OPA615: OTA to drive capacitive loads

Hello Tom,

  We apologize for missing your thread earlier. 

  The benefit of an dedicated OTA +SH is due to the integration of the required components on one chip. You are correct that you would be able to create one with discrete components. But, this device is useful for optimizing size, cost, ease of use, and precision (lower noise and lower propagation delay compared to using all discrete components). If you are looking for an amplifier that is configured as a voltage controlled current source, then the OTA portion of the device can be used. The SOTA portion is the sampling comparator which is most commonly paired with the OTA portion of the device for the properly configuration of the hold portion with the discrete output capacitor. We did release a new high speed voltage-controlled current output driver which is useful especially to drive lasers: LMH13000 and video. 

 If you just need to drive a capacitive load with a regular amplifier (voltage controlled voltage source). We do have many amplifiers that are able to drive capacitive loads.

 Which do you require for your design/application: an OTA or SH + OTA or a voltage controlled voltage source to drive a heavy capacitive load. 

Thank you,

Sima 

Hello Sima,

thanks for reply but you missed my question.

I am not primary interested in the SOTA part but in OTA topology in general.

I am referring to this quote from above: "A current mode amplifier would charge the hold capacitor using a constant current (ISRC), thus resulting in a linear increase of the capacitor voltage: dVCH dt = ISRC CH. Consequently, current mode (transconductance) implementations will have faster settling than voltage mode variants."

I would like to know, if this quote is still true for the year 2025, where voltage controlled voltage sources might have improved since the publication of the referring article.

"This can be achieved with either current or voltage mode amplifiers. The latter will provide a constant voltage to charge CH, resulting in a RC charge/discharge pattern: VCH = VSRC(1 − e − t RCH ), VCH is the voltage on the hold capacitor and VSRC is the constant voltage at the output of the amplifier."

What I get from this quote is that a voltage output that drives a capacity introduces a new pole or zero ( not so relevant for me right now). A current output seems not to. Will a good compensated voltage output behave like a current output?

Will a OTA be better suited to drive a 470nF capacity than a voltage output for signals up to 1 MHz and 10V amplitude?

Hello Tom,

  Thank you for the additional explanation, I understand the reasoning for the question now.

   We have been progressing with our power amplifiers (regular voltage feedback amplifiers, VFB) that are able to provide higher currents and support higher output voltages needed for your type of design. However, we still need to work through some math to determine if we have a good enough amplifier available to provide a voltage output to drive your necessary capacitor up to 1MHz at 10V amplitude. Otherwise, yes, you would need to use a VFB followed by a power discrete transistor or an integrated OTA.

  A 470nF capacitor at 1MHz is around 0.3Ohms resistive which is very heavy load to drive. This design would need a series resistor at the output of the amplifier for both adding to the load and isolation of the capacitor as you mentioned for stability.

   Slew rate needed would be around 63V/us for a 10Vp output at 1MHz sine-wave, which is not too high for our high-speed amplifiers. The issue is full-power bandwidth (slew-rate limited bandwidth) and claw curves (output voltage swing vs output current) which will be critical for choosing the correct amplifier. If using at unity gain, full-power bandwidth will be 1MHz (FPBW = slew/2*pi*Vpeak), and should pick an amplifier of around x2-x10 of the values so far for headroom. Also, since you will need an isolation resistor, your voltage after will drop due to the voltage divider at your load. In that case, you will need to recalculate the above values for an output voltage right at the output of the amplifier that needs to be set so after division from load you achieve 10V peak signal. The chosen resistor, and therefore total load, will also determine the amount of output current you need out of your amplifier. And, this will determine the amplifier that can support the output voltage swing needed directly at the output of your amplifier.

  Here are some threads for stability which will come up later in your design as you mentioned:

• First E2E thread
• Second E2E thread

  This thread is very useful to show this type of example: link to e2e thread here. This thread shows the steps Marek took to check if the amplifier can support the design or if needed to use discrete transistors at the output of the amplifier. For this case, they did have to use these transistors to support a 4A output current to drive a capacitance load at 28V peak at 200kHz. 

  For example, if a 50 ohm resistor to your load capacitance is enough to keep the amplifier stable, then you would need around 200mA of output current. Since this 10 Ohm resistor will create a voltage divider with your capacitor impedance of 0.3Ohms, then you are left with around 60mV. If your design can handle this value, then you can choose a power amplifier that can support 200mA, 63V/us slew rate at 1MHz if at unity gain. These values are easy to support with a VFB amplifier: link to power amplifiers filtered and sorted by GBW. Otherwise, you would need to use discrete transistors or an OTA since to obtain 10V after voltage divider would require around 1700V at output of amplifier which is not possible in this example.

Thank you,
Sima