I would appreciate some help designing a circuit to convert a voltage signal to a current signal. Here is my attempt at a schematic for the circuit.
I don't see a need to use the current limit function included in the diagram. If over current is a concern, I think I will use a fuse in the output.
Thanks to everyone who takes a look at this and helps me out.
It will help us to understand a bit more about what you are trying to do, so I have some questions:
The 150mV output from the Hall device... does it vary from 0V to 150mV?
Do you want the output current to vary 0A to 3A, proportional to the output from the Hall device?
Must the 0.05 ohm load be ground-referenced? Could it be floating?
Can you supply information on the Hall sensor? Does it have a buffered low-Z output?
The approach you have shown would dissipate (12V)(3A) = 36 watts. This is quite a lot of dissipation so the complete circuit will get HOT. There are two possible alternatives: 1) Run the op amp from a lower supply voltage, if available. 2) Use a class-D type amplifier.
In reply to Bruce Trump:
Hall Device voltage represents a current in a separate conductor of 5
Amps. That 5 amp current produces a
magnetic field in a gapped toroid core of about 30 G and the Hall device
sensitivity is 5 mV/G. The requirement
is to measure the current over a narrow range (4 to 6 A) very accurately (<0.1%). The Hall device is an Allegro A1321 and the
output is push-pull.
don’t know if it is possible to avoid the ground reference for the output
resistor. The current measuring
instrument is a Yokogawa WT 1600 power meter which has a current transformer
input stage with 25 milli-Ohms of resistance. The 50 milli-Ohm includes some
added resistance to achieve the correct output current.
think the output power will be 0.05*9 = 0.45 W but I don’t know what the input
will be. A heat sink will be required.
this point, I would like to stay away from PWM.
so much for your interest and I look forward to your further comments.
In reply to Rick Stockum:
It sounds as if the 0.05 ohm load is a winding of a current transformer, correct? It would seem that this load could then be floating and would not need to be ground referenced. This could simply the circuit used to drive the 3A current.
Your power calculation is the power dissipated by the 0.05 ohm load which is small by comparison. The power that must be dissipated by the amplifier drive circuitry is 36 watts at 12V operation. (Actually about 35.5 watts, accounting for the power delivered to the 0.05 ohm resistor.) This will require a large finned heat sink. I just want you to consider whether this is practical before we design a circuit.
Another comment... the spec on the Allegro Hall device shows an operating voltage of 5V with an absolute max of 8V. It should not be powered from 12V as you show it. Furthermore it is shown to be on last-time-buy status on the Allegro site.
the input CT has a winding resistance of 25 milli-Ohms on all ranges. If you can find a return for the current, the
output can float.
didn’t show the voltage divider that I use to decrease Vcc for the A1321 to
prevent clutter on the diagram. I have
A1324 units on order (delivery in 6 weeks) and a stock of A1321 devices on
hand. The Burr-Brown OPA548 operates on
any (single) voltage from +8 to +60 V and the performance is not significantly affected
over the full range of the supply voltage.
Power loss is not an issue.
Here is a possible circuit for your consideration. I agree with your selection of the OPA548. The 150mV signal is amplified by a gain of 40 and impressed on R3, making 6V on R3. The 3A in this resistor flows through your load, RL.
R3 is chosen to drop half the supply voltage on itself rather than the amplifier. This splits the 35.5 watts of dissipation between R3 and U1. Both will still get hot.
The accuracy and temperature stability of R3 will directly affect accuracy. The current flowing in R1 + R2 creates a tiny error but it's proportional to the signal value and can be neglected. Grounding technique will be very important due to the low value of R3. Kelvin-type connections of R2 to the high side of R3 and R1 and the input signal to the low side of R3 are critical.
I appreciate you efforts on this. I hope to breadboard it next week and will drop you a note on the results.
I would not use a fuse instead of the amplifier's internal current-limit circuit. With semiconductors, it is always a race between the silicon and the fuse to see which one blows first. The silicon almost always wins.
Regards, Neil P. Albaugh ex-Burr-Brown
All content and materials on this site are provided "as is". TI and its respective suppliers and providers of content make no representations about the suitability of these materials for any purpose and disclaim all warranties and conditions with regard to these materials, including but not limited to all implied warranties and conditions of merchantability, fitness for a particular purpose, title and non-infringement of any third party intellectual property right. TI and its respective suppliers and providers of content make no representations about the suitability of these materials for any purpose and disclaim all warranties and conditions with respect to these materials. No license, either express or implied, by estoppel or otherwise, is granted by TI. Use of the information on this site may require a license from a third party, or a license from TI.
TI is a global semiconductor design and manufacturing company. Innovate with 100,000+ analog ICs andembedded processors, along with software, tools and the industry’s largest sales/support staff.