"Myauk" <aungkokothet@gmail.com> wrote in message 
news:6c3cd791-0c2f-43b7-9f60-dd8554838f00@k22g2000prb.googlegroups.com...
> 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 (http://
> www.venable.biz/tp-03.pdf).
>
> 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

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. 

On Tue, 27 Dec 2011 21:44:25 -0800 (PST), Myauk
<aungkokothet@gmail.com> wrote:

>Dear All,
>
>I am now making a 48 V to 120V Boost Converter (with a single series
>560uH 10A inductor) using PWM IC SG3524,

How much current output do you expect?

Got a schematic?

John






 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 (http://
>www.venable.biz/tp-03.pdf).
>
>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

On Tue, 27 Dec 2011 21:44:25 -0800, Myauk wrote:

> 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 (http://
> www.venable.biz/tp-03.pdf).

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?

> 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.

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

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: http://www.ti.com/lit/ds/symlink/sg2524.pdf, 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

Myauk wrote:
> 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 (http://
> www.venable.biz/tp-03.pdf).
> 
> 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 :)
> 

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/

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 (http://
www.venable.biz/tp-03.pdf).

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