Agilent 3611A powersupply question - Page 1
Forum section: electronics / projects

author: opa627bm
author_rank: Regular Contributor
timestamp: March 11, 2014, 09:44:53 pm
content: Hello everyone, I am having hard time understanding the operation of the Agilent 3611A powersupply. It seems that there is no DC feedback path for the opamp but this is the opamp to drive the pass transistor.

author: tszaboo
author_rank: Super Contributor
timestamp: March 11, 2014, 10:02:53 pm
content: That is an integrator. Common technique for driving nonlinear stuff, the feedback comes from somewhere else, in a linear way.

author: opa627bm
author_rank: Regular Contributor
timestamp: March 12, 2014, 12:06:06 am
content: Here is the complete schematic and I don't see where the feedback comefrom

author: Rigby
author_rank: Super Contributor
timestamp: March 12, 2014, 01:33:04 am
content: the sense wire/terminal. noted by the "+S" on the schematic.

author: Rigby
author_rank: Super Contributor
timestamp: March 12, 2014, 01:56:43 am
content: here the voltage is being sensed from the output of the supply.

author: opa627bm
author_rank: Regular Contributor
timestamp: March 12, 2014, 02:27:30 am
content: Could you explain it in more detail? The sense line looks like the ground from the reference supply and I don't see how that will work.

author: Rigby
author_rank: Super Contributor
timestamp: March 12, 2014, 02:50:04 am
content: I wish I could. Someone else will have to do that.I just know that the +sense wire is half of how the supply knows what it is doing. the other half is -sense.The opamp compares the output of the supply to the voltage set at the panel, and adjusts until they equalize.That's my understanding, anyway. I could be wrong.

author: free_electron
author_rank: Super Contributor
timestamp: March 12, 2014, 03:21:01 am
content: These supplies actually measure the negative rail...The positive rail is used as a reference. Look where the supply of the opamp is referenced: The + 12 and -12 are centered around the +S signal.So the opamp actually senses the negative rail through R37. (the potmeter)the +5 volts is the balance reference. The opamp steers the pass transistor s in such a way that the midpoint of R37 and R15 is at +S levelChange the ratio of R37 versus R15 and you change the output voltage.this trickery sves em all kinds of scaling for the voltmeters and it allows for easy current mode steering : simply pull away the voltage control opamp outout ( that is what the diodes cr4 and cr5 do)

author: opa627bm
author_rank: Regular Contributor
timestamp: March 13, 2014, 04:24:43 am
content: I am very sorry but I still dont understand how it works. Could you explain what is the integrator circuit doing and how it corrects error and how did you get the DC gain to drive the pass transistor ?

author: quantumvolt
author_rank: Frequent Contributor
timestamp: March 13, 2014, 05:27:17 am
content: The HP supply works like this standard configuration dynamically. But this standard version samples a fraction of the positive output and compares it to a fixed reference.The HP supply on the other hand samples the positive output directly and compares it to a variable reference that is a linear combination of the +5 V reference supply and the negative output.The result is the same. Ask yourself which component goes in which direction (voltage up/down) for a small perturbance in output voltage.One important difference is that the HP dynamic reference version must have a load (the rightmost resistor over the output) in order to generate the ref voltage at the op amp input. It must also have a separate bias voltage source for the ref.The integrator is just working as a filter to "slow" high frequency correction of the output.

author: free_electron
author_rank: Super Contributor
timestamp: March 13, 2014, 07:09:59 am
content: The opamp U1 looks at +S ( the positive output ) through R8 (forget Cr6)It also looks at a voltage level at the R13 R15 Cr7 R37 node.The opamp will try to keep that node at the same voltage of +STo do that it will pull current from the base of Q2 through diode CR5 and R4. the more current it pulls the more Q2 conducts and the lower the gate voltage on the mosses becomes.This regulation mechanism works OPPOSITE from a normal regulator.Most regulators work on the principle of 'not enough , not enough , open the mos , open the mos.' the drawback of such a regulator is that, if the feedback disappears , output goes to max.This one works on the principle of 'too much , close the mos, close the mos.' here , if the feedback disappears the output goes to zero.

author: Rigby
author_rank: Super Contributor
timestamp: March 13, 2014, 12:43:43 pm
content: Holy battleballs, that made sense to me. Thank you.

author: Rigby
author_rank: Super Contributor
timestamp: March 14, 2014, 03:01:45 am
content: so, what about CR6 then? that will maximize the differential between opamp input pins, yeah? and maximize the amount of drain through Q2?

author: Kevin.D
author_rank: Frequent Contributor
timestamp: March 14, 2014, 07:31:22 pm
content: No CR6 just protects the inputs of the opamp from excess diff voltage .opa627bm heres a very simplified schematic to make it's operating principle clearer ,this is the basic supply with all the extras removed .This basic method is very common in Bench power supplies using both NFETS and NPN's .Think about the node at the non inv input as the feedback node , where it's V will be a fraction of-Vout which also has an +5V offset from the +5v Ref ,and the Fraction of Voltage fed back from -Vout will be dependent on the ratio of R1 /Vadj and so this ratio can be used to set the output Voltage .As for the the other bit's on the agilent supply I excluded .Q2 and associated resistors :simply provides a driver output stage ,since these large pass Nfets need a stronger drive than the opamp can deliver.Q4-Q5 : these NPN BJT's functions are a bit trickier to see.but they look like they are operating in REVERSE active region ,(if you look at the terminal voltages here you see that E>B>C thus emitter and collector current flow are reversed compared to normal npn). So that means they will form a emitter follower with R43 (which is shown in the collector) acting as an emitter resistor ,and so in the agilent supply they are acting as a (almost) constant current sink from fet Gates to -12 V.(I cant think why they chose to do it like this though except the reversed EB junction would also double as a mosfet gate protection zener (reverse EB junction breakdown ~7V) so saving parts.Neither Q4 or Q5 are essential to the basic operating of the power supply though so you can ignore these for your purpose .

author: free_electron
author_rank: Super Contributor
timestamp: March 14, 2014, 08:14:06 pm
content: remeber the opamp wants to see the same level on both its inputs.one input sits hard at +Vout levelthe other one is pulled towards +5 volt ABOVE vout level with the 10K resistor.to bring that point back down to Vout level ,the potentiometer pulls current towards -Voutso essentially the +5 volt and 10 k are a small , constant current source.that same current goes through the potmeter.

author: Rigby
author_rank: Super Contributor
timestamp: March 14, 2014, 08:43:33 pm
content: I thought I had it, now I'm lost. I'm missing something fundamental, and I don't know what it is.I'm still pretty new at this.I need to just continue playing with electronics. Eventually, I'll "get it" and then I can help explain to others that ask.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: March 16, 2014, 12:35:57 pm
content: Just a quick update If any one's interested in the function of Q4-Q5, they are for load equilizing between mosfets(because mosfets Vgs(th) has a negative temp co, one fet can have thermal runaway and end up hogging all the load current)After pondering these two for a bit you can see that they form a diff amp ( long tail pair) with R43 in the emitter (not obvious at first because they are npn's operating in reverse ) .Notice the base's will recieve the voltage from across the two low value current sense resistors R35,R36 which are in the fet source terminals . They will amplify any difference's between these two and apply it to the Fet gates (pulling down the gate Voltage of the Fet with higher current ) .You basically have whats below .It's a nice touch , since it allows you to drive multiple fets with a single driver and correct for quite large difference between fets using a very low value equilizing resistor in the source (which would need to be larger without the diff amp).

author: megajocke
author_rank: Regular Contributor
timestamp: March 16, 2014, 03:13:52 pm
content: I wonder if they are shown reversed in the schematic on purpose or if it is a typo. Other schematics from Agilent using the same circuit have them in the "right way" round.

author: dcsleep
author_rank: Newbie
timestamp: March 18, 2014, 02:49:02 am
content: I've been looking at this schematic for some time as well and appreciate the explanation of how the voltage control works.Now can someone explain how the current control works?