Electronic – PNP Transistor Not Working as Expected with Higher Voltage

The transistor may act as a switch or a variable resistor. If no voltage is applied to the base (more precisely: no current flowing into the base) then the switch is open. As base current is applied it gets amplified by the transistor into an N times larger collector current. The "N" is an important transistor parameter, called \$H_{FE}\$, and it defines the current amplification factor. For general purpose transistors this is often around 100.

So if you apply a base voltage (you need a series resistor!) so that there will flow, say, 1 mA, then there will be 100 mA collector current if the circuit allows it. That means that other components may limit that current to a lower value. Let's assume your LED has a 2 V voltage drop, that will be rather constant for that type of LED. Then assuming the transistor is fully conducting (no voltage drop between collector and emitter) you'll have 9 V battery voltage - 2 V LED voltage = 7 V across the resistor. If we choose a resistor value of 350 Ω then, according to Ohm's Law we have a current of 7 V/ 350 Ω = 20 mA through that resistor and therefore also through the LED. (20 mA is a typical current for an indicator type of LED.)
So, while the transistor would like to draw 100 mA, the resistor will always limit that to the lower 20 mA.

You don't say what the signal from the amplifier is. Is that a line level (500 mV) or a speaker output level (3 V for 1 W)? In the first case the voltage will be too low; a transistor's base has to be at 0.7 V minimum before current starts to flow. If you use the speaker output you can use a 1 kΩ resistor in series with the output to limit the base current.

Also place a diode (1N4148) in anti-parallel with the base: cathode to the base, anode to ground. This prevents too large negative voltages across the base, which would destroy the transistor.

The base-emitter junction of a bipolar junction transistor is a diode:

^{simulate this circuit – Schematic created using CircuitLab}

You have another diode in your circuit, the LED. And you obviously understand that you need a current-limiting resistor if you want to connect this diode to a 9V battery, because a diode has an approximately constant voltage drop that is less than your battery voltage.

The problem is the same with the transistor in your circuit. You have the base connected to the positive side of your battery, and the emitter connected to the negative side. You've done this, essentially:

^{simulate this circuit}

What will happen here? A whole lot of current will flow until the battery can supply no more, or something melts. If you are using a 9V battery, it can't supply a whole lot of current, and I bet if you measure the battery voltage in your circuit, it's about 0.65V, unless the battery is dead by now.

So what you need is a current-limiting resistor on the base of the transistor, like this:

^{simulate this circuit}

After you understand that, you might read Why would one drive LEDs with a common emitter?