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Explain this circuit: Digital linear PSU

Discussion in 'Power Electronics' started by Rixen, Apr 30, 2016.

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  1. Rixen

    Rixen

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    Feb 16, 2016
    Hi all

    I've been trying for some time now to build my own MCU (DAC) controlled linear PSU, instead of the good old potentiometer in the feedback loop.

    No success with this so far, nor do I have much luck finding sources which has information on how to do this, most I find are SMPS's..
    But I did find kerrywong.com who has built one, im hoping some of you guys can explain in more detail how his voltage control feedback loop works exactly, so I can apply it to my own project, ignore the current sense part for now.

    This is the one: [​IMG]

    For output voltage he gives the formula:
    [​IMG]

    I know it's the simple voltage divider formula.. but I dont understand how it applies here, because I dont understand how the OpAmp IC1A is configured here and how it actually works.

    Also, why is it multiplying the non-inverting input and not the inverting ?

    R8/C3 is a filter - R12/C4 is apparently also a filter? D2/D3 are protecting those inputs from voltage spikes? purpose of R9?

    He states that T6 "extends" the range of IC1A, how does that work and how is it calculated?

    I would really appreciate a thorough explanation :)

    This is the overall design im currently using, ignore all values for now as they are outdated.
    [​IMG]

    Thanks all!
     
    Last edited: Apr 30, 2016
  2. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    You should go to the original web site and read all three sections (I, II, and III) to gain a better understanding of this well-engineered dual-tracking ±30 V DC power supply.

    The op-amps are part of a negative feedback loop comprising transistors T1, T2, T3, T4, T5, and T6. Op-amp IC1A, driven by IC1B connected as a unity-gain buffer, compares the input control voltage, VControl (0 to 4 V) to the power supply output voltage +Vout as derived from voltage divider R10 and R11, and drives transistor T6 base through R7. T6 collector is connected through R5 to +Vin (40 V) to provide more voltage swing to drive T1, T2, T3, T4 and T5 connected as an NPN emitter-follower using the Sziklai pair configuration, which is a complementary form of the Darlington connection.

    The bases of T1, T2, T3, and T4 must be driven slightly more positive than the power supply maximum output of 30 V DC. T5 and T6 allow this to happen because the op-amp output cannot swing far enough to directly drive the bases of the power transistors, nor supply sufficient base current to control them at maximum current output.

    Your hand-drawn schematic eliminates the op-amp, removing the negative feed-back necessary to control the output voltage.

    It isn't multiplying anything. It's an operational amplifier, not a multiplier.

    These two networks affect the closed-loop gain and phase response of the circuit. They are there to prevent high-frequency oscillation because the author chose to use a 10 MHz bandwidth op-amp.

    No, they are there to protect the op-amp inputs from excessive differential input voltage, which can easily occur if the power supply output changes quickly. say, because of a rapid change in load impedance.

    R9 would normally be selected to minimize the differential bias current present at the input of the op-amp. It should present, in parallel with R19, the same impedance as R10 in parallel with R11 so the offset voltage caused by bias current from the inverting input is canceled by the offset voltage caused by bias current from the non-inverting input. It doesn't do this, however. The 100 kΩ value shown is much too large.

    Looks like you have some reading to do on op-amps and closed-loop negative feedback systems.

    Hop
     
    Last edited: May 5, 2016
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  3. AnalogKid

    AnalogKid

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    Jun 10, 2015
    There are two fundamentally different ways of designing a variable-output voltage power supply, and you are confusing the two.

    The first schematic is a fixed-gain DC amplifier driven by a variable input voltage. The gain equation is stated correctly right below the schematic. If you drive this circuit with the output voltage from a DAC, the power supply output voltage will vary linearly with the DAC input code.

    The second schematic is a variable-gain DC amplifier driven by a fixed reference voltage (Vz + Vbe(T5)). It works ok as a regulator, but not as well as the first schematic. The output will wander around a little as the temperatures of the zener diode and T5 change, and the single-transistor error amplifier (T5) is not nearly as high performance as a true differential amplifier. Inserting a DAC into the feedback loop as a variable resistance is more difficult than providing a simple input voltage as in the first schematic.

    In the first schematic, IC1A is the differential amplifier that runs the show. It is followed by two inverters in series (T6 and T5) and a non-inverting emitter follower (T1-T4), so the opamp + and - inputs function as marked.

    ak
     
    Last edited: May 4, 2016
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  4. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    @AnalogKid, thank you for correcting my analysis of the hand-drawn circuit that was posted above. It does provide negative feedback by virtue of the phase inversion at the collector of T5, and I failed to notice the change from a Szilaki pair (PNP and NPN) to a Darlington (NPN and NPN) emitter-follower.

    I agree that the hand-drawn circuit doesn't make for a very good power supply, but it does have the "advantage" of simplicity and no need for an op-amp and frequency compensation. As for inserting a DAC to control the output, a digitally programmable 10 kΩ potentiometer could do that, but it wouldn't be a linear output as a function of the digital "word" sent to the digital potentiometer. I like the first schematic a LOT better. Maybe use a more modern op-amp with less open-loop bandwidth, and SPICE model it to check for unconditional stability and response time to step changes in load and DAC command voltage.

    I like linear power supplies, despite the hit you take in efficiency, because they don't generate high-frequency noise that you somehow must deal with. My 100 watt linear amplifier for my KX3 transceiver uses a linear power supply (my choice) that weighs in at more than the transceiver and amplifier weights combined. A lot more. Of course, in the field, I use automobile battery power and leave the boat anchor at home. A solar panel would be even better, but 100+ watts panels are too expensive for my taste.

    I could afford a 10-watt folding solar panel, perhaps, and operate without the KXPA-100 in the field. That was the main selling point of the KX3 in the first place: portable, low-power (QRP) operation. I may pick a small folding solar panel during Hamvention® this year if the price is right. ARRL Field Day is coming up in June, so I may just join with some other hams and see what they do. Field Day is an annual contest to see how many contacts (QSOs) an amateur radio operator can make on all the ham bands over a 24 hour period "in the field" without recourse to "shore power". Unfortunately, portable generators ARE allowed, but you earn maximum "points" per contact by operating alone at less than 5 watts transmitter power, using batteries or solar power. The whole idea is get experience operating under "emergency" conditions. I attended only one Field Day (in 1966) with a bunch of hams from the Air Force base where I was stationed. They brought everything with them except for the kitchen sink... and a large tub filled with ice and beer stood in for the kitchen sink. A good time was had by all, but since I wasn't licensed yet, all I good do was observe... and drink some of the beer.:D
     
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  5. Rixen

    Rixen

    98
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    Feb 16, 2016
    Alright, apologies for the late replies, school and work doesn't always permit me to reply whenever I want and it takes me a while to make sure my post doesn't come out as complete rubbish :)

    We've had a grand total of 1.5 weeks in school with basic Op amp theory, it shows I know.. I know the basic Op amp rules and some of the standard configurations, I actually just noticed the feedback loop that you guys pointed out, where previously I had only seen the text book example with a single resistor, as for my hand drawn schematic there, I know I'm going to have to make changes to make it work, I just posted it so you had some idea where I was at.

    Yeah, wrong terminology sorry, I looked at his gain formula and in my head it was multiplying the DAC input.

    This is where I started getting confused, how are those resistor values calculated? (R10/R11) If we can make up some example and calculate them for some other output voltage?

    Once again, thanks for all the help you guys, it's really appreciated :)

    Oh and.. hi again Hop!
     
  6. AnalogKid

    AnalogKid

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    Jun 10, 2015
    It is. In the world of analog circuits, "multiply" can mean an analog multiplier, which is a very different circuit from an opamp gain stage. But the basic opamp gain stage you are familiar with *does* multiply the input by a constant. So in non-electrical-engineering English you are correct, but in EE terms you maybe misused a word.
    The resistors are in the standard non-inverting opamp gain equation. But in the first schematic, the "opamp" is IC1A, T6, T5, T4, T3, T2, and T1. All of that forms a power opamp with an asymmetrical, very high power output stage. But if you go back to your basic opamp course, R10 is the series feedback resistor, R11 is the shunt feedback resistor, and that's it.

    ak
     
    Last edited: May 5, 2016
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  7. Rixen

    Rixen

    98
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    Feb 16, 2016
    Alright, great.

    I hacked together a test out of whatever parts I had in my used parts bin, looks like it's working! :)
    Even though it looks like the gain is slightly lower than expected, could be lots of reasons for that I guess? Im also not sure that it's a rail-to-rail OpAmp, where do I see that in the datasheet?

    Datasheet also says to supply that TL072 with a maximum of 2.5mA per amplifier, it had some weird behavior when I did that and it needed more..

    Also, since T1-T6 is part of the feedback loop, their resistance must affect the gain too, is that something to worry about? I know they have very low resistance, offset with larger values on R10/R11?

    Hop, im trying to calculate some resistance values for R9 and R19 but im getting some crazy results, if I try to give them the same value as R10/R11 in parallel :eek:

    As always, thanks for the help guys!

    [​IMG]
    [​IMG]
    [​IMG]
     
    Last edited: May 6, 2016
  8. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    The problem is the parallel resistance of R10 = 22 kΩ and R11 = 3.3 kΩ is only about 2.9 kΩ. This is larger than the R19 = 1 kΩ so there is no way to choose a value for R9 that would make the parallel value of R9 and R19 equal to 2.9 kΩ. It isn't a problem with this circuit because the bias current maximum is only 1000 nA at maximum operating temperature, at least according to the ON Semiconductor datasheet for the NE5532 op-amp. This would represent a maximum offset voltage of about 2.9 mV even if the value of R19 was reduced to zero. It gets multiplied by the closed-loop gain, (1+R10/R11) or about 7.7 to create an offset of about 22 mV in the output. This is usually considered insignificant for a DC power supply.

    Hop
     
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  9. AnalogKid

    AnalogKid

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    Jun 10, 2015
    The source for the non-inverting input, the output of the reference circuit, has no DC load to ground to stabilize the output of its emitter follower T2.

    Wherever the input to the negative feedback voltage divider attaches to the circuit, that is the "output" as far as the composite amplifier is concerned. T1-T6 are not part of the feedback loop, they are part of the forward amplifier. As long as the actual opamp has enough output voltage swing, current, and bandwidth to drive all of that stuff, the feedback loop takes out such things as Vbe drift with temperature, part-to-part Vbe variations, etc.

    2.5 mA is the max current the device draws when it is doing nothing. Any output current comes through the power pins, because where else can it come from? So it power source must deliver it to the opamp, to deliver it to its load.

    The maximum input voltage range (sometimes called common mode voltage range) and output voltage swing are covered either in the parameter tables or in a plot. Post the actual datasheet you are using (they vary all over the place), and we can point it out.

    ak
     
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  10. Rixen

    Rixen

    98
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    Feb 16, 2016
    Heres the datasheet for the one I was playing around with.

    http://www.ti.com/lit/ds/symlink/tl072.pdf

    Apologies, there is a resistor there, I just checked my breadboard, it's just not on the circuit diagram, I think I added it as an "Oops" afterthought..

    Ugh, your right, "nothing" goes in or out of the inputs..
    I dont see any max output current or anything like it though and it says you can short the outputs indefinately, Im not sure really..

    There's a Vcm common-mode voltage in the datasheet of +/- 4V
    This is how close the inputs can be to the supply voltage?

    Theres also a maximum output voltage swing as you say, on +/- 10 - 13.5v depending on the load.

    Yeah okay, I dont think I need to worry about an offset that small, definately not for my purpose :)
     
  11. AnalogKid

    AnalogKid

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    Jun 10, 2015
    Max output current can be calculated from Figure 6.
    Input voltage range is in section 6.3. Note that the power for all of the stated parameters is +/-15 V. Performance at other voltages is in some of the plots.

    The input common mode voltage range is not +/-4 V. It is Vcc-4V to Vee+4V. So if the power rails are +/-5V, the input voltage range is +/-1 V peak.

    ak
     
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