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# THS3062: Problems when dealing with THS3062 based PI (Proportion-Integration) regulator

Part Number: THS3062

Dear TI expert:

I want to use an analog PI(Proportion-Integration) regulator based on op amp THS3062 in a system. Before this, I try to use this analog PI regulator in a relatively simple circuit as Fig1.

Fig1. Simple applications of PI based on op amp

L1’s voltage is summed with 2V at op amp’s + terminal, whose result is the input of analog PI’s. Obviously, this circuit can control L1’s current to decrease linearly with time at speed -2000A/s, and hence, L1’s voltage should stabilize at -2V. For the analog PI regulator based on op amp in Fig1, the proportion and integration coefficients KP and KI are as follows. The Cc is always set to 10pF.

KP=1+RC2/RC1  KI=1/(RC1*CC)

Firstly, I use a common op amp OP747 instead of THS3063 in the simulations. RC1=RC2=30Ω and KP is set to 2. With OP747’s spice model(downloaded from TI website), L1’s current and voltage is simulated as Fig2 and the control goal is met. When I set RC1=RC2=100Ω, 300Ω or 1kΩ, the simulation results are all as Fig2.

Fig2. OP747 based PI simulations results for L1 current(upper figure) and voltage(lower figure)

Then, I use the THS3062 based PI regulator in Fig1. Due to the simulations for OP747, KP is also set to 2 here and Cc is always set to 10pF. According to THS3062’s datasheet(Fig.3), when G=2, the recommended feedback resistor value is 560Ω. So RC1=RC2=560Ω is chosen for the analog PI regulator.

Fig3. THS3062 datasheet for recommended feedback resistor values

Hence, with THS3062’s spice model(downloaded from TI website), L1’s current and voltage simulations are as Fig4. The L1 voltage doesn’t stabilize at -2V, hence the fall speed of L1 current isn’t controlled to be -2000A/s.

Besides, the op amp’s inverting and noninverting terminal’s input current are very large in the range of several mA as Fig5. This is certainly not the correct result.

Then  increase RC1=RC2 to 1kΩ, and the simulations of L1 current, L1 voltage, THS3062 inverting noninverting terminal’s input currents(in the range of several hundred nA) are good now.

Fig4. THS3062 based PI regulator's simulation results for L1  current(upper  figure) and voltage(lower  figure)

Fig5. THS3062’s inverting and noninverting terminal’s input current

I have 2 problems:

1. In the simple example above, the chosen values of feedback resistor are very different from the datasheet’s recommendations, what about the more complex system? In practical experiments, how to choose the appropriate RC1 RC2 and Cc.
2. Why THS3062’s inverting and noninverting terminal has such a big input current(mA when RC1=RC2=560Ω in Fig.5) in this simple simulation? This will worsen the PI regulation. In my more complicated system’s simulations, I encounter a similar problem: In Fig1, the inverting terminal input current is as big as RC1’s current, then the feedback RC branch conducts nearly no current and the PI can then hardly function properly.I wonder why this big current arises and how to solve such a problem.

Best regards!

• Hi Yatao,

1. The datasheet feedback resistance values are chosen to provide a relatively constant bandwidth over gain. Increasing the resistance will decrease bandwidth and add phase margin. The relative values of Rc1, Rc2, and Cc depend on the PI coefficients desired.

2. The big input current could be caused by instability. In current feedback amplifier integrators, it is necessary to add a resistor (~1kΩ) between the inverting input and the integrating node, which in this case is between Rc1 and Rc2. It shouldn't affect the coefficient equations.

What are the supply rails of your amplifier? Is 2V the mid supply? The inverting input resistance is only 71Ω, not high impedance as for a VFA. Its current will be the same as Rc1, but if it is being held at ~2V, it should be connected by Rc1 to 2V instead of ground so you avoid a high current and more importantly get the differential error across your inputs, not 2V.

Best regards,

Sean
• In reply to Sean Cashin:

Hi Sean,

According to your reply, you think THS3062’s big input current is caused by 2 possible reasons. First is the unstable system; second is the mid supply voltage not equal to ground.

In this simple simulation, the supply voltage for THS3062 is ±15V, so I think the true cause is the unstable system.

I forgot to say, the simulation spice models are all downloaded from TI website, and the simulations are all conducted in cadence/Orcad Capture.

In all op amps’ datasheets and relevant materials, the stability analysis can be found only about op amps applied to inverting form, noninverting form and trans-impedance form and analysis about PI(Proportional-Integral) can hardly be found. Hence, I may have several problems to consult you.

1. I must first apologize that I say in my post that when I increase RC1=RC2 to 1kΩ with CC=10pF, then the THS3062 based PI works well in the simple simulation. This is not the case, and the simulation results are still bad, and much worse than the OP747 based PI regulator. I further increase RC1=RC2 to 10kΩ with CC=10pF, the simulation results are still bad.

2. How to analyze the stability and performance of op amp(including high-speed op amp and CFA) based analog PI regulator? Can you please offer some theoretical analysis or relevant materials?

3. Based on your reply, ‘it is necessary to add a resistor (~1kΩ) between the inverting input and the integrating node, which in this case is between Rc1 and Rc2. It shouldn't affect the coefficient equations’, is the configuration like Fig1(b)? But I tried the inserted resistor RR 1k, 10k, 100k, they still don’t work. And I’ve increase CC to 1nF, but the results are still bad.

Besides, I’ve never seen any materials say that ‘In current feedback amplifier integrators (Fig2 is from pp19/23 in THS3062’s datasheet, the configuration contains no additional resistor), it is necessary to add a resistor (~1kΩ) between the inverting input and the integrating node’. What’s the reason for doing this? And are there any other tips for using current feedback amplifier integrators?

(a)

(b)

Fig1 Analog PI implementation

Fig2. Integrator’s configuration in THS3062’s datasheet

4. Can you please offer me one simple simulation(best the same as mine) with THS3062 based analog PI regulator? And offer me with one simple simulation(best the same as mine) with a high-speed VFA op amp(±15V power supply, GBP not smaller than 350M) based analog PI regulator? And proportion coefficients KP is around 3, and integration coefficients KI is in the range of 10^7~10^8. I doubt that CFA can’t be applied to such an application.

Best regards!

Yatao

• In reply to Yatao Ling:

Hi Yatao,

You are using a non-inverting configuration so you should use the non-inverting circuit in Figure 57 of the datasheet. Here is a simulation for a Kp=2 Ki=10MHz circuit.

THS3062PIControl.TSC

Unfortunately, I am unaware of any current material on this subject to which to direct you. However, I hope using the circuit in Figure 57 will solve your issue.

Best regards,

Sean

• In reply to Sean Cashin:

Hi sean,

Firstly, I may now know why the circuit is unstable. It's probably due to the ideal output voltage amplifier, -10 is too big. And it may lead to the loop gain A*beta too big, resulting in negative phase margin. After I change -10 to -1, the circuit works. However, I still have no idea why the op amp's input current is so big when the circuit isn't stable?

Secondly, because THS3062's datasheet gives no plots about its transimpedance Z, can you tell me or are there materials about how to obtain the figure by TINA-TI? And how to obtain the Zout, Z(inverting node) by TINA?Finally, how to obtain the open-loop gain's magnitude and phase characteristics for high-speed VFA byTINA?

Best regards!

Yatao

• In reply to Yatao Ling:

Hi Yatao,

Transimpedance = -Vout/Iin. It is determined by the circuit, not the op amp.

Here is the simulation circuit for Zo, or closed loop output impedance. At 1MHz and G=1, it is 10^(-23.5/20)=.07 Ω, which is slightly less than the datasheet's .1Ω.

For Zin, connect a signal generator and a expected input resistor in series with the input and see if the mid point voltage signal is -6dB. You can find the capacitance from the -3dB point.

Aol

• In reply to Sean Cashin:

Hi Sean,

Thank you very much for your help.

But the trans-impedance and output impedance of THS3062 I’ve asked is as Fig1 depicted below. The rans-impedance is Z, input impedance ZB, output impedance ZOUT.Can you tell me how to measure them by TINA?

And please offer me materials about how to measure them, because I don’t quite understand why 1T capacitance and inductance are used in your last reply.

Fig1

Best regards!

• In reply to Yatao Ling:

The open loop transimpedance (Zol) measurement is

The inductor ensures that the amplifier has the correct DC operating point, and the capacitor is an AC short to ground.

Here is a precision labs video explaining this.

Best regards,

Sean

• In reply to Sean Cashin:

Hi Sean,

Thank you very much for your patient help. These will be my final questions.

1.Thanks for the website you recommend in your last reply. I see this slide Fig1(the resolution is poor) in that video. And they all break the loop from the op amp’s output. I wonder what are the right ways to measure the loop gain for CFA? And what are the principles to choose where to break the loop?

Fig1

2.Fig2 is from THS4631’s datasheet(pp5/34, Figure6). I test the closed-loop transfer function under Fig2’s condition with TINA-TI. And I find that with an isolation resistance 10ohms, the closed-loop transfer function will have no peaking even with 100nF load capacitor. Can you please show me the correct result?

Fig2

3.Feedback capacitors can stabilize VFA, which is not the case for CFA? Hence, if I want to use the THS3062 to build an analog PI regulator, how can I make the regulator on a PCB function well without oscillation or instability?

Best regards!

Yatao