Post Go back to editing

Hello, everyone

I'm currently designing a non-inverting amplifier circuit with AD8529. Its gain is 2 and it will drive a capacitive load. I want to aquire the open-loop model of AD8529 in order to estimate its transient response and to design appropriate compensation. I refered to the method of testing loop gain T in the book 'design with operational amplifiers and analog integrated circuits', Franco，S., 3rd edition, chapter 8. When simulating, I took off the inverting capacitor and capacitive load, so beta=0.5 and there was no zeros and poles in beta, and I considered T/beta as open-loop gain. Please refer to the attachment to see the simulation circuits and results.

Q1: When powered with +5V/-5V, the simulated open-loop gain curve is same as that of datasheet; When powered with +12V/0V, which is the real working condition, the simulated open-loop gain curve is absolutely different with that of datasheet; How to explain this? Will it be ok to take the simulated open-loop model powered by +5V/-5V as the 'real' model of AD8529, and design compensation with it?

Q2: Is the testing method right? Will it be ok to take 'Vout/(Vp-Vn)' as open-loop gain and test it? Is there any testing circuit recommanded by ADI?

Best regards

circuitsandresults.docx
Parents
• Hi Wang Xueliang,

I just wanted to give a quick follow-up for this thread. The reason your loop gain looked different for the single +12V supply than the ±5V supply was because the op-amp was saturated. The inputs and outputs in your circuit are biased to 0V and, per the datasheet, the output will be saturated below 80mV. From an AC standpoint, if your output is saturated and you input a small-signal, then your output will move very little or not at all. This is why you were seeing large attenuation for your +12V open-loop gain.

If you insert a dc voltage source of 6V, and use it to bias the AD8529 non-inverting inputs, R1SC, and R1OC, then you should get the response you expect.

A word of caution: be careful using the SPICE macro-model to predict an amplifier's stability with a capacitive load. We do our best to match the step response of the model to the datasheet, but there are often second-order effects and it is rarely a perfect 1:1 match. It is much better to do these evaluations and set the compensation in the prototype stage, preferably tuning the small-signal step response for the worst-case cap load at both extremes of your temperature range and output range.

As for your test method, it looks fine to me, but maybe a bit more complicated than it needs to be. With SPICE, you can use unrealistic component values in the circuit, which enables the open-loop response to be measured with a simpler circuit. See the following: (V1 is an AC 1 with a 6V dc offset).

I hope this helps.

Best regards,

Scott

• Hi Wang Xueliang,

I just wanted to give a quick follow-up for this thread. The reason your loop gain looked different for the single +12V supply than the ±5V supply was because the op-amp was saturated. The inputs and outputs in your circuit are biased to 0V and, per the datasheet, the output will be saturated below 80mV. From an AC standpoint, if your output is saturated and you input a small-signal, then your output will move very little or not at all. This is why you were seeing large attenuation for your +12V open-loop gain.

If you insert a dc voltage source of 6V, and use it to bias the AD8529 non-inverting inputs, R1SC, and R1OC, then you should get the response you expect.

A word of caution: be careful using the SPICE macro-model to predict an amplifier's stability with a capacitive load. We do our best to match the step response of the model to the datasheet, but there are often second-order effects and it is rarely a perfect 1:1 match. It is much better to do these evaluations and set the compensation in the prototype stage, preferably tuning the small-signal step response for the worst-case cap load at both extremes of your temperature range and output range.

As for your test method, it looks fine to me, but maybe a bit more complicated than it needs to be. With SPICE, you can use unrealistic component values in the circuit, which enables the open-loop response to be measured with a simpler circuit. See the following: (V1 is an AC 1 with a 6V dc offset).

I hope this helps.

Best regards,

Scott

Children
No Data