I didn't learn analog electronics well, and I know nothing about transistors as switching tubes.

西里古1992

I didn't learn analog electronics well, and I know nothing about transistors as switching tubes. [Copy link]

 

When the power supply voltage is 9V, the NPN transistor 2N2714 is turned on. How to assign the resistance values of R1, R2, and R3? The equivalent resistance of these three resistors is 170Ω. Is that the condition? I am really confused about transistors. I didn't learn them well? Attached is the 2N2714 datasheet. I don't understand many parameters. Vbeo means the on-state voltage between the base and the emitter is 5V? Which side of this circuit is the base resistor? Does the base current need to be calculated by the amplification factor and the collector current first?

西里古1992

How to choose the magnification

chunyang

I seriously suspect that the original poster even copied the question wrong. The only condition involves only R1, R2 and R3, which has nothing to do with the transistor. Even if the transistor is turned on at 9V, the value of R1-R3 will be very wide, as long as the voltage divider value of R2 and R3 is greater than or equal to 0.7V, and then an equation can be made based on the resistance value of the three in parallel. Due to the lack of sufficient limiting conditions, the solution obtained is a set, but it is still only a junior high school level question.

chunyang

Vebo is the reverse breakdown voltage between base and emitter.

chunyang

Sirigu1992 posted on 2018-9-14 10:33 How to choose the magnification

In engineering, you need to choose according to the lower limit.

elvike

Take out the analog electronic book and flip through it again

maychang

The given conditions are insufficient and cannot be calculated. "R1R2R3 forms an equivalent resistance of 170 ohms", there is no such algorithm.

topwon

In the saturated conduction state, 0.7V divided by R3 gives the current value I3 on R3, (9-0.7V) divided by R2 gives the current value I2 on R2, and I2-I3 is Ib. Ib multiplied by the minimum value of β is the minimum value of Ic. Ic multiplied by R4 is the voltage drop V4 on R4. (9-V4) is the maximum value of Vce. R1 seems to be useless.

maychang

topwon posted on 2018-9-14 13:40 In the saturated conduction state, 0.7V divided by R3 gives the current value I3 on R3, (9-0.7V) divided by R2 gives the current value I2 on R2, and I2-I3 is Ib. Ib ...

Yes, R1 is useless except for heat generation, and will only add burden to the 9V power supply. R1R2R3 is a triangle connection, which can be equivalent to a star connection. But there is absolutely no algorithm for "the equivalent resistance of three resistors is 170Ω".

在学习的路上

The original poster can take a look at the post posted by this gentleman https://en.eeworld.com/bbs/thread-459061-1-1.html

xiaxingxing

chunyang posted on 2018-9-14 10:48 Vebo is the reverse breakdown voltage between the base and the emitter.

Hello, what parameter determines the maximum forward voltage that the base and emitter of a transistor can withstand? Thank you!

maychang

xiaxingxing posted on 2018-9-26 11:07 Hello, what parameter should be used to determine the maximum forward voltage that the base and emitter of a transistor can withstand? Thank you!

What parameter should be used to determine the maximum forward voltage that the base and emitter of a transistor can withstand? There is no "maximum voltage that the base and emitter of a transistor can withstand". The voltage between the base and emitter of a transistor is roughly equivalent to the forward voltage that a diode can withstand, which will generate a forward current, and the forward current and the forward voltage are generally exponentially related. It is the forward current that burns the PN junction between the base and emitter of the transistor, and there will be no breakdown phenomenon when the PN junction is subjected to reverse voltage.

xiaxingxing

maychang published on 2018-9-26 11:17 What parameter should be used to determine the maximum voltage that the base and emitter of a transistor can withstand in the forward direction? There is no such thing as "the base and emitter of a transistor can withstand...

Hello, senior, "It is the forward current that burns the PN junction between the base and emitter of the transistor, and there will be no breakdown phenomenon when the PN junction is subjected to reverse voltage." The last half of the sentence should be changed to "there will be no breakdown phenomenon when the PN junction is subjected to forward voltage." Is that right??? Thank you!

maychang

xiaxingxing posted on 2018-9-26 11:25 Hello senior, "It is the forward current that burns the PN junction between the base and emitter of the transistor, and there will be no breakdown phenomenon when the PN junction is subjected to reverse voltage." ...

The last half sentence should be changed to "there will be no breakdown phenomenon when the PN junction is subjected to forward voltage." Is that right? ? ? Thank you! No. It is the original sentence, no need to change.

chunyang

xiaxingxing posted on 2018-9-26 11:07 Hello, what parameter should be used to determine the maximum forward voltage that the base and emitter of a transistor can withstand? Thank you!

The forward voltage drop of the emitter junction is basically constant, and there is no maximum voltage that can be normally withstood.

chunyang

xiaxingxing posted on 2018-9-26 11:25 Hello senior, "It is the forward current that burns the PN junction between the base and emitter of the transistor, and there will be no breakdown phenomenon when the PN junction is subjected to reverse voltage." ...

Since it is "forward", how can there be "breakdown"?

zhuyebb

Thank you for your selfless sharing and dedication

xiaxingxing

maychang posted on 2018-9-26 11:32 The last half sentence should be changed to "there will be no breakdown phenomenon when the PN junction is subjected to forward voltage." Is that right? ? ? Thank you! No. ...

Hello, "Vebo is the reverse breakdown voltage between the base and the emitter." When this VEBO is greater than a certain value, won't it be a reverse breakdown or even burn out the transistor? ? Moreover, "the voltage between the base and emitter of the transistor is roughly equivalent to the forward voltage of the diode, which will generate a forward current, and the forward current and the forward voltage are generally exponentially related. It is the forward current that burns the PN junction between the base and emitter of the transistor." What you explained in this paragraph is that the forward voltage is clamped across the two ends of the PN junction, and it will form a current and thus turn it on, right? At most, the current burns the PN junction... So my logic is that the forward voltage applied to the PN junction will make it turn on (even burn it under high current conditions), and the reverse voltage will break it down (when VEBO is greater than a certain value). I am a little confused. I would like to ask for your guidance. Thank you!

xiaxingxing

chunyang posted on 2018-9-26 14:48 Since it is "positive", how can it be "breakthrough"?

Maybe I have some confusion and don't understand the concept clearly. Thank you for your guidance!

maychang

xiaxingxing posted on 2018-10-9 21:57 Hello, "Vebo is the reverse breakdown voltage between the base and the emitter." When this VEBO is greater than a certain value, won't it be a reverse breakdown or even burn the transistor...

The so-called "breakdown" refers to the following phenomenon: when the applied voltage is small, the current is extremely small, and the current basically does not increase with the increase of voltage, but when the applied voltage gradually increases to a certain value, the current suddenly increases. When a forward voltage is applied to the PN junction, the current does not suddenly increase, so the forward current when a forward voltage is applied is not called "breakdown". The phenomenon of voltage increase and sudden and rapid increase of current is called "breakdown" only when a reverse voltage is applied to the PN junction. For modern ordinary low-power silicon NPN transistors, the voltage at which the emitter junction breaks down is roughly 5 to 7 V. The minimum voltage at which the manufacturer guarantees that the emitter junction will not break down is Vebo.