Hi everyone,

I’m having trouble understanding the basics of the second transistor schematic (the PNP circuit). My professor said I need to get familiar with how to control a bipolar transistor in an ideal setup.

1. The NPN transistor (circuit 1) is straightforward: the base–emitter voltage is driven by the voltage source Ue, and the base current is set by the series resistor. Easy to visualize and control.

2. The PNP transistor (circuit 2) is where I’m stuck. According to my professor, you can’t control it with a simple voltage source - you need a current source. But why?

• Control still happens between emitter and base, but the resistor between base and emitter is doing what? Is it acting as a pull up? Since the base is always tied to a current source, there shouldn’t be any floating node.
• How do I establish the base-emitter-voltage to control the transistor (besides controlling the base current via the current source)? A current source doesn’t fix the potential at 0.6 V below Vcc, so where does the necessary voltage drop come from?

Can someone help me understand this more clearly? My main goal is to see how we can use the transistor not just as a simple on/off switch, but as a controlled element in its linear region.

Thank you! Thor

 

There is very little difference between NPN and PNP transistor configurations. The only difference is the polarity of the supply voltages.



The circuit and bias is the same. Simply reverse the polarity of the voltage sources, VBB and VCC.

 

Hi everyone,

I’m having trouble understanding the basics of the second transistor schematic (the PNP circuit). My professor said I need to get familiar with how to control a bipolar transistor in an ideal setup.

1. The NPN transistor (circuit 1) is straightforward: the base–emitter voltage is driven by the voltage source Ue, and the base current is set by the series resistor. Easy to visualize and control.

2. The PNP transistor (circuit 2) is where I’m stuck. According to my professor, you can’t control it with a simple voltage source - you need a current source. But why?

• Control still happens between emitter and base, but the resistor between base and emitter is doing what? Is it acting as a pull up? Since the base is always tied to a current source, there shouldn’t be any floating node.
• How do I establish the base-emitter-voltage to control the transistor (besides controlling the base current via the current source)? A current source doesn’t fix the potential at 0.6 V below Vcc, so where does the necessary voltage drop come from?

Can someone help me understand this more clearly? My main goal is to see how we can use the transistor not just as a simple on/off switch, but as a controlled element in its linear region.

Thank you! Thor

View attachment 352919

It appears that your professor is concentrating on the very specific topology that is indicated by your drawings. There are specific points that need to be taken into account here.

The first is that the NPN circuit is driven from a source of some type that is connected to ground, and so is the PNP circuit. They are both driven from a source that is connected to ground.

Now imagine that the voltage Vcc is very high, even 500 volts DC just for example. The NPN circuit would probably have its emitter voltage close to ground, while the PNP circuit has its emitter at 500vdc.

Now how do you drive the PNP stage, as it would need a voltage that is around 500 minus 0.7 volts (and some AC) to drive it properly. One way to get this without going through the details of how to do it would be to use a current source. The current source will automatically provide the right voltage while also providing the right current.
Also, it does actually fix the voltage at 0.6 to 0.7 volts below Vcc no matter what that voltage is. I am not sure if you understand current sources that well yet.

I do not think these examples are complete enough though. They do not show the real reason for doing it this way in either circuit. As you can see from the other reply, if you power the PNP circuit with a voltage source that is common to Vcc instead of ground, you essentially have the same circuit in both cases. You might ask your professor about that.

If there was more detail in these drawings and explanations, we might be able to make this even clearer. One of the things that is really lacking is what voltage level is the emitter of the NPN stage subject to as we are assuming it is close to ground potential for now.

 

Hi everyone,

I’m having trouble understanding the basics of the second transistor schematic (the PNP circuit). My professor said I need to get familiar with how to control a bipolar transistor in an ideal setup.

1. The NPN transistor (circuit 1) is straightforward: the base–emitter voltage is driven by the voltage source Ue, and the base current is set by the series resistor. Easy to visualize and control.

2. The PNP transistor (circuit 2) is where I’m stuck. According to my professor, you can’t control it with a simple voltage source - you need a current source. But why?

• Control still happens between emitter and base, but the resistor between base and emitter is doing what? Is it acting as a pull up? Since the base is always tied to a current source, there shouldn’t be any floating node.
• How do I establish the base-emitter-voltage to control the transistor (besides controlling the base current via the current source)? A current source doesn’t fix the potential at 0.6 V below Vcc, so where does the necessary voltage drop come from?

Can someone help me understand this more clearly? My main goal is to see how we can use the transistor not just as a simple on/off switch, but as a controlled element in its linear region.

Thank you! Thor

View attachment 352919

It's a bit difficult to tell what your professor is really trying to get across. The two circuits shown are simply different -- and the difference has little to do with a some fundamental difference between NPN and PNP transistors. It really has more to do with noise rejection.

You don't show what the emitter of the NPN is connected to. Since the collector is hard-tied to Vcc, the circuit would have little point if the emitter was tied to ground. So the claim that the base current is set by Rb is misleading, because it is also influenced by what ever is connected to the emitter. Depending on the specifics, the base resistor might be the dominant factor, or it might be so insignificant that it can literally be removed and the source connected directly to the base.

In the PNP circuit, the resistor doesn't serve much of a functional purpose. Enough of the source current, Iq, has to flow through it to get the voltage drop across the resistor to be the necessary base-emitter voltage. Once that is achieved, all of the remaining current is pulled from the base of the transistor. When simulating the circuit, some simulators will have problems without the resistor in place because an ideal current source has infinite output resistance and the simulation may not converge to a stable initial operating point.

You can certainly use a voltage source to bias a PNP transistor. To do it in a way that is truly the analog of the first circuit, however, you need to have the external supply connected between the base resistor and Vcc, instead of between the base resistor and ground.

On paper, it doesn't matter. But in practice, it definitely can. In the NPN case, imagine that the value of Vcc varies -- maybe it's ripple or perhaps Vcc is load-dependent (as all voltage supplies are, to one degree or another). As Vcc changes, the situation at the base remains unchanged, and so the change in Vcc has very little impact on the transistor operating point (how much or how little depends on what is connected to the emitter).

But if a PNP is biased with the external voltage supply connected to ground, instead of Vcc, then changes in Vcc translate very heavily into changes in the transistor's operating point.

The measure of this is known as "power supply rejection ratio".