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What is up with this?

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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In a post in the homework section, @hevans1944 says:

The screen shot you posted is taken directly from Analog Devices Application Note AN-968 "Current Sources: Options and Circuits. I suggest that you read and understand the entire application note before your "team" proceeds. You will need to understand, as @Audioguru suggested in post #4, simple operational amplifier circuits to comprehend the concepts discussed in the application note.

I thought I would follow the advice to the op, but quickly found myself wondering...

On the bottom of the first column on page 4, when discussing the choice of mosfet for the circuit in figure 5 in the second column of page 3, they say:

Since heat dissipation of the MOSFET is also critical, a very important factor in determining a MOSFET is its RDS(ON) value. In this case the RDS(ON) value is 150 mΩ typically. For larger currents, consider using a RDS(ON) value of <20 mΩ, if possible.

It appears to me that the mosfet will not be switched hard on, so the Rds(on) is pretty irrelevant.
 

BobK

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The Rds(on) would limit the current it could source, perhaps? Or at least the compliance voltage range?

Bob
 

hevans1944

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Clearly, Rds(on) is irrelevant when the MOSFET is operating in the linear (or so-called "triode" mode), which it does in the Analog Devices Application Note.

I can only conclude that the statement concerning the use of a MOSFET with a smaller Rds(on) is a "Red Herring" someone threw in there as a generalization to be followed when designing switching MOSFET circuits for higher currents. Unless the circuit is (heavily) modified to perform pulse-width modulation of the load current, rapidly switching the MOSFET between cut-off and saturation, with the duty cycle controlled by the desired average current through the load, the specification of Rds(on) is meaningless.

To implement such PWM switching, using the example IRF640 with its 200 V Vds maximum rating and Idm (pulsed) maximum of 72 A, means the ability to control lesser average currents by PWM is limited by the rise in junction temperature caused by current through Rds(on). No mention is made of the compliance voltage a current source requires, but this is a very important design consideration. If the compliance voltage is made large, so that higher impedance loads are accommodated, then for the same current applied to a lower impedance load, more power must be dissipated in the series switching element. For the limiting case of a short-circuited load, the power that must be dissipated is the compliance voltage multiplied by the current being regulated.

That comment in the app note about Rds(on) isn't the only peculiar thing about this app note. Like, what is the purpose of the 10 meg-ohm resistor connected from the output of the control op-amp to its non-inverting input? The control op-amp appears to be acting as an integrator for an error signal that is the difference between the control input and the amplified current-sensing resistor output, but a 22 pF integrating capacitor seems to be rather small for that purpose. So what is the huge resistance from output to non-inverting input doing? Maybe this acts as a discharge path for the MOSFET gate capacitance? But then it also appears in Figure 4 with nary a MOSFET in sight:
upload_2018-1-27_23-3-45.png

I am tempted to grab a 2N7000 and a couple of op-amps and breadboard this puppy just to see what is going on... sometimes these application notes are somewhat lacking in accuracy.:confused: Maybe a true-blue analog engineer like @AnalogKid can shed some light on this... I'm at least fifty years removed from this sort of thing. But please don't tell me to LTSpice it!

I don't really expect the OP to understand any of the subtleties of this discussion, but it would be nice to know what they are trying to DO. Perhaps then we could offer some advice on how to do it. But, then, it is a school project so we are, perforce, limited in the scope of our answers by Electronics Point policy as Ian states in this thread.
 

(*steve*)

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so we are, perforce, limited in the scope of our answers

Which is why I moved this discussion. :)

Like, what is the purpose of the 10 meg-ohm resistor connected from the output of the control op-amp to its non-inverting input?

I expect someone like @AnalogKid or maybe @Ratch will chime in with a more authoritative answer, but my understanding is as follows:

The "output" is essentially the top of the current sense resistor, and the negative feedback limits the gain of the entire circuit to 1. The 22pF capacitor from the op-amp's output to its; non-inverting input is to limit the slew rate of the op-amp so that the entire circuit doesn't ring (or oscillate) when confronted with horrible loads.

The positive feedback is totally swamped by the overall negative feedback, causing an overall net negative feedback.
 

(*steve*)

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I have the feeling the answer is here and here.

But my google-fu is not good enough to find anything which explains it in more detail.
 
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BobK

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Clearly, Rds(on) is irrelevant when the MOSFET is operating in the linear (or so-called "triode" mode), which it does in the Analog Devices Application Note.
I think the statement that you want a lower Rds(on) for higher currents is correct. What was incorrect in the statement is that it is because of power dissipation. The power dissipation will be the same for any MOSFET that is regulating a constant current at a given supply voltage.

A smaller Rds(on) would give you a wider linear region making it work over a wider range of currents,

Consider that you are making a constant current source with a 10V supply.

If the Rds(on) of your pass transistor (a ridiculously high) 100Ω, the most current you could handle is 100mA before the entire supply voltage is dropped by the MOSFET. Now replace it with one with 1Ω. Now you can get currents up to 10A before that happens.

Bob
 

(*steve*)

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Now replace it with one with 1Ω. Now you can get currents up to 10A before that happens.

Yeah, but consider that they are suggesting a major difference between 20mΩ and 100mΩ. For a 10A current, you're looking at 200mV vs 1V.

Reducing the voltage overhead will have a far greater effect on dissipation, and by the time you need to consider the Rds(on), the regulator (whatever it is) has lost control already. If you chose a mosfet with an appropriate Id(max) and Vds(max) then Rds(on) is highly unlikely to be an issue -- especially when you consider that you'll probably have to choose a device with a far higher Id(max) to get one that can handle the constant Pd.
 

AnalogKid

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App notes - From back when IC vendors sent their apps guys out on tour to promote their latest parts, 2 things.

1. App circuits, and particularly demo board circuits *and layouts*, were heavily tweaked to make the part look as good as possible, and *any* deviation from what was published was at your own risk. This made questioning things like an unexplained 10 M resistor a bit tricky.

Seemingly contradicting that -

2. App note circuits never should be taken as gospel. At best they are detailed concepts; at worst, cocktail napkins. One of National's guys was surprisingly honest about this. Often an app circuit that was prototyped and verified for one part was lifted onto the datasheet of another part without verifying that no changes were needed. Also often, the circuits were not verified at all. They looked good on paper, and that was good enough. Granted, they looked good to some of the best analog designers on the planet, but still...

Of course, the ADI guys said they were "pretty sure" that all of *their* app circuits were completely verified, unlike "those other guys".

3. ADI had far more respect for National than for Maxim, and they *never* mentioned Linear Tech.

ak
 

AnalogKid

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The source impedance driving the negative feedback is the Thevenin equivalent of the current boost stage inseries with the load, in paralle with 1 ohm. 1 ohm and 22 pF is a corner freq of 7-point-large GHz, so an active integrator probably is out. Looks more like compensation to prevent ringing with a reactive load, and it very well could be specific to the opamp.

However - think about the equations for a Howland current source. At some (high) frequency, the ratio of 10 M and 2 K equals the ratio of 22 pF and 1 ohm.

ak
 

AnalogKid

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The site won't let me add on, so here is something else.

For a normal opamp the corner freq at which the pos and neg feedback loops balance is probably way too high for there to be any appreciable gain. However, if the opamp is a current feedback type, things make much more sense.

ak
 
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