Hi,
I need an amplifier circuit for driving an inductive load (Inductive coil). Inductance value range is 150mH to 200mH at 50Hz. Overall impedance range of the load is 15Ohm to 25Ohm. Circuit has to operate from a 24V single supply. . Excitation signal is a pure sine wave of low frequency (10Hz to 20Hz). The inductive load has to be drive at a constant current of 500mA. Please suggest a power amplifier IC with current limit.
Regards
ACR
Hi ACR,
The Improved Howland Current Pump (IHCP) that Kai suggested is the most logical circuit to apply, but there would be some issues that would have to be overcome. The IHCP is normally applied with dual supplies (+/-V) because the output load is referenced to ground. When driven by a 0V referenced sine wave input the IHCP sinks and sources current through the load. The current level it produces is proportional to the input voltage level. If a single +24 V supply is being used, then the sine input would have to remain positive throughout the entire waveform and the output would only be able to source current to a grounded load. Some sort of offsetting scheme would have to be devised for both the input and output which would increase the circuit complexity.
Additionally, the IHCP might unstable when driving an inductive load. The IHCP uses both negative and positive feedback involving two loops so compensation is not straightforward. Knowing the coil winding capacitance that Kai requested is necessary. That capacitance in conjunction with the coil inductance will create a parallel resonant load which further complicates compensation.
I show below an example of an OPA564 power op amp applied as a IHCP. In this example it is using dual supplies and driving an resistive load. The pump produces 0.5 Amps peak for a 1 V peak input. Note that this circuit hasn't been optimized and further work would be needed to improve this current pump's accuracy. I provide it as an example of the OPA564 power op amp configured as an ICHP.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hi ARC,
I haven't heard back from you so I continued to work on an alternate ac current solution for your application without having complete information. I believe I have a better solution than using an Improved Howland Current Pump. It is more predictable and tolerant of a complex load impedance. The voltage-to-current (V-to-I) solution I propose uses a current mirror approach. The TINA schematic and a graph are seen below.
I didn't have your coil details but based on the impedance information you provided, I calculated an approximate coil resistance for a 200 mH coil of 6.71 Ohms and assumed a 25 pF coil shunt capacitance. This V-to-I circuit is not especially sensitive to the load condition so if it isn't precisely defined the circuit is tolerant. The current output as shown is 500 mA for a +1 V input, and 1 Ampere peak for a +2 V input. If that peak current is too high it can be reduced to 500 mA by simply increasing Rs1 to 400 Ohms. The gain can be changed to accommodate a different input voltage range such as 0 to 5 V, etc.
A point of interest is that the drain voltage (V_Load) of T2 actually goes negative despite the use of the single +24 V supply. This can occur because of the inductor at the drain.
The TINA-TI file is included for your use.
Regards, Thomas
Precision Amplifiers Applications Engineering
Hi ARC,
I haven't heard back from you so I continued to work on an alternate ac current solution for your application without having complete information. I believe I have a better solution than using an Improved Howland Current Pump. It is more predictable and tolerant of a complex load impedance. The voltage-to-current (V-to-I) solution I propose uses a current mirror approach. The TINA schematic and a graph are seen below.
I didn't have your coil details but based on the impedance information you provided, I calculated an approximate coil resistance for a 200 mH coil of 6.71 Ohms and assumed a 25 pF coil shunt capacitance. This V-to-I circuit is not especially sensitive to the load condition so if it isn't precisely defined the circuit is tolerant. The current output as shown is 500 mA for a +1 V input, and 1 Ampere peak for a +2 V input. If that peak current is too high it can be reduced to 500 mA by simply increasing Rs1 to 400 Ohms. The gain can be changed to accommodate a different input voltage range such as 0 to 5 V, etc.
A point of interest is that the drain voltage (V_Load) of T2 actually goes negative despite the use of the single +24 V supply. This can occur because of the inductor at the drain.
The TINA-TI file is included for your use.
4454.OPA2192_ac_Hi_cur_src_01.TSC
Regards, Thomas
Precision Amplifiers Applications Engineering
