Analysis of the effect of resistance between the be electrodes of PNP transistors

shaorc

As shown in the figure below, if there is no resistor R3, then when the voltage at point A (i.e. Vin) is greater than (VT1+Vbe), the PNP will be turned on. If resistor R3 is added now, the voltage on R3 is equal to the Vbe voltage of the PNP tube. Then there is a calculation formula, Ur3=Vbe=（R3/(R2+R3)）*（Vin-VT1）, which means that adding R3 can affect the conduction of the PNP. [1] Is it necessary to add R3? [2] R3 can also prevent the leakage current of the PNP, that is, when there is no current input or output at the base, a small amount of current may flow through the PNP tube, and let this leakage current be consumed on R3?

cruelfox

Is this for overvoltage shutdown protection? R2 and R3 are voltage dividers, so R3 cannot be omitted. R1 can be omitted. PNP tubes do not have two states: saturation and cutoff. They also have a linear region. The same is true for the MOSFET behind.

shaorc

cruelfox posted on 2019-5-21 16:29 Is this for overvoltage shutdown protection? R2 and R3 are voltage dividers, so R3 cannot be omitted. But R1 can be omitted. PNP tubes are not in two states: saturation/cutoff...

R3 is used for overvoltage shutdown protection. This is clear, but R1 cannot be omitted either. It is the current limiting resistor of the voltage-stabilizing diode VT1.

cruelfox

R3 and R2 connected in series have the same function as R1

maychang

If there is no R3 (open circuit), then R2 is connected in series with the emitter junction of V1, and the current flowing through R2 also flows completely into the base of V1, which then generates the collector current of V1 (enlarging the current amplification factor of V1 by that many times). With R3, when the current flowing through R2 is not large, most of the current flows through R3 (the voltage-current relationship of the emitter junction of V1 is nonlinear, close to an exponential curve relationship). Only when the current flowing through R2 is large enough, will there be enough current flowing into the base of V1, reducing the current in Q1.

shaorc

maychang published on 2019-5-21 16:49 If there is no R3 (open circuit), then R2 is connected in series with the emitter junction of V1, and the current flowing through R2 also flows completely into the base of V1, thereby generating V1 collector current (expanding V1's...

So what you mean is that when the base current on R2 is not large, the PNP is not conducting, because the resistance of R3 is less than the resistance of the emitter junction of V1 at this time, but the current on R2 reaches a certain value, then the PNP will be turned on, reducing the current of Q1 and ultimately achieving the purpose of overvoltage shutdown. Then let R2 be directly connected in series with the emitter junction of V1, that is, don't have R3. Don't you see where the disadvantage is?

shaorc

cruelfox posted on 2019-5-21 16:40 R3 and R2 in series have already played the role of R1

Haha yeah I just discovered it

maychang

shaorc posted on 2019-5-21 17:14 So what you mean is that when the base current on R2 is not large, the PNP is not conducting, because the resistance of R3 is less than the resistance of the emitter junction of V1 at this time, but R2 is powered on...

"Then let R2 be directly connected in series with the emitter junction of V1, that is, without R3, don't you see where the disadvantage is?" Without R3 (open circuit), the collector current flowing through the transistor will change unexpectedly due to temperature changes, etc. With R3, the impact of temperature changes is much smaller. The impact of external interference is also relatively small.

shaorc

maychang published on 2019-5-21 18:08 "So let R2 be directly connected in series with the emitter junction of V1, that is, without R3, don't you see where the disadvantage is?" Without R3 (open circuit), then the flow through...

Well, knowing the role of R3, what are the calculation conditions for the value of R3? I can think of the following conditions First, R3 is connected in parallel with the emitter junction of PNP, and the voltage on R3 must be greater than 0.7V to turn on PNP. Secondly, the voltage of R3 divided by R2 and R3 (Vin-VT1) should be larger or smaller than R2?

maychang

shaorc posted on 2019-5-22 09:11 Well, now that we know the role of R3, what are the calculation conditions for the value of R3? I can think of the following conditions: First, R3 is connected in parallel with the emitter junction of PNP, ...

"First, R3 is connected in parallel with the emitter junction of PNP, and the voltage on R3 must be greater than 0.7V to turn on PNP." The input and output of the bipolar transistor are not in a corresponding relationship of switching nature. Unlike a comparator, the input will flip immediately when it reaches a certain value, but the output current will gradually change with the input.

maychang

shaorc posted on 2019-5-22 09:11 Well, now that we know the role of R3, what are the calculation conditions for the value of R3? I can think of the following conditions: First, R3 is connected in parallel with the emitter junction of PNP, ...

"Secondly, is the voltage R3 value of the voltage divider (Vin-VT1) between R2 and R3 larger or smaller than R2?" Solve the equations, because the relationship between the input current and input voltage of the bipolar transistor is nonlinear, and it is difficult to solve the equations. Usually, we can only draw the input characteristic curve and the resistance curve (the voltage-current relationship of the resistor is a straight line) to find the intersection to get the result.

shaorc

maychang posted on 2019-5-22 09:41 "Secondly, is the voltage R3 of the voltage divider (Vin-VT1) larger or smaller than R2?" Solve the equation, because the input current of the bipolar transistor...

Well, I will do it according to this