MOS drain and source

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MOS tube drives the motor, and the load is connected to the drain end; MOS + op amp constitutes a constant current source, and the load is also at the drain end. I would like to ask, can the load be placed at the source? What is the difference between the two?

maychang

The load can be placed at the source. However, if the load is placed at the source, the gate of the MOS tube needs a higher driving voltage. Taking the previous figure as an example, if the load is placed at the source, the gate voltage needs to be higher than the power supply voltage (at least higher than the power supply by the gate conduction voltage, plus the voltage required for MOS tube saturation). Usually the driving circuit cannot output a voltage higher than the power supply voltage, so if the load is placed at the source in the previous figure, the MOS tube cannot enter saturation conduction. The power consumption must be relatively large.

maychang

In the latter figure, if the load (at PIN2) is moved to the source, because the constant current circuit does not require the MOS tube to enter saturation, the LM358 does not need to output a voltage higher than the power supply voltage. However, if the load in the latter figure is placed on the source, the constant current range will be smaller than if the load is placed on the drain. The maximum output of the LM358 is lower than the power supply voltage, roughly the power supply voltage minus 1.5V. When the load is placed on the drain, the maximum output current of the MOS tube is the power supply voltage minus 1.5V minus the gate-source voltage required by the MOS tube minus the voltage drop on the 1-ohm resistor, divided by 1 ohm. When the load is placed on the source, the power supply voltage minus 1.5V minus the gate-source voltage required by the MOS tube minus the voltage drop on the 1-ohm resistor minus the voltage drop on the load.

maychang

It can be understood in this way: when the load is placed on the drain, it is a common source amplifier circuit (equivalent to the common emitter circuit of a bipolar transistor), and when the load is placed on the source, it is a common drain amplifier circuit (equivalent to the common collector circuit of a bipolar transistor). As we know, the voltage gain of the common drain amplifier circuit is slightly less than 1 and approximately equal to 1, while the voltage gain of the common source amplifier circuit can be very large. Since the voltage gain of the common drain amplifier circuit is very small, the driving circuit must output a larger voltage. When the driving circuit cannot output a larger amplitude range, the amplitude of the MOS tube output will inevitably decrease.

maychang

In the latter figure, for both the case where the load is placed at the drain and the case where the load is placed at the source, when the constant current value changes, VGS is actually the same. What is different is the voltage output by the op amp.

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maychang published on 2021-7-8 18:51 In the latter figure, if the load (at PIN2) is moved to the source, the constant current circuit does not require the MOS tube to enter saturation, so there is no need for the LM358 output to be higher than the power supply voltage...

I'm sorry, I was busy with other things these days and didn't reply to the teacher's post in time.

"When the load is placed at the drain, the maximum output current of the MOS tube is the power supply voltage minus 1.5V minus the gate-source voltage required by the MOS tube minus the voltage drop on a 1-ohm resistor, divided by 1 ohm. When the load is placed at the source, the power supply voltage minus 1.5V minus the gate-source voltage required by the MOS tube minus the voltage drop on a 1-ohm resistor minus the voltage drop on the load." You mentioned subtracting 1.5V here. What is this 1.5V? And why do we need to subtract the gate-source voltage?

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maychang posted on 2021-7-8 19:03 In the latter figure, when the constant current value changes, VGS is actually the same for the two cases of placing the load on the drain and the load on the source. The difference is the output of the op amp...

Unable to upload pictures

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Pingpiaoliu published on 2021-7-15 17:49 Sorry, I have been busy with other things these days and did not reply to the teacher's post in time. "The load is placed on the drain, and the maximum output current of the MOS tube...

The reason there is a 1.5V is because the op amp model you are using is LM358, and the maximum output of LM358 can only reach (approximately) the power supply voltage minus 1.5V.

The output power supply voltage of the op amp minus 1.5V is the gate voltage of the MOS tube. The gate voltage of the MOS tube minus the voltage between the gate and the source of the MOS tube is the voltage on the 1 ohm resistor. The voltage between the gate and the source of the MOS tube is the MOS tube conduction voltage plus the drain current divided by gm (transconductance).

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Pingpiaoliu published on 2021-7-15 17:49 Sorry, I have been busy with other things these days and did not reply to the teacher's post in time. "The load is placed on the drain, and the maximum output current of the MOS tube...

Calculate the voltage (maximum value) on the 1 ohm resistor and divide it by 1 ohm, which is the maximum drain current (the source and drain currents of the MOS tube are equal).

maychang

Pingpiaoliu published on 2021-7-15 17:49 Sorry, I have been busy with other things these days and did not reply to the teacher's post in time. "The load is placed on the drain, and the maximum output current of the MOS tube...

If the load is placed at the source, the maximum output voltage of the op amp will not change, but the maximum output current is equal to the power supply voltage minus 1.5V, minus the voltage between the gate and source of the MOS tube (MOS tube conduction voltage plus output current divided by gm), minus the voltage across the load, and the result is divided by the load resistance plus 1 ohm. Obviously, the maximum output current is much smaller when the load is placed at the source.

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maychang posted on 2021-7-15 18:43 The reason why there is a 1.5V is because the op amp model you use is LM358, and the maximum output of LM358 can only reach (approximately) the power supply voltage minus 1.5V. ...

"The voltage between the gate and source of the MOS tube is the MOS tube conduction voltage plus the drain current divided by gm (transconductance)." Is this MOS tube conduction voltage the threshold voltage?

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maychang posted on 2021-7-15 18:57 If the load is placed at the source, the maximum output voltage of the op amp will not change, but the maximum output current is equal to the power supply voltage minus 1.5V, minus the MOS tube gate...

I understand what you said. When the load is placed at the source, the voltage of the load is subtracted more than when it is placed at the drain. So the current is smaller. But why is the load current not the power supply voltage divided by (Rds+Rload)?

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Pingpiaoliu published on 2021-7-16 11:04 "The voltage between the gate and source of the MOS tube is the MOS tube conduction voltage plus the drain current divided by gm (transconductance). "This MOS tube conduction voltage is the threshold voltage...

"Is the on-state voltage of this MOS tube the threshold voltage?"

Yes. Usually recorded as V GS(th) .

maychang

Pingpiaoliu published on 2021-7-16 11:15 I understand what you said. When the load is placed on the source, the voltage of the load is subtracted more than when it is placed on the drain. So the current is smaller. But why is the load current not...

A purely resistive load is placed at the drain, and the resistance on the source of the MOS tube is removed. As shown in the previous figure, when the MOS tube is saturated and turned on, the current in the load is the power supply voltage divided by (Rds+Rload).

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Pingpiaoliu published on 2021-7-16 11:15 I understand what you said. When the load is placed on the source, the voltage of the load is subtracted compared to that on the drain. So the current is small. But why is the load current not...

In the latter figure, because of the strong negative feedback, the MOS tube and the operational amplifier form a constant current source, and the current in the load is determined by Vin and R4, which is Vin/R4. Changing Vin can change the current in the load.

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maychang posted on 2021-7-16 11:45 In the latter picture, because of the strong negative feedback, the MOS tube and the op amp form a constant current source, and the current in the load is determined by Vin and R4, which is Vin/R4. Changing Vi ...

I roughly understand that the analysis of MOS as a constant current source and driving a motor is different

maychang

Pingpiaoliu published on 2021-7-16 16:54 I roughly understand that the analysis of MOS as a constant current source and driving a motor is different

In the first picture on the 14th floor, if the load is a motor, when the MOS tube is saturated and turned on, the current in the load (motor) is not the power supply voltage divided by (Rds+Rload). For the motor, not only the resistance component of the load must be considered, but also its reactive component, and the back electromotive force of the motor must be considered.