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Need help with bi directional constant current source (±100mA)
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David Hess:
The LT1990 is an instrumentation amplifier and not a differential amplifier.  As shown, the two transistors boost the output current although this configuration is normally seen with operational amplifier.

Any instrumentation amplifier can be used in that configuration to make a Howland current pump.  The advantage is that the instrumentation amplifier has the precision resistor network used for the Howland current pump built in and will provide a known precision without trimming any resistors.
OM222O:

--- Quote from: David Hess on November 13, 2018, 12:00:03 am ---The LT1990 is an instrumentation amplifier and not a differential amplifier.  As shown, the two transistors boost the output current although this configuration is normally seen with operational amplifier.

Any instrumentation amplifier can be used in that configuration to make a Howland current pump.  The advantage is that the instrumentation amplifier has the precision resistor network used for the Howland current pump built in and will provide a known precision without trimming any resistors.

--- End quote ---

it's in the title: "±250V Input Range
G = 1, 10, Micropower,
Difference Amplifier" although it is also classified as a "instrumentation amplifier". I'm not sure why they classify it like that, but it certainly can be used as a differential amplifier. howland current pump is basically a differential amplifier with a shunt resistor to be honest.
David Hess:

--- Quote from: OM222O on November 13, 2018, 12:45:11 am ---Difference Amplifier" although it is also classified as a "instrumentation amplifier". I'm not sure why they classify it like that, but it certainly can be used as a differential amplifier. howland current pump is basically a differential amplifier with a shunt resistor to be honest.
--- End quote ---

A differential amplifier has a differential output which is unavailable on the LT1990 or any instrumentation (or difference) amplifier.  The only way to use it as a differential amplifier (with a differential output) is to use two with their inputs connected in anti-parallel which is sometimes done for exactly this.

To make things more confusing, in the past the term push-pull was used in place of differential but luckily that terminology stayed in the past.

I'm not exactly sure how they distinguish instrumentation and difference amplifiers.  It may be by the impedance of their inputs or lack of input dividers but there are parts which break these rules.
OM222O:

--- Quote from: duak on November 12, 2018, 05:04:44 am ---OM, here's an awful sketch of a current booster.

Ri handles the opamp's output current up to the point at which its voltage drop turns on one or the other transistors (~0.6 V / 100R = 6 mA).

D1 or D2 turn on when the inductor voltage tries to go beyond the power supply rail voltages.  If these weren't here, the inductor current  would try to force its way thru the opamp.

D3 and D4 are zener diodes that limit the power supply rail voltages when the energy in the inductor has to be dumped.  Since this is an actuator in a mechanical system, there may be an energy source may be external to the inductor (such as an arm) in addition to the energy stored in the magnetic field.  If there's a lot of energy, you might need some big zener diodes, or a more capable voltage limiter.

Do you know how to calculate the worst case power dissipation in the transistors?  It's not always obvious, but with a simple inductive load for the case of just holding, you can have maximum current with essentially zero voltage across it.  The transistors would then have to drop the rail voltage and so dissipate 100 mA * 24 V = 2.4 W.  If the actuator is moving, there will also be a back EMF impressed on the output voltage and further increase the power dissipation.

Tons of fun, eh?  Good luck!

--- End quote ---

Hi, I've been trying to do my own research and understand how this circuit works and I've read some articles about it as well.
for example this one: https://www.allaboutcircuits.com/technical-articles/how-to-buffer-an-op-amp-output-for-higher-current-part-2/


but I've had no luck with it so far ... sorry if I'm just being way too big of a noob but after running the simulations (dc sweep to see if I get the ±100mA depending on the control voltage), I'm hitting the output current limit of the op amp as you can see:


Have I done something wrong with the schematic? can you please help me figure out the mistakes? thanks
duak:
OM, you will have to increase the value of R1 to something like 100 ohms.  With R1 = 10 ohms, the diff amp would have to source 60 mA (0.6 V/10R) which, I'm sure, it cannot do.

BTW, what transistors are you using?

Cheers,
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