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Error Amplifier Design for SG3524

Dear All,

I am now making a 48 V to 120V Boost Converter (with a single series

560uH 10A inductor) using PWM IC SG3524, and I am a little bit confused with the use of error amplifier input pins (pin 1 and pin 2) and the compensation pin (pin 9). A typical example application circuit (see figure 13) divides the reference voltage 5V to 2.5V, and feeds it to pin 2 as reference. The output is also divided and feed to pin 1 setting it to give 2.5V (for example if the output is 15V, the divider will be 15k and 5k, giving 2.5V at steady state condition), there is no feed back resistor between pin 9 and pin 1. I assume that it is using the simplest type 1 amplifier configuration
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But what I am not sure is how exactly it comes to this, why is it the output is divided to give the 2.5V input to pin 1 which is nominally the same as reference voltage 2.5V at pin 2? Assuming that the Ramp voltage ranges from 1.2V to 3 V and the open loop gain is 10,000 as per application note, a 0.3mV differential voltage difference between pin 1 and pin 2 may drive the duty cycle to it's maximum.

Could anybody please confirm if my assumption is correct and explain to me how it works? And please let me know whether I should use the same configuration for my Boost Converter, to convert 48V to 120V, shall i just divide 120V nominal to become 2.5V to feed back to pin 1 while the reference at pin 2 is also 2.5V?

Pardon me if my questions sound stupid to you, I am going to test it out tonight anyway but I don't want to see smokes :)

Regards

Reply to
Myauk
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That's called the "error amplifier". An opamp tries to keep both inputs the same if you allow it to. For that it needs a means to control at least one of them and in a boost converter that is the switching action. It can pump more or less energy into the output capacitor and that determines the output voltage. So it will try to get the divided voltage to the same as the reference fed into the other pin. Which, I assume, is what you'd want.

--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

Dig through the data sheets a bit, and you will find that the error amplifier output impedance is high, meaning that compensation need only be put between pin 9 and ground. Look at your example again -- what's going from compensation to ground?

Well, if your output voltage is low, wouldn't you _want_ your duty cycle to be high, perhaps at maximum?

The action of the error amplifier is to do just what it says: amplify the error. If the output is a bit low, you want the duty cycle to go up to boost it. If the output is a bit high, you want the duty cycle to go down to let the output slip down.

The action of the compensation is critical: without it, you'll build a humongous power oscillator. You need a lead-lag network as described in here:

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but given the questions you're asking I'm not sure that you'll know enough to do the right thing.

The best suggestion I can give, without designing the whole thing for you, is to see if you can find an example schematic and copy it. Alternately, start with the values that TI mentions in their data sheet, then if it doesn't oscillate right off the bat, play with the values to get the most stable behavior as you switch loads into and off of the output.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply to
Tim Wescott

How much current output do you expect?

Got a schematic?

John

and I am a little bit

Reply to
John Larkin

Others have addressed your question about the feedback loop. I would like to know what power level you're going to be running at. I don't want to give acctual numbers, but as boost ratio and/or power level increases you want to use a transformer instead of a single inductor. The chip you are using is designed to provide a push-pull drive to a transformer and does a good job of it. Years ago I used the SG3524 to build a 12 volt to 300 volt converter using a transformer. Also, what frequency do you plan to run? And a word of warning before you go to the breadboard: don't leave pin 10 open.

Reply to
Michael Robinson

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