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Old 11-05-2009, 07:29 PM   #1
bobdole369
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Default SG2525 (SG3525) based PWM and power supply

OK, so I just finished building a DC-DC converter based on the SG2525 PWM and some IRFP240s. It works but I don't think I have a very good grasp of the Compensation, purpose of the opamp inputs (inverting and non-inverting inputs), etc.

Lots of circuits I see based on this just enable "shutdown" when voltage gets too high. Others modulate the non-inverting input with the inverting one tied to the Vref, still more use an extensive compensation network.

Why all the difference? Anyone an expert on the SG2525/3525 that cares to speak a bit on how the feedback is "intended" to work on this chip?

My circuit is 70kc, directly drives the gate of the IRFP240, that runs a forward boost converter - 100uh filter, MUR3060 diode (well half of it), into a 1000uF cap. Left to run wild it gets to 80 volts mighty quick.

My comp network is a simple transistor 2n3904 operating in the linear region biased so that 19V is towards the bottom of the compensation voltage, and thus drives the FETS at minimal duty cycle. As the power demand increases (and voltage decreases) the transistor increases the voltage at the comp pin to turn up the duty cycle.

SO why would someone use shutdown pin, and what is the purpose of the error amp? Best I could tell when non-inverting input was less than vref, the thing was on. When less - it wsa off, couldn't adjust duty cycle this way.

Anyone care to offer insight?
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Old 11-06-2009, 01:15 AM   #2
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Shutdown is not used for regulation, typically only for fault conditions (over voltage, over current, etc from component failures). If your knowledge of opamp behavior is lacking (or anything else applicable here), I really suggest you pick up a book to gain a basic understanding of something you're experimenting with.

Using the error amplifier to regulate the supply is the proper way to do this (in my book & in many others: it's the only way for any supply - I can assure you that you'll never find a book that advocates shutting down the IC to somehow provide regulation). All regulated supplies will invariably compare the output voltage at some point to a constant reference voltage in an error amplifier, which itself may be internal to the chip or external. This error amp output (and possibly others, such as in a current mode configuration) is then compared to a sawtooth waveform, which is generated by an internal oscillator & the external timing resistor + capacitor (or an external clock through the sync pin). The resulting comparison drives the outputs with a duty cycle dependent on the length of time that the fractional output voltage is above or below Vref (or whatever voltage is fed into the other error amp input terminal) - the duty cycle will also be a function of any addition error amplifiers used. The SG3525 also uses some digital logic to provide other useful functions, such dead-time.

Compensation exists for the purpose of limiting bandwidth and preventing oscillations of the supply, which result (largely) from the LC filter phase shift + 180º error amp phase shift. Compensation is implemented by an error amplifier with a complex gain - resistors & capacitors are used to place poles & zeros in calculated locations (sometimes a pole is used to cancel a zero that is inherent to the design, i.e. an ESR zero from the output capacitor). Current-mode & voltage-mode both have different compensation schemes, not to mention some topologies require different networks altogether (i.e. a discontinuous mode flyback). The result of placing these poles & zeros is the creation of a closed-loop system that is unconditionally stable, as defined by control theory stability criteria.

I would not attempt to further operate this unit without gaining a firmer understanding of what's going on here, especially if this is an off-line/non-isolated experiment (doubly so if there is no primary-secondary galvanic isolation!). Connecting anything other than feedback loop components (resistors & capacitors from the (-) error amp input) to the COMP pin is not recommended, as it is driven directly by the error amplifier - driving it externally could short out a transistor in the error amp & cause COMP to be permanently be pulled high or low, which makes the chip useless. It's always possible to design something that 'works', but designing something well will take effort to understand what's going on here, which I think you should look into (Pressman's Switching Power Supply Design provides all you will need to know & more on this topic). I suspect the only reason this hasn't blown up yet is the choice of a double output chip which will never reach 50% duty cycle with only one output used & the inherent dead time - if you had by chance used another chip that was capable of > 50% duty cycle on a single output, the transformer would saturate after several cycles, causing the FET to fail (likely short-circuit), and then whatever protection mechanism (if any) to trigger after that. Of course it's possible that your scheme is in some way providing regulation and failure wouldn't occur as I described, but until I see a schematic I'm doubtful.

Last edited by PM650; 11-06-2009 at 01:46 AM.
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Old 11-06-2009, 04:32 AM   #3
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Usually it's enough to have one input to vref, and the other to a voltage divider off the output (which will be equal to vref when the output is the proper voltage...)

If you're not terribly familiar with opamps, TI has an excellent book called "opamps for everyone"
http://focus.ti.com/lit/an/slod006b/slod006b.pdf
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Old 11-07-2009, 05:51 PM   #4
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Originally Posted by PM650 View Post
Shutdown is not used for regulation, typically only for fault conditions (over voltage, over current, etc from component failures). If your knowledge of opamp behavior is lacking (or anything else applicable here), I really suggest you pick up a book to gain a basic understanding of something you're experimenting with.

Using the error amplifier to regulate the supply is the proper way to do this (in my book & in many others: it's the only way for any supply - I can assure you that you'll never find a book that advocates shutting down the IC to somehow provide regulation). All regulated supplies will invariably compare the output voltage at some point to a constant reference voltage in an error amplifier, which itself may be internal to the chip or external. This error amp output (and possibly others, such as in a current mode configuration) is then compared to a sawtooth waveform, which is generated by an internal oscillator & the external timing resistor + capacitor (or an external clock through the sync pin). The resulting comparison drives the outputs with a duty cycle dependent on the length of time that the fractional output voltage is above or below Vref (or whatever voltage is fed into the other error amp input terminal) - the duty cycle will also be a function of any addition error amplifiers used. The SG3525 also uses some digital logic to provide other useful functions, such dead-time.
[/SIZE][/SIZE]

True indeed. But perhaps bobdole369 read about the following techniques : continuously mode of operation and discontinuously mode of operation.

continuously mode of operation :

The regulating pwm circuit will always be active.
The pwm signal will be altered when the output voltage is higher then the setpoint voltage.
This causes less ripple but consumes more current.
There will always be a current flowing through the inductor.

Advantages : smaller ripple on the output voltage.

discontinuously mode of operation.

The regulating pwm circuit will not always be active.
When the voltage is higher then the setpoint voltage, the regulator will stop the pwm signal untill the voltage drops under a certain minimum level.
The current through the inductor will drop to zero.

Advantages : Higher efficiency.

The LT1934 for example lacks an error amplifier but uses a comparator with hysteresis. This hysteresis determines the ripple on the output voltage.

Linear technology and renesas( former motorola, i always forget al those subdivisions)has some great examples and pdf about the how and why.
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Old 11-07-2009, 07:45 PM   #5
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Quote:
Originally Posted by William Gaatjes View Post
True indeed. But perhaps bobdole369 read about the following techniques : continuously mode of operation and discontinuously mode of operation.

continuously mode of operation :

The regulating pwm circuit will always be active.
The pwm signal will be altered when the output voltage is higher then the setpoint voltage.
This causes less ripple but consumes more current.
There will always be a current flowing through the inductor.

Advantages : smaller ripple on the output voltage.

discontinuously mode of operation.

The regulating pwm circuit will not always be active.
When the voltage is higher then the setpoint voltage, the regulator will stop the pwm signal untill the voltage drops under a certain minimum level.
The current through the inductor will drop to zero.

Advantages : Higher efficiency.

The LT1934 for example lacks an error amplifier but uses a comparator with hysteresis. This hysteresis determines the ripple on the output voltage.

Linear technology and renesas( former motorola, i always forget al those subdivisions)has some great examples and pdf about the how and why.


Transformers are never operated in continuous mode, the current always returns to zero after each switching period. Operating a transformer in continuous mode is analogous to maintaining a DC current bias through the winding, which will quickly cause the core to be driven into saturation & the switching transistors to be destroyed from the large currents. Or, if you did design it to work with a flux imbalance, winding losses would be huge & the hysteresis loop that the core traverses would be very small (low efficiency).

That LT1934 is a bit of an exception...with the majority of controllers around, error amps will not be anything special (no hysteresis) and max on/off times will not be fixed - as such, traditional compensation methods will be required. Not having a traditional oscillator & having preset delays + hysteresis to prevent oscillation being determined purely by propagation delay is hardly typical - this is why I wouldn't recommend shutting down the controller to provide some sort of regulation. No matter the mode with a typical controller, it is never shut down for anything other than fault conditions - discontinuous mode only implies that the inductor/transformer primary current falls to zero at the end of each cycle. The SG3525 is a universal dual-output type that can be expanded to have any number of error amplifiers & drive any output scheme from single forward to full bridge - the LT1934, however, is really built to do one thing and one thing only, which is not a fair comparison imo.
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Old 11-07-2009, 09:14 PM   #6
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Thanks for all the replies guys, esp PM650. I am no EE, don't claim to be. I am a tinkerer and have had a lot of schooling in electronics (US Navy NEETS, 2 years of voc/tech) so I'm no dummy. All the components I'm using are junkbin jobs, and as stated this project is for learning more than anything. If I do happen to get a 100% stable and safe circuit built - then I've got myself a 12V -> 19V DC-DC supply with an amp or so on the output. The idea here is to power a mixer that will provide an interface to a measurement mic that I'm using to measure the frequency response of the audio setup in my car. I also would use it as a float charger for some gel-cells, and of course those will need a charger circuit as well.

BTW this IS an isolated circuit. I'm supplying 12V from a 120-12V (10A) Meanwell din-rail mount supply. It's fused at 1A coming in to my circuit.

In any case - It's good to hear that I think I understood how it is "supposed" to work, as a lot of electronics stuff forums (specifically edaboards) seem very "results driven" and building and making stuff "just work".

Quote:
Using the error amplifier to regulate the supply is the proper way to do this (in my book & in many others: it's the only way for any supply - I can assure you that you'll never find a book that advocates shutting down the IC to somehow provide regulation).
Good - over and over again I see circuits built where the shutdown pin is pulled low when the output voltage gets too high. I understand how that works, but I did not think that was how it should be done. I imagine shutdown should be used in a big overvoltage, or some other unexpected state.

Quote:
Connecting anything other than feedback loop components (resistors & capacitors from the (-) error amp input) to the COMP pin is not recommended, as it is driven directly by the error amplifier
This was my original philosophy too. I intended to use the + input of the error amp connected via a 1K to the Vref output, to hold that one at 5.1V, and the - input to a voltage divider that at my desired output, would be 5.1V.

However when building the circuit I did so in stages. First I got the 2525 to start oscillating by connecting my intended components to softstart (1uf cap to ground), Rt, Ct, discharge, shutdown, and 4.7kohms on both outputs to ground. I was applying a voltage while watching on a scope, the output. The only real way I could get a duty cycle change was to apply external voltage between 0.8V and 3.6V to the comp pin. (via a potentiometer voltage divider and the supply voltage) Using the error amp input only seemed to turn off the chip as if I had activated shutdown.

So does it seem to you as if I have a bad 2525?

As it stands now I've built an inverting amp from a 2n3904 and biased it such that 22V input to the circuit saturates (0V output), and 17V input fully turns off the transistor (giving 4V output). I connected this up so that output goes into the comp pin. It works, and quite well too, but as you surmise, this is still not the right way to do things. I should be able to reconfigure my regulator input circuit such (perhaps as a voltage follower) to provide input to the - input on the error amp and THAT would be the way to go.

Quote:
(Pressman's Switching Power Supply Design provides all you will need to know & more on this topic)
The second time I've seen this book recommended. I WILL be picking it up.

Quote:
I suspect the only reason this hasn't blown up yet is the choice of a double output chip which will never reach 50% duty cycle with only one output used & the inherent dead time - if you had by chance used another chip that was capable of > 50% duty cycle on a single output, the transformer would saturate after several cycles, causing the FET to fail (likely short-circuit), and then whatever protection mechanism (if any) to trigger after that.
I'm not using any transformer, the output drives the gate of an IRFP240, which pulls +12V through a 100uH inductor that claims to handle 4A, and a half an MUR3060, then across a 1000uF cap with 4.7kohm bleeder, i.e. a boost converter. If I did happen to achieve >50% duty cycle, what exactly would happen in that case?

By what mechanism would the FET short if there was indeed a transformer involved?

Quote:
Compensation exists for the purpose of limiting bandwidth and preventing oscillations of the supply
So now I've obviously built something that "just works" and may have a bad part on my hands. No biggie I have lots more of these. So my next step is to understand the hows and whys of the compensation networks.

Thanks a ton, exactly why I sometimes turn to ATHT for help
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Last edited by bobdole369; 11-07-2009 at 09:21 PM.
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Old 11-07-2009, 09:24 PM   #7
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Originally Posted by PM650 View Post


Transformers are never operated in continuous mode, the current always returns to zero after each switching period. Operating a transformer in continuous mode is analogous to maintaining a DC current bias through the winding, which will quickly cause the core to be driven into saturation & the switching transistors to be destroyed from the large currents. Or, if you did design it to work with a flux imbalance, winding losses would be huge & the hysteresis loop that the core traverses would be very small (low efficiency).

That LT1934 is a bit of an exception...with the majority of controllers around, error amps will not be anything special (no hysteresis) and max on/off times will not be fixed - as such, traditional compensation methods will be required. Not having a traditional oscillator & having preset delays + hysteresis to prevent oscillation being determined purely by propagation delay is hardly typical - this is why I wouldn't recommend shutting down the controller to provide some sort of regulation. No matter the mode with a typical controller, it is never shut down for anything other than fault conditions - discontinuous mode only implies that the inductor/transformer primary current falls to zero at the end of each cycle. The SG3525 is a universal dual-output type that can be expanded to have any number of error amplifiers & drive any output scheme from single forward to full bridge - the LT1934, however, is really built to do one thing and one thing only, which is not a fair comparison imo.
True, transformers are never used that way. It would be no use as you described. I was thinking of coils. And really shutting down a complete pwm controller until a reset condition is applied, indeed is only done as a fault conditioning. The principle i was thinking of is described as in for example the LT1934.
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Old 11-07-2009, 09:48 PM   #8
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Quote:
Originally Posted by bobdole369 View Post
Thanks for all the replies guys, esp PM650. I am no EE, don't claim to be. I am a tinkerer and have had a lot of schooling in electronics (US Navy NEETS, 2 years of voc/tech) so I'm no dummy. All the components I'm using are junkbin jobs, and as stated this project is for learning more than anything. If I do happen to get a 100% stable and safe circuit built - then I've got myself a 12V -> 19V DC-DC supply with an amp or so on the output. The idea here is to power a mixer that will provide an interface to a measurement mic that I'm using to measure the frequency response of the audio setup in my car. I also would use it as a float charger for some gel-cells, and of course those will need a charger circuit as well.
I assume you will filter the output properly for the mixer use . The best no concession way is to make your switching voltage a bit higher and put a linear regulator behind it to totally clean up the ripples and spikes and other effects of passive filters. But if your mixer has a proper power supply there will be no use for that.

Quote:
BTW this IS an isolated circuit. I'm supplying 12V from a 120-12V (10A) Meanwell din-rail mount supply. It's fused at 1A coming in to my circuit.
In any case - It's good to hear that I think I understood how it is "supposed" to work, as a lot of electronics stuff forums (specifically edaboards) seem very "results driven" and building and making stuff "just work".

Good - over and over again I see circuits built where the shutdown pin is pulled low when the output voltage gets too high. I understand how that works, but I did not think that was how it should be done. I imagine shutdown should be used in a big overvoltage, or some other unexpected state.
Because as you already experienced so many things can go wrong with a pwm controller the chip is usually tucked with safeguard circuits as explained by PM650. But the shutdown pin can have 2 functions. Or it is a safeguard, or it is a pin to just switch off the powersupply when it is not needed. It depends on the chip. But to use a pin as a safeguard of overvoltage, it usually has a name that describes it's function. In my opinion shutdown is just that : Shutdown to not use the powersupply.
The sg2525 does not seem to have any internal safeguard circuits. Therefore the shutdown pin can be used to shutdown in case external circuits detect an fault situation or/and as just that, shutdown of the powersupply.


Quote:

This was my original philosophy too. I intended to use the + input of the error amp connected via a 1K to the Vref output, to hold that one at 5.1V, and the - input to a voltage divider that at my desired output, would be 5.1V.

However when building the circuit I did so in stages. First I got the 2525 to start oscillating by connecting my intended components to softstart (1uf cap to ground), Rt, Ct, discharge, shutdown, and 4.7kohms on both outputs to ground. I was applying a voltage while watching on a scope, the output. The only real way I could get a duty cycle change was to apply external voltage between 0.8V and 3.6V to the comp pin. (via a potentiometer voltage divider and the supply voltage) Using the error amp input only seemed to turn off the chip as if I had activated shutdown.
I would guess you need a voltage divider at your E-AMP inverting input.
and a reference voltage at your E-AMP non inverting input.

You scale down the output voltage with the resistor divider to match the ref voltage.
ref stands for reference. All you need is already in the ic itself. You only need to add resistors and capacitors.

EDIT: And a capacitor between the comp in and the inverting input !
Or your opamp will react allergic to any change in voltages .
1nF to 10nF will ususally do.
http://en.wikipedia.org/wiki/Operati...ons#Integrator


It is simple :

Output voltage
|
R1
|
- Feed back
|
R2
|
Ground


Feedback voltage = Output voltage * ( R2/ (R1+R2) )

Where the feedback voltage goes to pin 1.
And the ref goes to pin 2.

I assume you already have this :

http://www.st.com/stonline/books/ascii/docs/4286.htm

The error amp will do it's work then. Correcting the error ( is difference in voltage) between it's non inverting input and it's inverting input. That is the sacred rule of op amps : The op amp wants the voltage on both pins to be the same.


Detailed :
http://en.wikipedia.org/wiki/Operational_amplifier
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Old 11-07-2009, 11:20 PM   #9
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Quote:
Originally Posted by William Gaatjes View Post
True, transformers are never used that way. It would be no use as you described. I was thinking of coils. And really shutting down a complete pwm controller until a reset condition is applied, indeed is only done as a fault conditioning. The principle i was thinking of is described as in for example the LT1934.

My apologies, I just realized that this is a boost converter. When he said 'forward boost converter', I focused on 'forward' as the topology, and 'boost' as the overall purpose of this design. Now I know this is only a boost converter - no transformer involved, which is a nice piece of information to have. You were quite right to mention continuous vs discontinuous operation in this case.

Quote:
Originally Posted by bobdole369 View Post
BTW this IS an isolated circuit. I'm supplying 12V from a 120-12V (10A) Meanwell din-rail mount supply. It's fused at 1A coming in to my circuit.

In any case - It's good to hear that I think I understood how it is "supposed" to work, as a lot of electronics stuff forums (specifically edaboards) seem very "results driven" and building and making stuff "just work".
Yes the diy world is in a sad state of affairs in the more complex areas, many just want something to work instead of understanding the design process.
Quote:
Good - over and over again I see circuits built where the shutdown pin is pulled low when the output voltage gets too high. I understand how that works, but I did not think that was how it should be done. I imagine shutdown should be used in a big overvoltage, or some other unexpected state.
Generally it's connected to something with a latched or timered-latch/hysteresis to keep it shutdown until it is manually reset, or the timer ends. Sometimes you'll find those giant projection TVs that have a fault somewhere and make a 'flub-flub' noise - this is the smps attempting to start up, encountering a fault, shutting down, and restarting after an elapsed time after the fault trigger - the process keeps repeating until the TV is unplugged.
Quote:
This was my original philosophy too. I intended to use the + input of the error amp connected via a 1K to the Vref output, to hold that one at 5.1V, and the - input to a voltage divider that at my desired output, would be 5.1V.

However when building the circuit I did so in stages. First I got the 2525 to start oscillating by connecting my intended components to softstart (1uf cap to ground), Rt, Ct, discharge, shutdown, and 4.7kohms on both outputs to ground. I was applying a voltage while watching on a scope, the output. The only real way I could get a duty cycle change was to apply external voltage between 0.8V and 3.6V to the comp pin. (via a potentiometer voltage divider and the supply voltage) Using the error amp input only seemed to turn off the chip as if I had activated shutdown.

So does it seem to you as if I have a bad 2525?
It's great news that you have a scope, we won't be wasting time guessing at things here. The error amp (presumed working) will provide a high output whenever the (+) input exceeds the (-) input - so you should be able to get an adjustable duty cycle by connecting (+) to vref and (-) to some adjustable voltage. You're right that the duty cycle should be changing in response to a voltage that appears on the COMP pin - the problem here is that the error amp takes the difference between its two inputs and amplifies it by of a factor of several thousand (its open-loop gain is 7500 typically). This means it will only put out 0V or VCC - you need to add negative feedback around the error amplifier (to reduce gain) for the porposes of testing this 3525 in an open-loop configuration (open-loop = duty cycle not determined by the actual output voltage). For closed-loop regulation you may not need to reduce the gain of the error amp for stable operation, depending on the type of compensation scheme used.
Quote:
As it stands now I've built an inverting amp from a 2n3904 and biased it such that 22V input to the circuit saturates (0V output), and 17V input fully turns off the transistor (giving 4V output). I connected this up so that output goes into the comp pin. It works, and quite well too, but as you surmise, this is still not the right way to do things. I should be able to reconfigure my regulator input circuit such (perhaps as a voltage follower) to provide input to the - input on the error amp and THAT would be the way to go.
You shouldn't need any active components just to get this chip working in isolation. Just need to configure the error amp properly.
Quote:
The second time I've seen this book recommended. I WILL be picking it up.
Yes, the 3rd revision just came out in March. Unfortunately, I don't think the late Dr. Pressman will be writing any future books.
Quote:
I'm not using any transformer, the output drives the gate of an IRFP240, which pulls +12V through a 100uH inductor that claims to handle 4A, and a half an MUR3060, then across a 1000uF cap with 4.7kohm bleeder, i.e. a boost converter. If I did happen to achieve >50% duty cycle, what exactly would happen in that case?

By what mechanism would the FET short if there was indeed a transformer involved?
My mistake, when you said 'forward boost,' I immediately assumed a forward converter. With regards to smps topologies, 'forward' in general implies power is transfered to the load when a/the transistor is conducting (exactly opposite of how a boost converter behaves). The ‘>50% duty cycle danger’ was in reference to this being a forward converter. A forward converter must be allowed some time for the core to ‘reset’ (primary current returns back to zero, generally through a ‘catch diode’ that isn’t conducting when the transistor is). This reset is generally guaranteed by limiting the duty cycle to somewhat under 50%. Not allowing a complete reset will cause a DC bias current to be present on the primary, which in the best case, will only cause higher copper losses (windings dissipate more heat). This DC current bias also moves the core further up into the hysteresis loop, which causes higher current draw nearer to saturation. The extra current draw is purely extra core loss – whatever magnetic material used for the transformer core will heat up more as a result. If the DC bias current is high enough and the supply continues to operate, it will approach saturation. Saturation is a function of the magnetic properties of the transformer core material, it is the point when an inductor begins to act like a piece of resistive wire (equal to the very low DC resistance of the windings). A saturated inductor in a forward converter, as well as in a boost converter, will generally destroy the switch if no protection is present.
Quote:
So now I've obviously built something that "just works" and may have a bad part on my hands. No biggie I have lots more of these. So my next step is to understand the hows and whys of the compensation networks.

Thanks a ton, exactly why I sometimes turn to ATHT for help
No problem.

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Old 11-08-2009, 06:45 AM   #10
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This website might give some assistance.

http://schmidt-walter.eit.h-da.de/smps_e/smps_e.html
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Old 11-08-2009, 11:56 PM   #11
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Originally Posted by bobdole369 View Post
OK, so I just finished building a DC-DC converter based on the SG2525 PWM and some IRFP240s. It works but I don't think I have a very good grasp of the Compensation, purpose of the opamp inputs (inverting and non-inverting inputs), etc.

Lots of circuits I see based on this just enable "shutdown" when voltage gets too high. Others modulate the non-inverting input with the inverting one tied to the Vref, still more use an extensive compensation network.

Why all the difference? Anyone an expert on the SG2525/3525 that cares to speak a bit on how the feedback is "intended" to work on this chip?

My circuit is 70kc, directly drives the gate of the IRFP240, that runs a forward boost converter - 100uh filter, MUR3060 diode (well half of it), into a 1000uF cap. Left to run wild it gets to 80 volts mighty quick.

My comp network is a simple transistor 2n3904 operating in the linear region biased so that 19V is towards the bottom of the compensation voltage, and thus drives the FETS at minimal duty cycle. As the power demand increases (and voltage decreases) the transistor increases the voltage at the comp pin to turn up the duty cycle.

SO why would someone use shutdown pin, and what is the purpose of the error amp? Best I could tell when non-inverting input was less than vref, the thing was on. When less - it wsa off, couldn't adjust duty cycle this way.

Anyone care to offer insight?

Wow, you actually have quite a bit of design work to if you want anything reliable from the converter... I'll try to explain the whole concept here but you're really trying to understand about a year's worth of academics here.

The shutdown pin is used to keep the whole supply "off" by a seperate reset logic circuit or a microcontroller and has nothing to do with compensation of the converter.

Internally, the SG3525 basically takes the difference between the output voltage and reference voltage (inputs to the error amplifier) and feeds that into a comparator, which then feeds into (I'm guessing) a sawtooth waveform to generate the PWM.

The SG3525 has a dedicated compensation pin to help keep the converter stable. This is especially important in boost converter designs (Vout > Vin) where there is an inherent unstable pole that requires compensation for stability. For buck converters (Vout < Vin), compensation is a little easier since there are no unstable poles (all poles are in left hand plane).

Now back to the error amplifier... if you set Vref to 1V and want your output to be 10V, your feedback network will be a ratio R1/R2 (and perhaps maybe additional capacitors for compensation, but lets keep it simple for now) so that the voltage between the two resistors will be 1V, which is the same as your Vref.

If your Vref is fed into the inverting input of the error amplifier and the feedback voltage is fed into the non-inverting input, the output of the error amplifier will be Vfb - Vref, which is the ERROR between the desired voltage and the actual voltage.

If the output voltage is lower than 10V, Vfb will be less than 1V and the output of the error amplifier will be negative, which will in turn cause the duty cycle to increase to correct for the lower output voltage. If the output is greater than 10V, well you get the idea.

The whole circuit is a closed-loop system that may or may not be stable depending on the topology. However in order to guarantee that the system will be stable under the specified conditions (ie, max. input voltage, min input voltage, max load current, etc etc) and that it will response as quickly as possible to changes, you need to add "compensation" to the network and its values are determined by the caps, inductors, and resistors used in the controller.

Good luck with this! If you need some more help, PM me the circuit and I'll whip up some analysis for you.

Last edited by uclabachelor; 11-09-2009 at 12:01 AM.
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Old 11-17-2009, 09:34 AM   #12
bobdole369
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Update:

It's working.

I've managed to figure out how the error amp works, I believe its now properly set up, and it regulates the voltage!

When I was just poking around I didn't really understand that there was actually what amounts to a real op-amp inside, exposing its inputs and outputs. According to the datasheet (which when carefully read - actually shows you these things), the output from the op-amp connects directly to the Comp output pin, which connects internally to part of the PWM. THis is where I had been injecting my 1-3V signal to adjust duty cycle.

I missed the bit about an op-amp with no negative feedback being essentially infinite gain, so microvolt level changes would make wild swings on the duty cycle. This is why I couldn't figure it out!

So I took all that circuitry out, and put in a simple voltage divider with a PCB-mounted pot to fine-tune the output voltage.

Have some assorted bits, but basically I've got it set up as it should be as a differential amp with reasonably high gain. vref goes to the noninverting output, the output from my voltage divider (meant to be 5.1V when the output voltage is at 19.5V) goes to the inverting input. I have 40K for feedback, which should limit the gain to about 30db. It works. I applied power, and wham - 19.44V on the output. My pot lets me swing the output up or down a volt either way. I apply a bit of a load and watch the output voltage dip and its back at 19.44 within a half second or so.

I'm good for 150ma, but that due to my inductor choice (100uH) and output filter (1000uF). I'll work on that bit next.
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