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schematic

simulate this circuit – Schematic created using CircuitLab

So the BJT's B-E diode is turned on with 0.7V, and beta in active mode is 100 and beta in saturation mode is 10. I'm trying to find which BJT is on and find Ib, Ic, and Vout.

I'm under the impression that the top pnp is off since if it is on, the Vb is 1.8V and that means current flows from the 2.5V source to the Vb (high to low) but I think the base current in a pnp flows out of the base. I'm not too sure though.

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    \$\begingroup\$ Suppose R1 and V2 were taken away. What do you think, then? Would current come out of the PNP base and then go into the NPN base? \$\endgroup\$ – jonk Mar 29 at 6:07
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    \$\begingroup\$ Are you sure that circuit is correct? You have 2 Vbe diodes in series dropping a total of 2.5V. \$\endgroup\$ – Math Keeps Me Busy Mar 29 at 6:11
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    \$\begingroup\$ Aren't the npn and pnp swapped around? \$\endgroup\$ – fifi_22 Mar 29 at 6:37
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    \$\begingroup\$ "beta in saturation mode is 10"? What do you mean by that; beta has no meaning in saturation mode? \$\endgroup\$ – Hearth Mar 29 at 15:39
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If you simulate your circuit, you will get this result:

enter image description here

As you can see, the current through the bases of the transistors is almost 40mA and the current through the 1k resistor is less than 1mA. The is because the base emitter paths of the two transistors are appearing across V1. Because of this, the full 2.5V appears across the base-emitter paths of the two transistors and results in a very large current flowing through them. Now, because there is current flowing through the bases of both transistors, both transistors are actually on and are shorting V1, which is why you are only getting a sub 1mA current flow through the 1k resistor.

If you take a look at the current flowing through the collectors of the BJTs, you will find that it is nearly 700mA!

enter image description here

So your configuration of the transistors shorts the power supply (V1) irrespective of what the control signal is doing (V2).

But if you swap the positions of the two BJTs, you will get a buffer. Also, such a configuration can be used to drive MOSFETs.

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    \$\begingroup\$ The simulation is faulty. If the transistors survived, and each transistor has a Vbe of 1.25V, there would be more than 40mA flowing through the bases, and more than 700mA through the collectors. But the transistors will not survive, or at least one of them won't. \$\endgroup\$ – Math Keeps Me Busy Mar 29 at 13:51
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The two transistors as originally wired will short out the power supply via their respective base-emitter junctions so you should arrange for this type of modification: -

enter image description here

This circuit can now be driven correctly and, with V2 at 2.5 volts, Q2 is off and Q1 is activated.

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