# How does this Level Shifter with PNP Transistor Circuit Work

Discussion in 'Circuit Simulation & PCB Design' started by Cool Name11, Dec 4, 2016.

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1. ### Cool Name11New Member

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Voltage Source V2 is managed between 0 and 5V,So how does the voltage vary at point 4 or across the 3 ohm resistor

2. ### Les JonesWell-Known Member

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When V2 is zero volts (Or between 0 and 0.5 volts) Q2 will not conduct as the voltage on its base will not be enough to make the base emitter junction conduct. This means that there will be no base current to Q1 so it will not conduct. As is is not conducting the voltage at point 4 will be zero. If V2 is above about 2 volts Q2 will be turned fully on so a current of about 3 mA into the base of Q1 will turn that darilington transistor fully on givving an output voltage of about 11 volts at point 4. The exact transition point will depend on the hfe of the two transistors.

Les.

3. ### Ian RogersSuper ModeratorMost Helpful Member

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I hate these diagrams that show a fixed voltage... Misrepresentation... The V2 should be shown to pulse 5v~0v~5v...

If a noob looks at this they will come to this conclusion..

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5. ### Cool Name11New Member

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When i run the Simulation for the Voltage v2~ 0V, how come there are 12V at the base of Darlington transistor TIP127G, Please Explain

6. ### Ian RogersSuper ModeratorMost Helpful Member

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When I simulate this circuit... I get 8v pulse across R2... with 11.2v / 11.8v on the base of the Darlington However!! I haven't the TIP127G model... only TIP127..

7. ### Les JonesWell-Known Member

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Because there is nothing to pull it down. Even when Q2 is conducting Q1 base will only be about 1.2 to 1.4 volts less than 12 volts. (2 base emitter diode forward volts drop.)

Les.

8. ### specWell-Known MemberMost Helpful Member

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Take a look at the TIP127 data sheet: Q2 (TIP127) is a Darlington transistor pair, but it also includes a resistor connected between the base and emitter of each transistor. These two resistors serve three functions:
(1) Conduct leakage current away which may turn the second TIP127 transistor on.
(2) Vastly speed up the turn-off time of the transistors, especially the second transistor, which is a high power type (the TIP127 is still pretty slow at 50KHz though).
(3) Increase the break-down voltage of the two transistors in the TIP127, especially the second transistor

In your circuit, the base of the TIP127 is connected to the 12V supply line by two resistors (8K Ohm and 220 Ohm) in the TIP127. This is why your simulator, quite correctly, says that, with Q1 turned off, the base of the TIP127 is at 12V.

When Q1 (BC547) collector current increases, the two resistors in the TIP127 will conduct current and Q2 base will drop linearly, until the TIP127 base voltage reaches around -0.5V from 12V supply line, when the first transistor of the TIP127 will conduct and turn the second transistor on. At that point, Q2 base will snap down to around -1.2V from the 12V supply line. After that, the base voltage of Q2 will not drop much further as the collector current of Q1, and hence the current through the TIP127, increase.

spec

(crossed posts Les )

DATASHEETS

Last edited: Dec 4, 2016
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9. ### AnalogKidWell-Known Member

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You are starving Q1; it does not have enough base current. The TIP127 has a minimum current gain of 1000, and needs a base current of 12 mA to get lower saturation voltages. Decrease R1 to 1.0K and report your results.

ak

10. ### Ian RogersSuper ModeratorMost Helpful Member

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I said it works.... I had already dropped my simulation base resistor to 470 ohm, but I tested the OP's circuit and it still worked ( just ) using the 4K7...... Q1 has a good gain so 1mA is enough..