as the peak, the amp is under compensated, and adding an integrator cap in the feedback is just a bandage on a sucking chest wound. get to the root cause, increase the compensation cap, or increase the diff amp current or both. i don't know why the first response of most experimenters is to add an integration cap when the problem lies elsewhere. two root causes of instability exist, 1) compensation, the dominant pole cap is too small, or the diff amp )LTP) current is too low, or both, and 2) grossly mismatched input stage and output stage slew rates ( for a detailed discussion of this go here: http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1154,C1009,C1028,P1219,D4138

the whole app note is very good reading, but if you want to cut to the chase, read appendix C. there's also a very good appendix detailing the application of murphy's law to electronics (who says engineers have no sense of humor? they actually put this in an app note...)

 

[Rafael L]

actually it's there for the same reason as the peak, the amp is under compensated, and adding an integrator cap in the feedback is just a bandage on a sucking chest wound. get to the root cause, increase the compensation cap, or increase the diff amp current or both. i don't know why the first response of most experimenters is to add an integration cap when the problem lies elsewhere.

Yes, you are correct

two root causes of instability exist, 1) compensation, the dominant pole cap is too small, or the diff amp )LTP) current is too low, or both, and 2) grossly mismatched input stage and output stage slew rates

I am working with the topology Symasym amp:
http://www.diyaudio.com/forums/showthread.php?s=&threadid=60918
If you look at the scheme, the input of transistors VAS are not equal (they have rolloff different ) that was what was causing the peak, I believe have solved with two Cdom.
....Complicated topology....

( for a detailed discussion of this go here: http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1154,C1009,C1028,P1219,D4138

Error link, what is the title of the article?

Attached is my simulations (Middlebrook and closed loop), the closed-loop the response in frequency to 1.2 Mhz -3dB, This brute rolloff , I think that this frequency is very high for an power-amplifier, I would have problem?

Thanks

 

[unclejed613]

the link works..... try a search on linear.com for AN47 or Application Note 47 (for some reason, searching for AN-47 with a dash in it brings up the link, but far down the list of search results)