Pure hardware AC constant resistance load and INA related issues

呜呼哀哉

Pure hardware AC constant resistance load and INA related issues [Copy link]

 

As shown in the figure above, after the input voltage is divided, it enters the OP common-phase terminal. The given resistance value is multiplied by the sampled current value I and subtracted from the actual voltage to obtain a constant resistance.

Is my overall idea correct?

Now a few details:

1. The result of directly using the op amp's inverse-phase inputs will be poor, right?

2. INA itself is a closed-loop device, and usually gives a data graph of closed-loop gain and bandwidth (equivalent to a Bode plot of a low-pass filter). The question is now that INA is in the feedback loop, how does it affect the larger closed loop? Will there be no phase delay through INA?

3. The S of MOS tube stands for gain, A/V, and is also in the feedback loop, even without INA, as shown in the following figure

Is the actual Aβ approximately equal to A*S*R of OP? When S*R is greater than 1, it is equivalent to raising A. Will the originally unit-gain stable op amp become unstable?

I hope all forum friends can give me some advice, thank you.

呜呼哀哉

Dear forum friends, please tell me after reading this. The schematic diagram has been designed, but when I look at it in reverse, I am confused. I need your guidance.

呜呼哀哉

Forum friends, please give us your thoughts. I don’t need to be comprehensive, just give us some points.

maychang

Alas, published on 2022-2-17 18:02 Dear forum friends, please tell me after reading this. The schematic diagram has been designed, but I am confused when I look at it from the other side. Please give me some guidance.

The polarity of this tube is P-channel, and it can only work linearly when the source is positive and the drain is negative.

maychang

"Is my overall idea correct?"

The overall idea seems to be fine.

maychang

Alas, published on 2022-2-18 09:27 Forum friends, please give some advice. I don’t ask for comprehensiveness, just some points.

I don't know what your AC constant resistance is for, nor do I know what accuracy you require for your AC constant resistance.

If your AC constant resistance requirement is not too high, a rectifier bridge plus a DC constant resistance circuit will form an AC constant resistance circuit. However, because the rectifier bridge resistance is ignored, the accuracy is not high. But the structure is very simple and you can use the ready-made circuit.

呜呼哀哉

maychang posted on 2022-2-22 12:27 I don’t know what your AC constant resistance is for, nor do I know what your AC constant resistance accuracy requirements are. If your AC constant resistance requirements are not too high, ...

In fact, when AC is applied to the back-to-back MOS tubes, one of them will always be turned on through the body diode, and will also have a resistance value equivalent to the diode.

But I don't quite understand why after adding a rectifier bridge, how can a DC constant resistance be used as an AC? After all, although the voltage after the rectifier bridge is DC, it is also changing, and the current still has to change to be equivalent to resistance. How can the structure be simple?

maychang

Alas, published on 2022-2-22 13:37 In fact, when the back-to-back MOS tubes are connected to AC, there is always one that is turned on through the body diode, and there will also be a diode equivalent resistance value. However...

The DC constant resistance circuit originally works under the condition of voltage change. It is because the voltage at both ends of the DC constant resistance circuit changes and the current flowing through it also changes that constant resistance can be achieved. The working state of the ordinary DC constant resistance circuit is exactly the same as the DC constant resistance circuit in the AC constant resistance achieved by the rectifier bridge plus the DC constant resistance circuit.

Of the two power MOS tubes "back to back", one of them is always turned on through the body diode. But it should be noted that this body diode is within the feedback loop, so its resistance can be compensated.

maychang

Alas, published on 2022-2-22 13:37 In fact, when the back-to-back MOS tubes are connected to AC, there is always one that is turned on through the body diode, and there will also be a diode equivalent resistance value. However...

"But I don't quite understand why after adding a rectifier bridge, how can the DC constant resistance be used as AC?"

The two ends of the AC input of the rectifier bridge are the AC constant resistance input ends. The two ends of the DC (output ends) of the rectifier bridge are the DC constant resistance input ends. The rectifier bridge is added with a DC constant resistance circuit, in which the DC constant resistance circuit works on DC. Looking from the two ends of the AC input of the rectifier bridge, its resistance is equal to the resistance set by the DC constant resistance circuit plus the on-resistance of the two diodes.

maychang

Alas, published on 2022-2-22 13:37 In fact, when the back-to-back MOS tubes are connected to AC, there is always one that is turned on through the body diode, and there will also be a diode equivalent resistance value. However...

Two "back-to-back" MOS tubes are often used for AC switches. However, AC switches can also be realized by using a rectifier bridge plus a MOS tube. As long as the MOS tube is connected to the DC ends of the rectifier bridge, the two AC ends of the rectifier bridge are AC switches, and the conduction and shutdown of the MOS tube can be controlled to control the conduction and shutdown of the AC ends.

maychang

Alas, published on 2022-2-22 13:37 In fact, when the back-to-back MOS tubes are connected to AC, there is always one that is turned on through the body diode, and there will also be a diode equivalent resistance value. However...

"How can the structure be so simple?"

Your original design uses at least two power MOS tubes. The sources of these two power MOS tubes are connected together or connected together through current sampling resistors. The gates must be driven separately, and the drains cannot be connected together due to different potentials. Since the drains cannot be connected together, a heat sink cannot be used. The two heat sinks must be insulated from each other, or the two power MOS tubes must be insulated from the heat sink separately.

If a rectifier bridge plus a single power MOS tube is used to form an AC constant resistance circuit, only a single power MOS tube is needed, and only a single drive is required, so the circuit is obviously simpler. Only one heat sink is needed. Even if multiple MOS tubes are connected in parallel because the current capacity of the MOS tube is not large enough, the drains of each MOS tube can be connected together without insulation.

maychang

Alas, published on 2022-2-22 13:37 In fact, when the back-to-back MOS tubes are connected to AC, there is always one that is turned on through the body diode, and there will also be a diode equivalent resistance value. However...

"How can the structure be so simple?"

From the perspective of the power MOS tube drive circuit, if a rectifier bridge plus a DC constant resistance circuit is used, the MOS tube only needs to be driven in one direction, and each op amp can use a single power supply. If the circuit you designed is used, the two MOS tubes must be driven separately, and each op amp must use a dual power supply. A single power supply is always simpler than a dual power supply.

呜呼哀哉

Thank you for the careful reply, thank you very much.

I feel that I am still lacking in theoretical knowledge. The original plan had reached the circuit design stage, but suddenly I found that the forward transmission parameter S of the MOS tube was a problem. In addition, the voltage on the current sampling resistor must be amplified by INA before entering the multiplier, which all have gain. So the switching gain of the circuit composed of the operational amplifier, MOS, and INA in front will not be stable with unity gain. However, this value must be fed back to the non-inverting input of the operational amplifier, and the voltage cannot be divided, so this circuit will not be compensated.

maychang

Alas, posted on 2022-2-22 15:23 Thank you for the careful answer, thank you very much. I feel that I still lack theoretical knowledge. The original plan has reached the circuit design stage, and suddenly...

Yes, all amplifiers, including MOS tubes, have gain. But what do you want to do? I can't see. The post says "no compensation", so I guess it's for feedback stability?

maychang

Alas, posted on 2022-2-22 15:23 Thank you for the careful answer, thank you very much. I feel that I still lack theoretical knowledge. The original plan has reached the circuit design stage, and suddenly...

"But this value must be fed back to the op amp's non-inverting input, and it cannot be divided."

It can be divided. Voltage division is nothing more than adding a link with a gain less than 1. In other words, voltage division is equivalent to reducing the gain of your amplifier to the original gain multiplied by the voltage division ratio.

呜呼哀哉

maychang posted on 2022-2-23 16:25 "But this value must be fed back to the op amp's in-phase input, and it cannot be divided." It can be divided. Voltage division is just to add a gain less than 1...

The reason is that an analog multiplier is used, which is limited by the power supply. After the voltage is divided at this point, either the multiplier input is too small or exceeds the power supply range.