Need help with bi directional constant current source (±100mA)

[OM222O]

Hello
I need to design a dual channel constant current source which can drive an inductor to ±100mA. the inductance is 3.2H and I want as good of a transient response as possible, so I chose the ADA4522-2ARZ-RL which is a precision differential amplifier that can be driven with upto 60 volts. I will be using a 48V supply which is divided to give ±24V output. I will also use a DAC (LTC2607-1) and a 2.048V reference (ADR4520B) to produce the control voltage. I will also use a voltage divider followed by a buffer amplifier(TLV9001) to produce a 1.024V reference and will connect it to one of the inputs of the differential amplifier. therefore if the output of the ADC is greater than 1.024 I will get -100mA and vice versa. I came up with this circuit which works great:
http://tinyurl.com/y7grgwwp


The only issue is that the differential amplifier can't source or sink that much current (about 14mA at best!) so I somehow need to boost it's output using BJTs or Mosfets. I saw this circuit which was showcased in the datasheet of LT1990 (a single channel differential amplifier) which seemed like the perfect choice as they also use a 10ohm shunt and want to produce 100mA!


after trying to simulate the circuit, it seems to not work for me ... the op amp is still the element that's passing all the current!
http://tinyurl.com/y9ylnbec


as far as I can tell, this is just the basic howland current pump with 2 pass elements. I'm not sure if I'm making a mistake in my simulation or what, but I can't seem to get this circuit to work. please help me fix it.

[JackJones]

Your falstad simulation doesn't quite match the circuit in the datasheet. The 1k resistors are supposed to go to the power supply pins of the opamp. I'm not sure if falstad allows to do this though. If not, I'd suggest trying LTspice or some other simulator that allows that.

[OM222O]

Your falstad simulation doesn't quite match the circuit in the datasheet. The 1k resistors are supposed to go to the power supply pins of the opamp. I'm not sure if falstad allows to do this though. If not, I'd suggest trying LTspice or some other simulator that allows that.

unfortunately it doesn't allow me to do that, but I also noticed another difference. They have connected the - input to ground and used the "ref" pin instead! the dual channel differential op amp (ADA4522-2ARZ-RL) does not have such a pin. are there any other methods that can provide the boosting effect?

[SiliconWizard]

Why not directly use Linear's application schematic instead?

[OM222O]

I can't find the same part in the dual op amp package and I'm fairly space limited on the PCB (17.8 x 35mm) and it's mostly populated. I need to change other parts of the circuit if I have to get 2 of these chips. If all else fails, I have to use the same schematic and change my circuit around
________________________________________
EDIT:

I have tried everything I could and gave up on making the ADA4522-2ARZ-RL work. I want to use the same schematic that was provided by the LT1990 and use two of them. now the issue is that I need the response time to be as quick as possible, so the voltage that the op amp can produce is important. the LT1990 can only be used with ± 15v which is nowhere near as quick as I want it to be. it takes about 100ms to go from -100 to 100mA for my inductor.


I tried connecting the transistors to ±24V supply separately but it had no use and there were some strange 1A current draw initially.


at this point I honestly have no idea what to do please help.
is it possible to use a low voltage differential amplifier, followed by the ADA4522-2ARZ-RL in unity gain to get the higher voltage range for the desired result?

[duak]

I would ditch the common emitter booster stage.  The 1 A current pulse on startup is due to the charging of the capacitor between the bases of the booster transistors multiplied by their betas.  You can reduce the size of the cap, but you'll never really eliminate the current pulse.

I would replace it with a complementary emitter follower current booster placed between the opamp's output and the junction between the sense (shunt) and feedback resistors.

I'd also add a couple of catch diodes between the driven end of the load and the two power supply rails.  These diodes handle the compliance voltage developed by the inductor when the current magnitude is reduced too quickly.  Note also that your power supply voltage may pump up when this occurs and exceed the voltage ratings of anything connected to them so you may have to add some zener diodes to the rails to limit the voltage rise.

Does this make sense?  If not, I can explain further.

Cheers,

[OM222O]

yes, I can understand what you mean, but can you also include some schematics as well? Do I need zeners across the inductor only, or do I need it across each rail (GND to +24 and GND to -24 and -24 to +24) to stabilize the voltage? I have also not heard of a "complementary emitter follower current booster". I did some searching but I'm not sure if I understand it correctly. Does that circuit allow me to operate at the higher voltage of ±24V across the transistors even if the amplifier is supplied with±15? I also saw "Low offset follower" which I think is the setup that I need. or maybe even a bi directional current mirror connected to the higher voltage? 

[SiliconWizard]

The following works. I think your schematic contained an error (look at how the transistors and opamp are connected to the load).

Note that I attached a simulation with a 3.2µH inductor to assess fast transients. With a 3.2H inductor, it works fine, but it can't be nearly as fast as this! Current will take a much longer time to settle. This is an inductor after all!

Also note that I lowered the resistors' value around the opamp. With 1Meg resistors, this would be extremely slow to settle. I don't know how fast you are after. I simulated here with 1k resistors, if you change them to 10k you will notice the difference already, but it may be acceptable for your use. 1 Meg makes it much slower, but you would have less ringing. Up to you.

[OM222O]

I would rather use an integrated differential amplifier due to high common mode voltages and resistor matching becoming an issue. I was wrong and thought that the ADA4522 was a differential amplifier but it in fact is not! most common differential amplifiers have built in 1Meg resistors which is why I chose that value. I can't seem to find a formula that relates the resistor tolerance to the CMRR of an amplifier. would 0.05% be good enough?
I think if I go with the 1K resistors, I would also need another op amp as a virtual ground as the low value resistors can pass mAs of current which creates an issue.Please watch the video linked below

[Lee Leduc]

Check out figure 15 in the LM675 datasheet.
http://www.ti.com/lit/ds/symlink/lm675.pdf

[SiliconWizard]

I would rather use an integrated differential amplifier due to high common mode voltages and resistor matching becoming an issue. I was wrong and thought that the ADA4522 was a differential amplifier but it in fact is not! most common differential amplifiers have built in 1Meg resistors which is why I chose that value. I can't seem to find a formula that relates the resistor tolerance to the CMRR of an amplifier. would 0.05% be good enough?

As for the CMRR, you can start here: http://www.analog.com/media/en/reference-design-documentation/design-notes/dn1023f.pdf

You will have a heck of a hard time finding an integrated instrumentation amplifier that is fast enough AND can take +/-24V. The opamp you selected seems to be a good fit actually.

Are you really intending on driving a 3.2H inductor with a constant current source though? As I hinted above, 3.2H is HUGE. At those values, the dynamics of the current source may not matter that much anyway, as any real (not ideal) current source will take a long time to settle...

I think if I go with the 1K resistors, I would also need another op amp as a virtual ground as the low value resistors can pass mAs of current which creates an issue.Please watch the video linked below

What do you call "virtual ground"? The 1.024V reference you made out of the 2.048V one? In your original schematic, it's already buffered by an opamp follower (LTV9001) so that shouldn't be a problem whatsoever.

[OM222O]

they are using pots to calibrate out errors. I need 12 of these in different parts of the robot and I don't want to fiddle around with pots to get it working. I think if I use 0.05% matched resistors with 15PPM/C temp co, I will have a maximum of 3% error at worst case scenario which is not really ideal ... if I don't have any other choice, I will go with a differential amp (like the LT1990) followed by the ADA4522 as unity gain. it seems like the best option in terms of accuracy so far. I'm just waiting for "duak" to explain more about the follower circuit.


again maybe I'm calculating the common voltage wrong ...

[OM222O]

I would rather use an integrated differential amplifier due to high common mode voltages and resistor matching becoming an issue. I was wrong and thought that the ADA4522 was a differential amplifier but it in fact is not! most common differential amplifiers have built in 1Meg resistors which is why I chose that value. I can't seem to find a formula that relates the resistor tolerance to the CMRR of an amplifier. would 0.05% be good enough?

As for the CMRR, you can start here: http://www.analog.com/media/en/reference-design-documentation/design-notes/dn1023f.pdf

You will have a heck of a hard time finding an integrated instrumentation amplifier that is fast enough AND can take +/-24V. The opamp you selected seems to be a good fit actually.

Are you really intending on driving a 3.2H inductor with a constant current source though? As I hinted above, 3.2H is HUGE. At those values, the dynamics of the current source may not matter that much anyway, as any real (not ideal) current source will take a long time to settle...

I think if I go with the 1K resistors, I would also need another op amp as a virtual ground as the low value resistors can pass mAs of current which creates an issue.Please watch the video linked below

What do you call "virtual ground"? The 1.024V reference you made out of the 2.048V one? In your original schematic, it's already buffered by an opamp follower (LTV9001) so that shouldn't be a problem whatsoever.

as I mentioned it will be used to drive a voice coil actuator and yes, the inductance is really large. it's more like a transformer coil than an inductor and it cannot be swapped out for another one.

The transient response is also important as it will be controlling parts of the robot that need to react fast to the environment. The cost is not a big issue as it's supported by university (that should be a given when I'm using a 16 bit DAC, high end voltage references,etc  )

by virtual ground I meant this point in the circuit:


as you see the current changes from 1mA to 1.5mA when he removes the op amp and connects that point directly across the load resistor. then he increases the value of the 4 resistors to 100k to get rid of that error. using more op amps won't be an issue as the ADA4522 comes in a quad op amp package as well! 2 of the can be differential amplifiers and 2 can be buffers to keep the value of the resistors down for the better transient response. I'm just worried about the 24V common mode voltage. (It's actually 0V - 24V - 48V )I'm not sure if it's considered ground so we can dismiss it and consider the common mode voltage to be 1.024 or if it's 25.024  but I think we're getting to a good solution here.

I actually tried to use the analog devices error budget calculator but there seems to be an issue with it as it raises the error for the ADA4522 and says the supply difference exceeds 36 volts although it should be rated for 60V operation ...
here's a link if you want to have a look:
https://www.analog.com/en/design-center/interactive-design-tools/opamp-error-budget-calculator.html

[duak]

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!

[Alex Nikitin]

Here is a suitable circuit. The opamp, transistors and diodes types choice is purely for the simulation, you should choose suitable for your application, including voltage and power dissipation limits. If you use appropriate devices the maximum output voltage can be boosted to 100V or more. As shown the circuit will produce +/-100mA output current for +/- 5V input, that also could be changed according to your needs.

Cheers

Alex

[OM222O]

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!

thank you very much. I just had one final question: shouldn't there be current limiting resistors in series with all diodes? D1 through D4. the entire transient will take about 28ms so I'm not sure if the diodes can dissipate that power as it's just a short burst (actuator won't go from -100 to +100 rapidly, I just need the response time to be as quick as possible to react to the environment, not to rapidly move all the travel range) something like alex recommended seems like a better choice but is still missing resistors for D3 and D4?

I'm also not sure why he is using 4 transistors ... which one is dissipating the major amount of power?

[Alex Nikitin]

thank you very much. I just had one final question: shouldn't there be current limiting resistors in series with all diodes? D1 through D4. the entire transient will take about 28ms so I'm not sure if the diodes can dissipate that power as it's just a short burst (actuator won't go from -100 to +100 rapidly, I just need the response time to be as quick as possible to react to the environment, not to rapidly move all the travel range) something like alex recommended seems like a better choice but is still missing resistors for D3 and D4?

The energy in the coil is not that great, so any mid-sized diode should be OK.

I'm also not sure why he is using 4 transistors ... which one is dissipating the major amount of power?

The output pair obviously dissipates more than the previous stage. The opamp and the first pair of transistors (Q1 and Q4) make two voltage controlled current sources - one for each polarity, and the output pair (Q2 and Q3) together with resistors and diodes work as two current mirrors with a combined output. The ratio is 10:1 , so for 10mA through Q1 or Q4 the output will be 100mA through Q3 or Q2 respectively. This circuit is convenient as you essentially limited only by the size and voltage handling of transistors. The circuit as is is working in Class B so is not very linear in the area of low output currents. If that matters, just add a resistor between bases of Q2 and Q3, providing about 0.5mA current (so for +/-40V output stage supply it should be ~150K), however that may create a small offset current due to an asymmetry of two current mirrors. It could be trimmed down if required though.

Cheers

Alex

[OM222O]

Thank you very much for the detailed explanation. as I mentioned before size is a concern on my PCB so I will probably go with the circuit that duak provided. if it proved to be unsuitable after testing, I will give your circuit a shot as well.

[Alex Nikitin]

Thank you very much for the detailed explanation. as I mentioned before size is a concern on my PCB so I will probably go with the circuit that duak provided. if it proved to be unsuitable after testing, I will give your circuit a shot as well.

You should not forget that in either circuit output devices may dissipate considerable power so should be sized accordingly and placed on a heatsink (100mAx40V = 4W for example).

Cheers

Alex

[duak]

OM, you will not want to put resistors in series with the clamp diodes (D1 & D2) as they are trying to bypass currents that could damage the transistors or opamps.

Putting resistors in series with the zener diodes will allow the rail voltages to rise further.  Are you able to quantify how much mechanical energy may have to be dumped?  You could implement a more capable energy dumper by augmenting the zener diodes with some transistors and resistors.  If you have a number of drivers & actuators, only one energy dumper for each rail is needed.

Best o' luck,

[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]

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.

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]

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.

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]

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!

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,