Basic idea

The OP's circuit is a single-supply AC complementary emitter follower where the input voltage is shifted upwards and the output voltage is shifted downwards by half the supply voltage. Both transistors smoothly change their conductivity in opposite directions, with one of them being non-conducting at any given moment.

CircuitLab experiments

The simulations below can help find the answer.

DC simulation

To get a good intuitive idea of circuit operation, we can apply the following three tricks:

• By replacing capacitors with equivalent DC voltage sources, we can simplify the analysis of this AC circuit and treat it as a DC circuit. This approach enables us to readily observe voltages and currents by the help of CircuitLab DC Live Simulation tool (hovering the mouse over the circuit elements).

• Then we can replace the AC input voltage source by three DC voltage sources for each case - 0, 1 and -1 V.

• Finally, we can replace the 1 kΩ load by a voltmeter with the same resistance.

Vin = 0 V: The "capacitor" C1 "lifts" the zero input voltage to 6.7 V, and the "capacitor" C2 to 5.3 V. Thus each of the two base-emitter junctions is biased with 0.7 V (1.4 V between the bases in total). The voltage at the midpoint between the diodes and the output common emitter voltage is equal to 6 V. The capacitor C3 "moves" this voltage down to 0 V across the load.

^{simulate this circuit – Schematic created using CircuitLab}

Equivalent circuit: Another trick to imagine what the two transistors Q1 and Q2 do is to think of them as a voltage divider of two resistors R1 and R2. The next trick is to determine their resistance by reading the voltage and current across the transistors in the schematic above and calculating the resistance. Note that this is an instantaneous resistance because the collector-emitter sections of the transistors represent non-linear resistances.

^{simulate this circuit}

Conclusion: At Vin = 0 V, both transistors are semiconducting with equal resistance. So the "divider's gain" is 0.5, and the emitter voltage is Vcc/2 = 6 V.

Vin = 1 V: If the input voltage increases, this change is transmitted through the capacitors C1 and C2 to the bases of the two transistors.

^{simulate this circuit}

It forces Q1 to reduce its "resistance" (R1) and Q2 to increase it (R2). The divider's "gain" and accordingly the emitter voltage, increases.

^{simulate this circuit}

Vin = -1 V: Now the voltages change in the opposite direction, downward.

^{simulate this circuit}

R1 increases, R2 decreases

^{simulate this circuit}

DC Sweep Simulation: Let's now sweep (change linearly) Vin to see the graphs of voltages...

... and currents.

AC simulation

Two "shifting" capacitors: Finally, let's replace the "shifting" voltage sources with capacitors and the DC input voltage source with an AC voltage source, and explore the real AC circuit.

^{simulate this circuit}

One "shifting" capacitor: It is interesting that the circuit can be implemented by only one "shifting" capacitor connected between the input source and the middle point between diodes.

^{simulate this circuit}

It can be even connected to some of the diode ends - to D2's cathode...

^{simulate this circuit}

... or to D1's anode.

^{simulate this circuit}

I suppose that the 2-capacitor circuit can provide higher base currents that is important in the case of heavy load.

Answering the OP's questions

The time has come to give concrete answers to the OP's questions.

1. Every book and video out there... says when one transistor is on the other will be off...

The research above shows that this is not always true.

2. I thought the voltage difference between bases and emitter is 1.4 for both and 0.7 for single transistor and kept constant, so how?

Your reasoning is correct. The two transistors are almost always conducting but to different degrees, and are complementary. Figuratively speaking, their "resistance" cross-fades from one to the other. The higher the load current, the greater this difference, and at some point one of the transistors is almost off. The other transistor never enters saturation because its collector-emitter voltage cannot approach zero.

3. So how is the lower transistor(PNP) switched off? For a transistor to be off, its voltage difference between base and emitter should be less than 0.7 volt and this is not happening because the base of the PNP is -0.7 to its emitter.

In order to definitively clarify this conceptual phenomenon, I propose modeling it with a simpler electrical circuit where the base-emitter junctions of the transistors are substituted with voltmeters having a 1 kΩ internal resistance, as well as the load. Also, to simplify and tidy up the circuit, let's use a bipolar power supply.

Vin = 0 V, RL=100 kΩ: Let's first apply zero input voltage or simply remove the input voltage source, and make the load high resistance (100 kΩ).

^{simulate this circuit}

What we and the OP see does not surprise us. The total 1.4 V bias voltage Vbb across the circuit of two diodes in series is distributed equally (0.7 V each) on the two base-emitter junctions (the voltmeters Vbe1=1k and Vbe2=1k). This is because the common emitter voltage (across the load RL=100k) is 0 V.

Vin = 1 V, RL=100 kΩ: Now let's increase with 1 V the input voltage. The total 1.4 V voltage across the diodes and base-emitter junctions does not change, only "moves" up. Since the load has high resistance, the common emitter point follows Vin, and Vbe1 = Vbe2 = 0.7 V as above.

^{simulate this circuit}

Vin = 1 V, RL=1 kΩ: Now let's significantly decrease the load resistance (RL = 1 kΩ). The total 1.4 V voltage does not change but now it is distributed unequally across the two base-emitter junctions. As you can see, Vbe2 is only 367 mV.

^{simulate this circuit}

Vin = 1 V, RL=100 Ω: If we further decrease the load resistance, Vbe2 will even become negative, and the Q2's base-emitter junction will be backward biased.

^{simulate this circuit}

Nonlinear resistors (PN junctions): As per the OP's comment below “is it ok to have voltage drop greater than diode voltage drop”, let’s replace the linear resistors Rbe1 and Rbe2 with nonlinear ones (diodes acting as base-emitter junctions). Since they decrease their resistance when the current increases, the voltage drop can’t be “greater than diode voltage drop”.

^{simulate this circuit}

With the exception of your claim (from the diagram) that the junction of the two diodes is 0V (it's not, it is +6V), I think your reasoning is good.

In fact, what you have observed is a fundamental flaw in this design, that both transistors are indeed switched on, at least partially. That will naturally lead to a (fairly) short circuit between the two supply rails via the transistors. That's a big problem with this design, and is evident in a simulation:

^{simulate this circuit – Schematic created using CircuitLab}

I've reduced R1 and R2 to exaggerate the problem, but the point is that there's a large current flowing down through Q1 and Q2 even without any load connected. The usual solution is to add "ballast " (emitter degeneration) in the form of resistances between the emitters:

^{simulate this circuit}

Since each of these resistors R3 and R4 will develop a small voltage across them, due to quiescent current, they have the effect of reducing the potential difference \$V_{BE}\$ between the base and emitter of their respective transistor. Consequently, the transistors tend to switch off somewhat, reducing quiescent current.

Obviously this technique increases the output resistance of the system, which is not ideal, but as long as the load impedance is significantly larger in comparison, it's a small price to pay for reduced quiescent current.

Often we employ negative feedback in a closed loop to overcome this shortcoming. For illustration I'll replace the diodes with a fixed 1.4V source, because that's what R1, R2, D1 and D2 are there to do:

^{simulate this circuit}

I've added a heavy load, R5, and this is a plot of \$V_{IN}\$ (blue) and \$V_{OUT1}\$ (orange, left circuit open loop):

As you can see, the output is an attenuated version of the input, so gain is a little less than unity due to the emitter resistors.

On the right though, by closing the loop, the op-amp is able to compensate for the undesired attenuation. I won't plot it here, because the input and output waveforms are identical, but you can simulate it for yourself.

TS --"H is the lower transistor switched off? For a transistor to be off, its voltage difference between base and emitter should be less than 0.7 volt and this is not happening because the base of the PNP is -0.7 to its emitter."

It's only around -0.7V when the transistor is on (looks rather like a forward biased diode). When it's off, the base-emitter voltage can be anything down to 0V or positive voltage (within its reverse bias limits).

The thing to keep in mind is that transistors are current controlled devices. They turn on when current flows across the base-emitter junction.

But there's a catch: the base-emitter junction is a diode with forward voltage drop (Vbe) of 0.7V. For a signal to turn on the transistor, it needs to be shifted by one Vbe voltage.

That's what the two diodes do. They introduce a bias to the AC coupled input to overcome the transistor Vbe voltages. The two diodes together offset the bases by 2*Vbe, or 1.4V.

The input signal is AC coupled to the bases. It can be coupled to just one base, or between the two diodes with practically equal effect.

Biased this way, we now have a set of voltage followers that conduct alternately, depending on the signal polarity:

• Idle midpoint: circuit self-biases to half voltage. Neither transistor is "on", except for a bit of idle current
• Signal positive: upper transistor conducts, output follows it up
• Signal negative: lower transistor conducts, output follows it down

Here's a sim to try (simulate it here):

Pay particular attention to what the base currents are doing. That's the 'pushing' and 'pulling' action, activating each follower in turn as each base-emitter sees forward bias.