# current follower

Discussion in 'Electronic Design' started by john, Jun 8, 2007.

1. ### johnGuest

Hi,

Can anybody give some designing hints to make a current follower
circuit using opamps or transistors. I do not know whether it is
possible or not... I do know the voltage follower using opamps...

John

2. ### Winfield HillGuest

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
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.

3. ### johnGuest

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
John

4. ### Spehro PefhanyGuest

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

5. ### JoergGuest

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.

6. ### Spehro PefhanyGuest

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

7. ### Rich GriseGuest

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

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

8. ### Spehro PefhanyGuest

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

Best regards,
Spehro Pefhany

9. ### Winfield HillGuest

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.

10. ### D from BCGuest

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

11. ### Frank MilesGuest

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

-f
--

12. ### JasenGuest

Common base ampliifier?
Current mirror ?

Bye.
Jasen

13. ### Tony WilliamsGuest

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.

14. ### WinfieldGuest

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

15. ### Fred BartoliGuest

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)

<-- 1mA
1K ___
.-----|___|-+-<+Supply
/| | _ |
/+|--GND /| >| / \ |
.--+-< | ___ /+|--- GND |--(- +)--+
| | \-|-+-|___|-+< | /| \_/ |
| .-. \| | | \-|-. | |
| | | | | ||| | | _ |
| | | --- '-||--+ >| / \ |
| '-' --- || | 47p |--(- +)--'
| | | 1u .-. || /| \_/
| '------' | | .--||----+ 5V
| | | | || |
| '-' /| | ___ | ___ ||
| | /+|-+-|___|--+-|___|-||----> high Z
| A <-+-< | 100K | || AC out
| | \-|-. | 2uF
| 10u 20K | \| | |/ film or ceramic
| || ___ '------' .---|
| IN >------||-|___|--. | |> || ___
| || | | +--||-|___|-< 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).

16. ### johnGuest

Hi,

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

John

17. ### Fred BartoliGuest

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.

18. ### Tom BruhnsGuest

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

19. ### WinfieldGuest

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

20. ### johnGuest

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