What is the voltage when the voltage is broken from left to right?

小太阳yy

What is the voltage when the voltage is broken from left to right? [Copy link]

 

If the voltage is input from point A and the circuit is disconnected and connected through an LED, what is the voltage at point B? I am a little confused

maychang

"I'm a little confused."

It depends on what instrument you use to measure. Different instruments will produce different results.

When measuring with a common multimeter, the positive probe (red) must be connected to point B, and the negative probe (black) must be connected to the ground. The result of the multimeter measurement is slightly less than 3V. A poor pointer multimeter may measure less than 2V, and a good pointer multimeter may measure more than 2V. A digital multimeter may measure 2.8V.

maychang

It should be noted that when a multimeter, oscilloscope or other instrument is measuring the voltage at point B, the circuit is no longer "open".

小太阳yy

maychang posted on 2022-3-29 14:40 It should be noted that when a multimeter, oscilloscope or other instrument measures the voltage at point B, this circuit is not "open circuit".

Actually, if it is in the open circuit state, it should be 3V, right? I measured it with an oscilloscope and it is quite different. Logically, the internal resistance voltage range of the oscilloscope and the digital multimeter should be at the megohm level, right? If I add a MOS tube at the back to form a loop by controlling G, when the MOS is not turned on, will there be leakage current?

maychang

小太阳yy posted on 2022-3-29 15:03 In fact, if it is in an open circuit state, it should be 3V, right? I measured it with an oscilloscope and it is a bit different. It stands to reason that the oscilloscope and the digital multimeter...

"It should actually be 3V, right? I measured it with an oscilloscope and it's a bit off."

What is the value measured by the oscilloscope? There must be a numerical value, right?

maychang

小太阳yy posted on 2022-3-29 15:03 In fact, if it is in an open circuit state, it should be 3V, right? I measured it with an oscilloscope and it is a bit different. It stands to reason that the oscilloscope and the digital multimeter...

"If I add a MOS tube at the back to form a loop by controlling G, will there be leakage current when the MOS is not turned on?"

This has nothing to do with whether the MOS tube is turned on. In fact, most MOS tubes are already slightly turned on when Vgs reaches 3V.

maychang

小太阳yy posted on 2022-3-29 15:03 In fact, if it is in an open circuit state, it should be 3V, right? I measured it with an oscilloscope and it is a bit different. It stands to reason that the oscilloscope and the digital multimeter...

"If I add a MOS tube at the back to form a loop by controlling G, will there be leakage current when the MOS is not turned on?"

The input of the MOS tube is a capacitor, whose leakage resistance is larger than the reverse resistance of the diode. In the short time after contacting point B, this capacitor is charged and there is current. But as time goes by, the capacitor is close to full and the current tends to zero.

小太阳yy

maychang posted on 2022-3-29 16:26 "If I add a MOS tube at the back to control G to make the circuit form a loop, when the MOS is not turned on, there will be leakage...

When the voltage at point G is 0, will there be leakage current flowing through the MOS? . 10V upper end

maychang

Xiaoyangyy posted on 2022-3-29 17:58 When the voltage at point G is 0, will there be leakage current passing through the MOS? . 10V upper end

When the gate voltage is zero, the MOS tube obviously has leakage current. As for how much current there is, it depends on the MOS tube model. For high-power MOS tubes, the leakage current can reach 10uA or even higher when the drain voltage is 10V. The leakage current of low-power MOS tubes is relatively small. The leakage current of MOS tubes in analog switches is even smaller, generally in the nA level.

小太阳yy

maychang posted on 2022-3-29 14:38 "I am a little confused. " It depends on what instrument you use to measure. Different instruments will have different results. When measuring with an ordinary multimeter, the multimeter must be correct...

I measured the internal resistance of the voltage range of the digital multimeter and it was 11MΩ. Then I measured the internal resistance of the two LEDs and it was 21MΩ. . . .

maychang

小太阳yy posted on 2022-3-29 18:23 I measured the internal resistance of the voltage range of the digital multimeter, which is 11MΩ, and then I measured the internal resistance of the two LEDs, which is 21MΩ. . . .

The forward voltage-current relationship of a light-emitting tube is a strongly nonlinear relationship and should not be measured using resistance.

If you use a pointer multimeter with a high resistance setting to measure, you will find that the "resistance" of the LED is not that high, and it can even make the LED glow slightly.

chunyang

When the circuit is open, the voltages at points AB are the same, but your test depends on your test method and equipment, so it may not be considered an "open circuit" at that time.

小太阳yy

chunyang posted on 2022-3-30 15:49 When the circuit is open, the voltages at points A and B are the same, but your test depends on your test method and equipment, and it may not be considered a "circuit break".

So how do I measure its actual voltage? If I use a multimeter to measure it, it is much less than 3V.

chunyang

小太阳yy posted on 2022-3-31 19:16 So how do I measure its actual voltage? If I use a multimeter to measure it now, it is much less than 3V

If you have to test it, you need to use a voltmeter with extremely high internal resistance. The higher the internal resistance, the more accurate it is. You might as well think about why. But the question is, why do you have to go to such a dead end? Don't you believe in theory?