I need help with circuit knowledge

林聲不知处

I need help with circuit knowledge [Copy link]

 

 

I recently watched the company's engineers review the schematic diagram and learned that a resistor should be added to the base of the PNP transistor. It can work, but Ube12v is too high. Should I add a resistor to divide the voltage? The problem is that the IB current controls the transistor, so when used as a switch, the base current should enter from B. Where does this IB come from? If this resistor is not added, how much base current will be?

林聲不知处

I need help from a senior, I feel so useless, the company is so busy and I am a bit afraid to ask such basic questions

Nubility

Without this resistor, it should burn out. 12V is directly added to the PN junction.

maychang

[The question is that the IB current controls the transistor, so when used as a switch, the base current should enter from B. Where does this IB come from? ]

For PNP transistors, the base current flows out from the base.

This base current flows from the positive end of the DC power supply through R33, the emitter junction of the PNP transistor (note the direction of the arrow of the emitter symbol of the PNP transistor), Q10, and flows into the negative end of the DC power supply.

maychang

[What would be the base current if this resistor is not added]

The DC power supply voltage minus the PNP transistor emitter junction voltage drop (about 0.7V) is divided by R33 plus the MOS tube on-resistance.

maychang

[It is said that it can work but Ube12v is too high, should I add a resistor to divide the voltage? ]

This is your guess, right?

林聲不知处

maychang posted on 2024-9-14 11:51 [It is said that it can work but Ube12v is too high, add a resistor to divide the voltage? ] This is your guess, right?

Emm, the R&D engineer said: "Add a resistor here at the base", but I don't know where to add it at the base and how much ohms it should be.

林聲不知处

maychang posted on 2024-9-14 11:48 [The problem is that the IB current controls the transistor, so when used as a switch, the direction of the base current should enter from B. Where does this IB come from? ] For PNP transistors...

Sorry, I described the base current the wrong way. I don't understand it because it flows out from the B pole. Does it mean that the base current of the PNP tube is the current generated by 12V from R33 to the PN junction EC? [DC power supply voltage minus the emitter junction voltage drop of the PNP transistor (about 0.7V), divided by R33 plus the on-resistance of the MOS tube] Isn't this the current value that passes through the switch tube after it is turned on? For N tubes, the base current is basically controlled by a DC power supply and a current limiting resistor at the base. This is easy to understand, but it is hard to understand why the P tube flows out.

maychang

林聲不知处 posted on 2024-9-14 13:28 Sorry, I described the base current the wrong way. It is precisely because it flows out from the B pole that I don’t understand. It means that the base current of the PNP tube is generated by 12V from R33 to...

[For N-tubes, the base current is basically controlled by adding a DC power supply and a current-limiting resistor to the base. This is easy to understand, but the outflow of the P-tube is hard to understand.]

What's so confusing about this? PNP and NPN tubes are opposite in both voltage and current directions. Of course, PNP tubes also control the base current by adding a DC power supply and a current limiting resistor to the base. In the picture you posted, the negative end of the DC power supply passes through the MOS tube and the red R is the current limiting resistor. The MOS tube controls the base current of the PNP tube. It's just the opposite direction of the NPN tube.

Dxiaobai

R33 is the current limiting resistor

se7ens

A resistor must be added to the base level to limit the base current of the PNP tube.

In addition, the placement of R23 is incorrect. It should be placed between BE of PNP.

led2015

If no base resistor is added, the base current will be directly determined by the power supply voltage and the forward conduction voltage (Ube) of the base-emitter junction of the PNP transistor. Since Ube is usually small (for silicon PNP transistors, Ube is about -0.6V to -0.7V), if the power supply voltage is high, the base current may be very large and even exceed the rated current of the transistor, causing damage to the transistor if no resistor is added to limit the base current.

addfly

I am so lucky that I can learn drawing review from Z even though I have little knowledge of circuits.