Electronic load circuit allways full voltage on gate of the mosfets - Page 1
Forum section: electronics / projects

author: teslinius
author_rank: Regular Contributor
timestamp: June 03, 2022, 04:18:54 pm
content: So i am testing an electronic load circuit just like it's in the schematic, dac output to non inverting input of the opamp (lm324), inverting to point between the 0.1ohm resistor and mosfet source that acts as a feedback, and a 100k pulldown resistor for the mosfet, 1k gate series resistor for protection in case it goes short circuit, 10k pulldown on the output of the opamp and 220k and 1nF to prevent ringing, or so i read in some articles. So the problem is that when the dac output is 0V (actually 600uV but ok) and no load connected so 0V also on the inverting input there is 9.5V on the mosfets which is the maximum of the opamp since it's not rail to rail, which is weird, i'm not exactly sure why it does this, only thing that makes sense it that it is an non inverting amplfiier circuit with 100mohm and 220k resistor so te gain is massive. Any ideas what's wrong?

author: fourfathom
author_rank: Super Contributor
timestamp: June 03, 2022, 04:40:56 pm
content: Looks like it's doing what it should be doing. Assuming 600 uV at the (+) input (and no input offset voltage/current), the opamp output is being driven high. Since there is no current flowing through the mosfet the feedback voltage is nearly zero (about 13 uV, via the gate resistors and the 0.1 Ohm shunt). Apply some voltage to the mosfet drain and things will probably look more like you expect.

author: MarkF
author_rank: Super Contributor
timestamp: June 04, 2022, 01:43:30 am
content: There is nothing wrong with your circuit. I have the same issue, although my feedback circuit is different.The problem is with the DUT powered off or disconnected, there is NO voltage across the sense resistor. And with a single power supply, you can not set the input reference voltage negative (below the sense resistor voltage).I added a resistor to always force a small current across the sense resistor which allows you to set the reference below the sense resistor voltage turning off the MOSFET. It only introduces a few mA current error. AND IT ONLY WORKS IF YOU SET YOUR INPUT REFERENCE TO ZERO. Any non-zero reference will drive the MOSFET gate to max once again.My full circuit:https://www.eevblog.com/forum/beginners/a-load-off-my-mnd/msg2151265/#msg2151265@Jay_Diddy_B circuit:His load reverses the op-amp inputs. It requires a dual-power supply and doesn't seem to have the same issue. (See my question about this near the end of his topic.)https://www.eevblog.com/forum/projects/dynamic-electronic-load-project/Edit:I'm using 620Ω and 5.6KΩ simply because I was using those values in the ÷10 resistor divider. Better values would probably be 100Ω and 1KΩ respectfully. Although the higher values tend to protect the op-amp if the MOSFET fails. I wouldn't go much lower than 5nF on the feedback capacitor. Too low of a value might cause oscillations. The 10nF allowed me to drive the load with a square wave up to about 5kHz. Don't hold me to that. It's been a long time ago.

author: teslinius
author_rank: Regular Contributor
timestamp: June 04, 2022, 05:57:37 pm
content: Oh ok so the offset is doing this, i have double supply +12V and -12V, i tried connecting the load terminals to a voltage source and it drew 160mA constantly even with varying voltage so that's ok, after the dac output i have an op07 opamp also supplied with positive and negative in voltage follower configuration and with offset nulling potentiometer, but it doesn't work. I am using another op07 in differential amplifier configuration and the offsetting works there but it doesn't work here maybe it doesn't work for voltage followers?

author: fourfathom
author_rank: Super Contributor
timestamp: June 05, 2022, 01:30:13 am
content: Is there some reason you care what the gate voltage is when there is no power applied to the drain? With no feedback (because there isn't any drain current) the gate voltage is going to be either positive or negative max. You might be able to trim it for zero gate voltage, but that balance won't last over time/temperature. It really doesn't matter, as long as the gate van handle the voltage.With 600 uV DAC output at the opamp input, when power is applied to the drain you will see about 60 uA drain current. If you want to zero that out, one method is the bias resistor shown in MarkF's post.I am curious about the function of R26 and R28 in your original schematic. They don't hurt, but I don't see that they do anything useful.

author: MarkF
author_rank: Super Contributor
timestamp: June 05, 2022, 02:36:27 am
content: You care what the gate voltage is when the DUT is turned OFF because it completely turns ON the MOSFET and will present a short to the DUT when it is turned ON.Resistors R26, R28 and R29 should not be there at all. (You could leave R29 to insure the MOSFET is OFF if the +/- power supply is OFF.)Resistor R27 limits the op-amp output current to the MOSFET gate capacitance and prevents oscillation.You want to add a resistor in series with the MOSFET source and op-amp (-) input to limit its current and so that the load current goes through the sense resistor and not the op-amp virtual ground.

author: fourfathom
author_rank: Super Contributor
timestamp: June 05, 2022, 07:06:43 am
content: Quote from: MarkF on June 05, 2022, 02:36:27 amOK, I can see the benefit of having the FET default to the OFF state. But you still have that issue if the current-set voltage is greater than zero before you connect or turn on the DUT.Quote from: MarkF on June 05, 2022, 02:36:27 amI would add that resistor to equalize the input resistors to minimize input current -caused offset voltage. But that virtual ground is still a very high impedance and only has an input bias current of under 150 nA (LM358). Look at the components in the feedback loop -- there is no current flowing into the virtual ground.

author: strawberry
author_rank: Super Contributor
timestamp: June 05, 2022, 07:26:26 am
content: how about negative feedback?

author: MarkF
author_rank: Super Contributor
timestamp: June 05, 2022, 08:53:53 am
content: Quote from: fourfathom on June 05, 2022, 07:06:43 amI guess I forgot to say that my fix was a kludge because I had already had a PCB made and didn't want to do another one. What you really want to do is monitor the DUT voltage and turn OFF the MOSFET when the voltage goes below a predetermined value. That does assume you can decided what value you would like it to be.And yes. You do have to remember to zero the current set voltage before powering up the DUT. Not a big deal for me and not worth all the extra parts or even a micro-controller to do it automatically.A little side-by-side comparison might be helpful here (R6 is the sense resistor):To tell the truth, I really haven't given the series feedback resistor much thought before. I just automatically put one in.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 05, 2022, 09:16:35 am
content: The inversing amplifier normally works a little different: with the OPs output to the full chain, and the feeback from the divider.The resistor in series in the feedback path is needed to make the capacitor to slow down the OP work. So the resistor is correct and missing in the circuit of the 1 st post.Sensing the voltage over the DUT is a good start. One can turn off the current source all together and relay on starting with zero current.The alternative that is not much more effort is limiting the set voltage to a fraction (like 20% or 50%) or the votlage at the MOSFET / DUT output. This does not need a µC or much effort. A relatively simple OP-amp can do. One may however need a negative supply. The somewhat tricky part can be to make the limiting circuit stable and avoid oscillation in that extra loop.Limiting the set voltage to a fraction of the output voltage results in a kind of constant resistance mode, that takes over at low voltage. So the "resistance" of the load would not drop below a certain limit, that should be a bit higher than the shunt + MOSFET on resistance.

author: teslinius
author_rank: Regular Contributor
timestamp: June 05, 2022, 11:43:01 am
content: So basically i just want to get rid of the 160mA current it's drawing even when the dac output is 0 set in the code, for the 220k resistor i found it was used in some example circuits instead of the 1nF cap so i added it for reserve, for reference i took also these 2 schematics. One is from a very old magazine or something.

author: MarkF
author_rank: Super Contributor
timestamp: June 05, 2022, 12:12:16 pm
content: Are saying your load draws a minimum of 160mA from the DUT even when the current setting is zero?You really haven't posted enough of your circuit to tell what's going on. However, the circuit in your first post is just wrong.The circuit I posted can control the current down to a few mA. It starts getting touchy down below 5mA. If I remember right the lowest setting I obtained was 2mA.If you look at the link to my full circuit, you will see that my controlling voltage is 0 to 5V which produces a 2.5A max load. The 10:1 divider puts 0.5V at the (+) op-amp input which in turn drives the MOSFET until a 0.5V voltage is across the sense resistor. With your 0R1 ohm sense resistor, that would be 5A. Which would be the max I would go for the IRFP250. If your DAC voltage range is other than 5V, modify the resistor divider to get 0.5V at the op-amp (+) input. Just feed your DAC output into my circuits Ext. Input and forget the left half with the 10 turn pot.If you want a good reference circuit, here is the schematic for the HP-6060 Electronic Load:

author: teslinius
author_rank: Regular Contributor
timestamp: June 05, 2022, 01:14:12 pm
content: Yes the problem is that the load draws 160mA even when current setting is zero. The full schematic is in the attachment.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 05, 2022, 01:38:32 pm
content: The OP offset can provide some minimum current. The LM324 can have some 5 mV (depending on the version may be up to 10 mV) of offset and with 0.1 ohms as the sens resitor this would be up to 50 mA per channel. To be sure to get control down to very small currents one could add some extra signal to the feedback. In the HP circuit this are R25, R26, ...For low currents one could also turn off 3 or the 4 channels totally and use only 1 active channel.

author: teslinius
author_rank: Regular Contributor
timestamp: June 05, 2022, 02:02:58 pm
content: Shouldn't there be none offset voltage since i am powering it from double supply? (+12V and -12V)

author: teslinius
author_rank: Regular Contributor
timestamp: June 05, 2022, 10:23:37 pm
content: After i added 1k resistors between inverting input and the point between the mosfet source and load resistor the voltage on the gate of the mosfets dropped to 280mV and now it doesn't draw any current anymore so that's good. Why did they use a capacitor in the hp6060 schematic (1800pF 100v C4) instead of a resistor?

author: MarkF
author_rank: Super Contributor
timestamp: June 06, 2022, 12:14:55 am
content: I believe all the capacitors around both op-amps in the HP-6060 are to smooth/slow down the response and also to eliminate oscillations.Like I said earlier, the 10nF is to prevent oscillations and smooth the response to transients from the DUT. I would not go with a lower value unless you want to push the switching frequency to a load (i.e. you want to test a DUT current draw while being toggled ON and OFF at some frequency). In which you would need to do extensive testing to insure your circuit doesn't oscillate.Reference Op-amp Integrator:https://www.electronics-tutorials.ws/opamp/opamp_6.html

author: boB
author_rank: Frequent Contributor
timestamp: June 06, 2022, 01:03:01 am
content: Quote from: MarkF on June 05, 2022, 12:12:16 pmThat schematic is interesting in that HP used 62 mA FUSES in series with the FET GATES ! I have never seen that before !I have used lots of Transistor Devices linear Dynaloads before.boB

author: magic
author_rank: Super Contributor
timestamp: June 06, 2022, 07:30:34 am
content: Quote from: teslinius on June 05, 2022, 10:23:37 pmThe main feedback path was through R33 and the differential amplifier built around OP270.C1,C49 apparently bypassed that path at high frequencies to override phase lag of the diff amp circuit.C71 bypasses the whole bunch, including the power transistor.You don't have that differential current sensing stuff so the circuits posted by MarkF are the usual solution.At DC feedback is through resistor from the current sensor. At AC it's through the capacitor (like C71). The transition is at the corner frequency of those RC elements: f=1/(2πRC).---The problem of the MOSFET being turned hard on in absence of DUT voltage still remains. Next time I build something similar I plan to experiment with protection circuits, for example like the one below. Here, when drain voltage decreases, D1 and D2 turn on Q1, which feeds current into R4 and makes the opamp think that the set current has already been reached. Then the opamp reduces gate voltage.This gradual turn-off needs to be tuned to happen before running into MOSFET's RDS(on) limitation, which has the opposite effect of the opamp rising gate voltage in futile effort to get more current going. On the other hand, if it happens too early the ability to test low voltage PSUs with high current will be compromised. Thermal drift of MOSFET's characteristics may make getting optimal performance more difficult in practice than in SPICE; a BJT could be better in this regard.