Author Topic: How are some Buck Converters able to output below Vref?  (Read 3619 times)

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Offline thejoggingmat

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How are some Buck Converters able to output below Vref?
« on: August 29, 2015, 11:59:05 am »
I was playing with my newly arrived B3603 module when I was curious: How was it able to achieve voltages below Vref?

The B3603 is a fancy digitally controlled power supply module based on the LM2596. By defult, the LM2596 can't go below its Vref of ~1.25V (so do most of the other common buck converters that I know of).

Looking at the LM2596's datasheet,


Even if one were to adjust R2 to 0R, the Vout can never fall below Vref.


What puzzled me more was the B3603's schematic (source: https://github.com/baruch/b3603/raw/master/docs/B3603_BottomBoardSchematics.pdf)


It did not seem to indicate how it could go below Vref.

Historically, I have also seen (though I can't remember from where) LM2596-based buck converters that have claimed to be able to go belo ~1.25V.


So, how do these buck converters achieve voltages below Vout?
« Last Edit: August 29, 2015, 12:06:40 pm by thejoggingmat »
 

Online Marco

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Re: How are some Buck Converters able to output below Vref?
« Reply #1 on: August 29, 2015, 12:37:44 pm »
The feedback for the LM2596 is simply banged to the rails (with an added Schottky drop and a 1/2 resistor divider). This makes the internal reference voltage meaningless.

The actual reference voltage used for control is formed by a PWM signal combined with the low pass filter input on the "PI controller"
« Last Edit: August 29, 2015, 12:40:26 pm by Marco »
 

Offline codeboy2k

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Re: How are some Buck Converters able to output below Vref?
« Reply #2 on: August 29, 2015, 03:32:05 pm »
The output of a PI controller is a slow moving voltage, in response to the error signal input.  It will settle slowly and respond slowly to set-point changes, and it will react slowly to transient changes.  It won't be bang-bang against the rails.   The 2 diodes back-to-back before it gets to the feedback input is just a diode OR, to make sure that only one of the 2 controllers is in control at any given time (Voltage or Current). 

The internal reference is still being used, and the voltage from the PI controllers (whichever one is in control) will do whatever it needs to do, and reach whatever voltage it needs to reach, within it's abilities, in order to make the reference input voltage = 1.235 Volts. So, working backwards from the reference input, you see that Vref is 1.235 V. Then there is the voltage divider, so the top of the voltage divider will be Vref * 2 = 1.235*2 = 2.470.  Then there is a diode drop to the output of either PI controller, which means the PI controller output wants to stay at 2.470+0.2V = 2.67 V. With error feedback from the OUT+ and OUT- points into the first AMP U2A (which is a weighted difference amp), and the PWM setpoint from the micro going into the PI controller's setpoint input, this will be what determines the output path as it settles to 2.67 V.

The output voltage is measured at OUT+ relative to OUT- , not at OUT+ relative to GND.  So the controller can reach 0 Volts output because it's not actually at 0 Volts relative to GND, it's going to be at 0 Volts relative to OUT-.  So as long as OUT- and OUT+ are EQUAL, it's considered 0 Volts output.

I didn't do the math, but you can analyze the U2A difference amplifier using supposition to get the transfer function, and determine what value is required for OUT+ equal to OUT- that gives Vout = 0 for that amplifier. Note that R13, R14 and R11 inject a 50mV voltage source into the amplifier too.  So your supposition calculations will have 3 voltage sources, 50mV, OUT+ and OUT-.  Supposition means set all voltage sources to 0V except 1, calculate the transfer function for 1 voltage source only, then repeat for the others.  The final transfer function is the sum of the 3 transfer functions. When you figure out the final transfer function you can set OUT+ equal to OUT- and simplify it further.  Then solve for Vout = 0 Volts.   This 0 volts is what will be matched with the setpoint PWM signal, and I am assuming that 0 volts PWM = 0 volts output desired.    Once solved, you will probably find that be OUT+ = OUT- = 1.235 V :-)  ( or maybe slightly higher )..


 

Offline Chris C

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Re: How are some Buck Converters able to output below Vref?
« Reply #3 on: August 29, 2015, 04:22:40 pm »
Simplest possible answer:

The LM2596 simply tries to keep the voltage at the feedback pin at Vref.  You can't get it to go lower than Vref with a voltage divider, or even a direct connection between Vout and feedback.

But if you amplify Vout by 2X (like with an op-amp), and run that into the LM2596 feedback pin, then Vout will be Vref/2.  Or 1.23/2=0.615V.  (Whether the LM2596 will be stable connected in such a way, for simplicity's sake, is a separate topic.)

Notice there are op-amps driving the feedback pin in your controller.  They can amplify Vout, making operation at lower than Vref possible.

Now I gotta study that schematic and [codeboy2k]'s answer.  I've never looked closely at a schematic for a PID controller before. ;)
 

Offline thejoggingmat

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Re: How are some Buck Converters able to output below Vref?
« Reply #4 on: August 29, 2015, 05:55:51 pm »
The output of a PI controller is a slow moving voltage, in response to the error signal input.  It will settle slowly and respond slowly to set-point changes, and it will react slowly to transient changes.  It won't be bang-bang against the rails.   The 2 diodes back-to-back before it gets to the feedback input is just a diode OR, to make sure that only one of the 2 controllers is in control at any given time (Voltage or Current). 

The internal reference is still being used, and the voltage from the PI controllers (whichever one is in control) will do whatever it needs to do, and reach whatever voltage it needs to reach, within it's abilities, in order to make the reference input voltage = 1.235 Volts. So, working backwards from the reference input, you see that Vref is 1.235 V. Then there is the voltage divider, so the top of the voltage divider will be Vref * 2 = 1.235*2 = 2.470.  Then there is a diode drop to the output of either PI controller, which means the PI controller output wants to stay at 2.470+0.2V = 2.67 V. With error feedback from the OUT+ and OUT- points into the first AMP U2A (which is a weighted difference amp), and the PWM setpoint from the micro going into the PI controller's setpoint input, this will be what determines the output path as it settles to 2.67 V.

The output voltage is measured at OUT+ relative to OUT- , not at OUT+ relative to GND.  So the controller can reach 0 Volts output because it's not actually at 0 Volts relative to GND, it's going to be at 0 Volts relative to OUT-.  So as long as OUT- and OUT+ are EQUAL, it's considered 0 Volts output.

I didn't do the math, but you can analyze the U2A difference amplifier using supposition to get the transfer function, and determine what value is required for OUT+ equal to OUT- that gives Vout = 0 for that amplifier. Note that R13, R14 and R11 inject a 50mV voltage source into the amplifier too.  So your supposition calculations will have 3 voltage sources, 50mV, OUT+ and OUT-.  Supposition means set all voltage sources to 0V except 1, calculate the transfer function for 1 voltage source only, then repeat for the others.  The final transfer function is the sum of the 3 transfer functions. When you figure out the final transfer function you can set OUT+ equal to OUT- and simplify it further.  Then solve for Vout = 0 Volts.   This 0 volts is what will be matched with the setpoint PWM signal, and I am assuming that 0 volts PWM = 0 volts output desired.    Once solved, you will probably find that be OUT+ = OUT- = 1.235 V :-)  ( or maybe slightly higher )..



Wow, that's a lot to take in.  :scared:
I need a while to study this.
This is honestly the first time I have heard of 'PI Controller'. Before this I always thought PID Controller was just a fancy term for the cheapo temperature controllers on eBay (and didn't really look into it).

Simplest possible answer:

The LM2596 simply tries to keep the voltage at the feedback pin at Vref.  You can't get it to go lower than Vref with a voltage divider, or even a direct connection between Vout and feedback.

But if you amplify Vout by 2X (like with an op-amp), and run that into the LM2596 feedback pin, then Vout will be Vref/2.  Or 1.23/2=0.615V.  (Whether the LM2596 will be stable connected in such a way, for simplicity's sake, is a separate topic.)

Notice there are op-amps driving the feedback pin in your controller.  They can amplify Vout, making operation at lower than Vref possible.

Now I gotta study that schematic and [codeboy2k]'s answer.  I've never looked closely at a schematic for a PID controller before. ;)


Apparently, a brief Google and a quick glance at the datasheet for the MCP6001 (http://ww1.microchip.com/downloads/en/DeviceDoc/21733j.pdf) looks like a generic opamp. Though I might be wrong as I didn't glance past page 4.
 

Offline thejoggingmat

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Re: How are some Buck Converters able to output below Vref?
« Reply #5 on: August 29, 2015, 06:05:44 pm »
Hate to digress, but this board is more and more of a mystery as every minute passes. Can't believe a $10 power supply board could host such a vast tissue of conundrums and paradoxes.

The B3603 also has a top board, which hosts the 7segment LEDs + 2 shift registers and a STMicrocontroller (with a 10bit ADC). Basically the logic of the supply. I am not sure why the circuit needs to have a lowpass filter as well, leading to the opamp.
 

Offline codeboy2k

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Re: How are some Buck Converters able to output below Vref?
« Reply #6 on: August 29, 2015, 06:18:41 pm »
the MCP6001 is a generic op-amp, yes.  But it's configured as a PI (Proportional-Integral) controller.

A standard op-amp configuration (inverting or non-inverting) is a Proportional controller.  With the addition of the capacitor in the feedback loop, it becomes a PI controller.

The low pass filter is for the PWM square-wave from the microcontroller on the top-board.  The low-pass filter removes the high-frequency components of the PWM signal and you get a DC voltage out. This DC voltage is applied to the opamp input as a set-point voltage.  It will be compared against the error signal from the other opamp (U2A for voltage and U2B for current)

 

Offline Rick Law

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Re: How are some Buck Converters able to output below Vref?
« Reply #7 on: August 31, 2015, 06:03:16 am »
Hate to digress, but this board is more and more of a mystery as every minute passes. Can't believe a $10 power supply board could host such a vast tissue of conundrums and paradoxes.

The B3603 also has a top board, which hosts the 7segment LEDs + 2 shift registers and a STMicrocontroller (with a 10bit ADC). Basically the logic of the supply. I am not sure why the circuit needs to have a lowpass filter as well, leading to the opamp.

There is a lengthy discussion of this board, including discussions of attempts with various improvement.

Your "schemetic source" in github by Baruch was originally posted by Flex on this thread:
https://www.eevblog.com/forum/reviews/b3603-dcdc-buck-converter-mini-review-and-how-the-set-key-could-be-fatal/msg614243/#msg614243

but there are (at least) two versions of that PCB board  (reply #180).  You may want to check that thread out.

Note that the reply I linked to with the schematic was also trying to pin down the PWM frequency.  The 1.293KHz  frequency on that reply is wrong.  In later discussion, you will find  1.945KHz (13bit pwm) and the rest of the folks on that conversation all reconfirmed and agreed in a later reply that 1.945KHz (13 bit pwm) is the right one.
« Last Edit: August 31, 2015, 06:07:13 am by Rick Law »
 


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