current follower

A current follower, as I understand it, might better be called a current booster. You can make one from two resistors and a power opamp. Given there's already a current source, you just add a series resistor (which won't affect the load since it's being driven from a current source, after all) and an opamp follower with another resistor to the load. The ratio of the resistors determines the output current, Io = Ii (1 + R1/R2). There'll be an output current error from the opamp's offset, Vos/R2.

Like the Howland and other types of opamp current-source circuits, it becomes a voltage source at high frequencies, as if it was a current source with lots of parallel capacitance.

Hi,

I designed the constant current source and it has some leakage current so I put an inductor acorss it. I got rid of the dc but its impedance changes with frequency and sometimes the impedance goes below the load impedance which is not desirable for my application. So, I thought that if I can make and add a current follower in series with the inductor. So for DC current the inductor will be shorted and the unity gain current follower will dump the charge into ground. and for AC whatever the inductor offers the impedance it will be in series with the very high impedance of the opamp . So, the total impedance will be higher than my load impedance. John

It depends a lot on the specs... maximum input resistance, output compliance, unipolar or bipolar, common ground or not, available power supply, etc.

One simple bipolar method would be to put a load resistor and voltage follower ahead of a Howland current source (not that I'm a particular fan of that configuration).

Best regards, Spehro Pefhany

Or it becomes an oscillator as one guy on the German NG just found out. So I pointed him to AoE because you guys were honest and called it a "Textbook current source ..." and at the end "... is not widely used".

Anything that relies on tightly toleranced components gives me the goose pimples and it has the potential to give a CFO a heart attack.

It might have exactly the same output current as input, but have more voltage compliance, isolation or other useful characteristics. Such things are common in process control.

Yes, that seems to be important in this case. And it will only work up to a certain load resistance when you run out of voltage headroom.

Best regards, Spehro Pefhany

Guys, help me out here. I searched for almost 5 minutes on variations of "howland current source", and the closest I've gotten to a schematic so far looks like an LM317 in battery charger mode. ??:-/

Anybody got a schematic of this Howland circuit, and maybe, a preferred circuit alongside it?

Usually, when I need a current source, I actually make a current sink, with some constant Vb and some emitter resistor on an NPN. Admittedly, not too tight on the tolerance, but it got the job done. :-)

Thanks, Rich

I have actually seen it (Howland) used in a commercial product, but I don't like it.

Best regards, Spehro Pefhany

If you go to Google books, and search with the term, "howland current source" you'll get the discussion in The Art of Electronics, right at the top of the list.

Below that you'll see where it's discussed in a dozen other books plus some journals, etc. Happy reading!

Right.

Foggy memory.. :P Isn't the Howland with 2 transistors.. Ic1 = Ic2 Something like that.. The bases are tied together... D from BC

Do you need unipolar or bipolar? How accurate? Tempco? Speed? Input/output impedances?

-f

Common base ampliifier? Current mirror ?

Bye. Jasen

Is this related to your other thread in sed, entitled "Inductive Circuits"? There you defined your cc-source as 0-600uA, 10 to 32KHz, (is it 10Hz or 10KHz?), and Rl = 20k to 400k.

Several people remarked that 600uA into 400k requires a 240V compliance, which I'm sure you do not intend, but did not see a clarification from you.

If the voltage compliance is within 12VPk then the LM13700 might be interesting. For higher voltages it will be an opamp plus transistors, several examples of which have been discussed here over the years.

With Tony's ASCII circuits shining, we might add.

Tony Williams a écrit :

Also 400k at 32kHz is just 12pF, which isn't much. Surely not achievable with a Holland circuit and will require naturally constant current with low parasitic feedback output circuits, like a cascoded current source.

The ticket is probably something along those lines: (hope it will not wrap)

high Z | A ------||-|___|--. | |> || ___ | || | | +--||-|___|-< from A | (5V/mA) ___ | |\\ | | || 47K '---------|___|--------+-----|+\\ | |/ | | >-|---| C adj (pF range) 10K | .--|-/ | |>

| | |/ | | | '-------|-----+ | | | .-. | .-. | | /+\\ | | 10K| | 5V( ) | |1K '-' \\-/ '-' | | | '----------+-----+-< -Supply

The cascode might not be required for the impedance you target, but it is so easily added that you don't want to delete it.

The low frequency DC servo loop might be interesting to adjust over all the load impedance ranges if the low freq corner is 10Hz, and will lead to very low transient response. Alternatively, it might be interesting to make the first integrator a S/H circuit with somewhat higher GBW product, make the following opamp a simple inverter, and servo the output voltage only during the rest periods (if the working duration can be made not too long).

The C adj capacitor is adjusted to compensate for (almost) all the output parasitics and will start rolling off at 340kHz (assuming a 10pF compensation).

Hi,

I saw the following link

and found the howland current source. I am still confused about all the disscussions. My question is that if I use the current source ( that I built and cannot go back and change the design) in parallel with an inductor who is in series with the howland current source can solve the problem of saving the load from the DC leakage current and let the ac signal appear acorss the load completely. Please advice!

John

john a écrit :

Your link doesn't work

I can see this.

I don't undestand what you mean. Once you place the inductor in // then you place it in series. I can't make sense of what you say. Maybe you could sketch something and post it somewhere. Maybe too you can tell us what you're trying to achieve with this current source you want. Maybe you can clear the questions you were asked instead of letting us guess and shoot in the dark.

BTW, given the frequency bands and the impedance values you gave us, counting on inductors to do the job is totally hopeless.

This sounds like the same problem I posted a reply to recently. If you want to get rid of the DC into the load, just put a capacitor in series with the current source. Make it large enough to have a reactance low enough to not have unacceptable voltage drop at the lowest frequency of interest; but if driven by a true current source with infinite compliance, it will even pass DC. I think for your application, something in the area of 10uF should do well; I suggest polypropylene dielectric, though polyester/Mylar would probably do fine too. So anyway, you need some place for the DC to go: just add a resistor from the current source output to ground, or to a voltage such that the current source output DC voltage is what you want. You had earlier mentioned a low DC current, something in region of 100 nanoamps as I recall. So you can use a very large resistance and have a small DC drop across the resistance. That's MUCH better than trying to get an inductor to do the job. Go look up your earlier thread, and look at my posting there. There are some additional caveats to consider.

Cheers, Tom

That's a good suggestion, Tom, I hope John sees it.

Hi,

The capacitor will discharge through the load which is not desirable in my application. I put an inductor in parallel with the current source and it did take care of the DC leakage current but for low frequency AC signals, inducroe does not offer high impedance. So, I was hoping to find a circuitry that can short the DC and offer high impedance when current is AC.

John