Regarding the issue of using NPN transistors for high voltage conversion output.

tongshaoqiang

Regarding the issue of using NPN transistors for high voltage conversion output. [Copy link]

 

As shown in the figure below, when looking at the information of a sensor conditioning chip, it is recommended to use JFET or NPN transistors to build a power supply voltage conversion circuit in the typical circuit. The purpose is to obtain a stable low voltage from a wide range of voltage input.

The circuit is quite ingenious and I didn't understand it at the moment. Please help me understand it. Thank you!

Test prototype, input voltage at point C is 20v, control voltage at point B is 5.6v, output voltage at point E is 5v, if control voltage is set to 3.9v, output voltage at point E is 3.3v. I don't understand how BCX5610 is used if it is not used as a switch. The resistor in the figure is 51K

chunyang

The transistor is an emitter follower here, so as long as the base potential is stable, the emitter output voltage is the base potential minus a PN junction voltage drop.

yet

chunyang posted on 2021-2-22 23:04 The transistor is an emitter follower here, so as long as the base potential is stable, the emitter output voltage is the base potential minus a PN junction voltage drop.

The answer makes people laugh

The question is about the case where 5610 is not used as a switch. You are talking about PN junction. If it is yes, then give a straight answer. If it is no, don't mislead others.

tongshaoqiang

yet Published on 2021-2-23 08:17 The answer is hilarious. The question is about the case where 5610 is not used as a switch. You are talking about PN knots. If it is yes, then give a straight answer. If it is no, don't mislead others...

I think what Chunyang said makes sense. 5610 is not used as a switch here.

tongshaoqiang

chunyang published on 2021-2-22 23:04 The transistor here is an emitter follower, so as long as the base potential is stable, the emitter output voltage is the base potential minus a PN junction voltage drop.

Thanks to teacher chunyang for the guidance!

maychang

This is a linear voltage regulator circuit.

There is a rectangular box in the figure, and a triangle is inside the box, and the left side of the triangle is marked with "+" and "-". Obviously, the triangle is an op amp, which has two input terminals and one output terminal. When the op amp works linearly, the two input terminals are "virtually shorted", and the two input terminals must be at the same potential. In other words, the voltage at point E is equal to VREF after being divided by two resistors.

maychang

The linear voltage regulator circuit is not just the field effect transistor BSS169 or the bipolar transistor BCX5610, but more importantly the operational amplifier (because the voltage gain of the operational amplifier is much larger than that of the transistor).

Precisely because the voltage gain of the op amp is very large, the circuit can achieve voltage regulation regardless of whether it uses field-effect transistors or bipolar transistors, and the output voltage is equal to VREF divided by the voltage-dividing ratio of the two resistors (only in this way can the potentials of the two input terminals of the op amp be equal).

okhxyyo

tongshaoqiang posted on 2021-2-23 08:43 I think what chunyang said makes sense. 5610 is not used as a switch here.

Teacher Chunyang answered the function of 5610 here, I think it is OK. The previous netizens have different opinions, you just need to understand Chunyang's answer here~

chunyang

yet Published on 2021-2-23 08:17 The answer is hilarious. The question is about the case where 5610 is not used as a switch. You are talking about PN knots. If it is yes, then give a straight answer. If it is no, don't mislead others...

Even if the answer is wrong, laughing can only reflect your low quality. Why don't you tell me what is right? What does it mean to not mislead others? People who can only troll but not help others are just complete garbage!

tongshaoqiang

chunyang posted on 2021-2-23 11:55 Even if the answer is wrong, laughing can only reflect your low quality. Why don't you tell me what is right? What does it mean to not mislead others? You can only be a troll but not help...

Chunyang, I would like to ask another question. The resistor between the collector and base of the transistor in the figure is the resistor that sets the input impedance of the transistor. This resistor is set larger to ensure that the transistor works in the emitter-follower working state, right?

tongshaoqiang

chunyang posted on 2021-2-23 11:55 Even if the answer is wrong, laughing can only reflect your low quality. Why don't you tell me what is right? What does it mean to not mislead others? You can only be a troll but not help...

Chunyang, I would like to ask another question. The resistor between the collector and base of the transistor in the figure is the resistor that sets the input impedance of the transistor. This resistor is set larger to ensure that the transistor works in the emitter-follower working state, right?

tongshaoqiang

maychang published on 2021-2-23 09:31 The linear voltage regulator circuit is not only the field effect tube BSS169 or the bipolar transistor BCX5610, but more importantly the op amp (because the voltage gain of the op amp is higher than that of the triode...

Thanks to teacher maychang for his guidance.

The internal op amp and the external transistor should jointly build this voltage stabilization circuit, and the transistor works in the emitter-follower state. My understanding is that point E is the feedback input of the internal op amp, and VREF is the internal configuration voltage. Is the voltage at point B the input voltage divider (the resistor voltage divider between the collector and base of the transistor) or the output voltage of the op amp? I think it is the input voltage divided by the resistor, but after adjusting the voltage divider resistor through the experiment, the voltage at point B is unchanged (5.6v), and it should still be the output voltage of the op amp (equal to VREF). Is this understanding correct?

maychang

tongshaoqiang posted on 2021-2-24 11:48 Thanks to teacher maychang for his advice. It should be the internal op amp and the external transistor that jointly build this voltage regulator circuit. The transistor works in the emitter-follower state...

Let’s first talk about the resistance between the collector and base of the transistor.

I haven't found the NSC2860 manual, so I don't know if this resistor is necessary. If the NSC2860 op amp output is weak pull-up, then the op amp may not be able to provide the transistor base current (flowing out of the op amp and into the transistor base), so this resistor is necessary. If the op amp has a push-pull output and can provide the transistor base current, then this resistor is not necessary, and the entire circuit will work normally after removing it (open circuit).

maychang

tongshaoqiang posted on 2021-2-24 11:48 Thanks to teacher maychang for his advice. It should be the internal op amp and the external transistor that jointly build this voltage regulator circuit. The transistor works in the emitter-follower state...

Let’s talk about the voltage at point B (op amp output).

Simply looking at the two input terminals of the op amp, the voltage at point B is uncertain. Because the op amp works in an open loop, the voltage amplification factor is very large. However, after passing through the bipolar transistor or MOS tube, the voltage at point E must be equal to VREF after being divided by two resistors inside the NSC2860. At this time, the voltage at point B is equal to the voltage required to make such a large current (this current is determined by the load) flow through the bipolar transistor or MOS tube.

maychang

tongshaoqiang posted on 2021-2-24 11:48 Thanks to teacher maychang for his advice. It should be the internal op amp and the external transistor that jointly build this voltage regulator circuit. The transistor works in the emitter-follower state...

The internal operational amplifier and triode or MOS tube of NSC2860 form a large loop negative feedback. This negative feedback makes the voltage at point E stable, but the voltage at point B is not stable.

Because the voltage at point B is uncertain, the output can work normally whether it is a bipolar transistor or a MOS tube. Most MOS tubes start to conduct when the gate is about 3V higher than the source (the current in the MOS tube is very small, usually specified as 0.1mA), and the current will increase only when the voltage between the gate and the source increases. For the MOS tube to be fully turned on, the gate voltage must be 5 to 7V higher than the source or even more. When the bipolar transistor is turned on, the base is about 0.6V higher than the emitter. Even if the emitter current increases to the maximum current allowed by this type of tube, the base is only about 1V higher than the emitter.

maychang

tongshaoqiang posted on 2021-2-24 11:48 Thanks to teacher maychang for his advice. It should be the internal op amp and the external transistor that jointly build this voltage regulator circuit. The transistor works in the emitter-follower state...

Due to the large loop negative feedback, whether using bipolar transistors or MOS tubes, although the voltage at point B and the voltage at point E of the two tubes are quite different when they are working normally, both tubes can work normally.

chunyang

tongshaoqiang published on 2021-2-24 11:37 chunyang I would like to ask again, the resistor between the collector and base of the transistor in the figure is the resistor that sets the input impedance of the transistor. This resistor is set to a large value to maintain...

This resistor is the bias resistor of the base. In addition to what maychang said above, there is another possibility - at the power-on transient, when the built-in op amp has not yet worked properly, the bias resistor can turn on the BCX5610, which helps the output voltage to build up quickly. Using an oscilloscope, when there is a bias resistor, the rising edge of the output voltage should be steeper than when there is no bias or the bias is weaker. Whether this is the case, you need to read the device manual to determine.

chunyang

tongshaoqiang posted on 2021-2-24 11:48 Thanks to teacher maychang for his advice. It should be the internal op amp and the external transistor that jointly build this voltage regulator circuit. The transistor works in the emitter-follower state...

In steady state, the potential at point B is obviously determined by the output voltage of the op amp. At this time, the op amp output can be regarded as equivalent to a voltage source (of course, there are conditional restrictions). Changing the resistance value of the bias resistor within a certain range will naturally not affect the potential at point B.

Gen_X

The 5610 is not really a switch here, it can be considered as a typical series voltage adjustment circuit, which is completed together with the op amp