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OPA192: Improved Howland Current Pump for High Inductance, High Frequency Operation

Nyasha Masuka
Prodigy 245 points
Part Number: OPA192
Other Parts Discussed in Thread: OPA549, OPA567, OPA569

Good day everyone,
I need to design a current amplifier circuit that drives approximately 220 mH coil, 2 Ohm series resistance with an alternating current of 25-50 kHz, 5V amplitude and a current of approx 1A.

I have referred to this forum for some research, and some users suggested using an "Improved Howland Current Pump" (IHCP). Someone even provided its design design principles as shown in the following link.


I have followed those steps, and have managed to develop a current source in the mA range. However, when I try to increase the frequency or inductance, the output current and voltage waveforms become distorted. I have included my TINA TI simulation files.

OPA192 Howland.TSC

What are the design steps that I can follow to optimize this circuit for:

  1. Higher inductance ~220 mH
  2. Higher frequency 25-50 kHz
  3. Higher output current ~1A
  • 0
    TI__Guru 67601 points

    Hi Nyasha,

    Since the impedance of an inductor is proportional to it magnitude and frequency, ZL=2*Pi*f*L, you must first consider ZL value to see if what you ask for is physically possible on +/-15V supplies used to power OPA192.

    For L=220mH and f=1kHz, ZL=6.28*1000*220e-3 = 1381ohm, so the Vout_peak = 25mA*1381ohm = 34.54 V - see below. This alone is not possible on +/-15V supplies.

    For L=220mH and f=50kHz, ZL=6.28*50,000*220e-3 = ~69kohm, so for 25mA output current the Vout_peak = 25mA*69kohm = 1,727 V - see below.

     

    Likewise, for L=35uH and f=50kHz, ZL=6.28*50,000*35e-6 = ~11ohm, so for Iout=1A  the Vout_peak = 1A*11ohm = 11 V - see below.

    This means that for  L=220mH, f=50kHz and Vout=1A, Vout_peak = 1A*69kohm = 69,080V ! - thus, of course, this is not possible on +/-15V supplies.

    Also, since the maximum output current of OPA192 is limited by its short-circuit current of +/-65mA, you would have to use another power op amp to deliver +/-1A peak current. 

  • 0 in reply to Marek Lis
    Prodigy 245 points

    Thank you for your feedback, it's really insightful. I have a follow-up question. I have tried simulating with several high-current opamps ( OPA569, OPA549, OPA567 etc). I have noted that they give good current output at low frequency and low inductance. However, the output current diminishes as frequency and inductance are increased, even when I increase the supply voltage.

    In your response, you mention that this alone is not possible on +/-15V supplies. How do you suggest I achieve this, even if I reduce the frequency and inductance requirements, to say (f=25KHz, I=1A and L=50mH)?

  • 0 in reply to Nyasha Masuka
    TI__Guru 67601 points

    Nyasha,

    Since under the condition mentioned above, ZL=6.28*25,000*50e-3 = 7.85kohm, this means that on +/15V supplies the most current you could source is less than 2mA (Vout_peak = 2mA*7.85k >15V). 

    This means that with Iout=+/-25mA, the minimum supplies would have to be close to +/-200V - see below.

    Similarly, this means that in order for your circuit to work with Iout=1A, you would need to have op amp capable of supplies above 1A*7.85kohm or 7850 V - an impossible task.