I have some work to do in which I must implement a circuit in a circuit simulator and then again on a breadboard. There are a few concepts I need clarified. This is the circuit:


If I understand correctly, $V_\text{in}$ and $V_\text{out}$ are used to represent where the voltage enters and leaves to simplify circuits.

How is the circuit closed past the $V_\text{in}$ and $V_\text{out}$ and the $V_\text{cc}$? I need to be able to implement this on both a simulation and in real hardware. How would I do this?

Also, from what I know, $V_\text{cc}$ is the voltage supply to the collector, but, in general, how can anything supply the collector with voltage, when current specifically flows out from the collector (at least with NPN)?

From what I can tell, in this case the current would flow from $V_\text{cc}$ to $V_\text{out}$, but that doesn't make sense since it is called $V_\text{cc}$.

Please explain in simple terms. I've seen some answers but none that make it clear how to implement $V_\text{in}$ and $V_\text{out}$ and how they become a closed circuit from what is not shown on the diagram.

  • $\begingroup$ I would like to help with this post, but it is difficult because the post asks several questions. It's much easier to answer when the post asks one single question. Can you please edit this post to ask a single focused question, and then make additional separate posts with your other questions? $\endgroup$ – DanielSank Aug 2 '17 at 22:01
  • $\begingroup$ I gather from your questions that you are new to the tradition of showing signals (voltages) at points in a circuit with the understanding that the the other reference point is the common or ground connection, even though that other connection is not explicitly shown. There isn't really any need to consider currents flowing on any paths other than those actually shown in the circuit to understand how it works. I recommend Horowitz & Hill's The Art of Electronics to learn more. $\endgroup$ – user55515 Aug 2 '17 at 22:30
  • $\begingroup$ @user55515 yeah I wasn't familiar with that, thanks. I needed to understand what happens beyond them because I needed to implement the circuit irl on a breadboard. From what I've figured out, I'm fairly sure I just need to measure them. The specific questions are here (Questions), if you can add anything that would help I'd appreciate it. I'm not going to be able to answer anything beyond this for now. $\endgroup$ – user3444061 Aug 3 '17 at 0:14

How is the circuit closed past the Vin and Vout and the Vcc?

Vin: You'll connect some kind of signal source between ground and Vin to provide an input signal. (for example, a microphone)

Vout: You'll connect some kind of load from Vout to ground in order to make use of the amplified signal.

Vcc: You'll connect a DC voltage source (for example, a benchtop power supply or a 9 V battery) from ground to Vcc in order to provide power to the circuit.

Also, from what I know, Vcc is the voltage supply to the collector, but, in general, how can anything supply the collector with voltage, when current specifically flows out from the collector (at least with NPN)?

You are confused. Conventional current flows in to the collector of an NPN transistor when it is in forward active mode.

If you are confused about the difference between electron current and conventional current, there have been multiple questions here about it.

If you're confused about the operation of NPN transistors, there are numerous online and offline textbooks that explain it.

From what I can tell, in this case the current would flow from Vcc to Vout, but that doesn't make sense since it is called Vcc.

Current can flow in multiple paths. In this case current can flow from the supply connected to Vcc through both the collector of the BJT and into the load connected to Vout.

Depending on the nature of the load, the current into the load might always be positive, or it might sometimes be positive and sometimes negative (flowing back into the BJT collector when the BJT is operating to pull the output voltage below the bias point).

The name of the node in a circuit is just a name. It's not a guarantee of what voltage might be found there or what direction current will flow in or out of the node (In fact, by Kirchoff's current rule, the total current into a node is always equal to the total current out of it).

"Vcc" is a common name used for the most positively biased node in a circuit containing BJT's, but you should analyze the circuit based on the components in it, not based on the names given to its nodes.

  • $\begingroup$ Conventional current flows in to the collector of an NPN transistor when it is in forward active mode. I understand this much. I know the difference between electron and conventional current and I'm pretty sure I understand how NPN transistors work. From searching online, I've found out that the Vcc is the voltage that is supplied to the collector.What I have a problem with is how the voltage is being supplied to the collector from the above circuit when current doesn't flow in the direction of the collector. $\endgroup$ – user3444061 Aug 2 '17 at 22:39
  • $\begingroup$ 1. Current does flow from the Vcc node, through Rc and the LED, into the collector of the BJT. Why do you think it doesn't? 2. In this circuit, Vout is the voltage that's applied to the collector. Vcc is the voltage applied to the node labeled "Vcc". You need to analyze the circuit in front of you, not worry about what other people named nodes in other circuits that aren't what you're looking at. $\endgroup$ – The Photon Aug 2 '17 at 22:42
  • $\begingroup$ For that matter, the circuit you linked doesn't show Vcc connected directly to the BJT collector either. It shows it connected to a resistor that's connected to the collector. The only difference in the circuit in your question is that there's an LED also. But the LED doesn't prevent current flowing "downward" through the resistor and into the BJT collector. $\endgroup$ – The Photon Aug 2 '17 at 22:45
  • $\begingroup$ So you're saying current would flow into the collector? From what I've learnt, the direction of current is from the emitter to the collector.. at least from the diagram we were shown in class. i.imgur.com/g7USPjQ.png $\endgroup$ – user3444061 Aug 2 '17 at 22:46
  • 1
    $\begingroup$ You need to supply an input signal to Vin if you want the amplifier to have something to amplify. If you're just building this to learn how amplifiers work, you're correct the load might not be anything but a measurement device. A voltmeter will work to see the DC and very low frequency behavior. If you want to amplify AC signals (like audio, for example) you'll probably want an oscilloscope instead. $\endgroup$ – The Photon Aug 2 '17 at 23:03

So lets look at Vout first. Suppose the transistor is cut off and not conducting at all. Then Vout will be Vcc minus the whatever the forward voltage on the diode is. (You don't specify it.) Now lets put the transistor into strong conductance so it essentially has no resistance. Vout will now be basically 0. (This is measuring it relative to ground.)

Will any current flow out of Vout? We don't really want it to. We can assume that the measuring device is very high resistance, so it doesn't change the circuit when it measures.

So what is Vin? That's there to keep the transistor biased into the conductance regime. Assume the function generator varies between +1 and -1 volts relative to ground. If we make Vin be 1V, then the voltage on the base varies between +2 and 0. So it will vary between cutoff and conductance. Without known a lot more about the circuit, we can't figure out what voltage would make the transistor saturate and not conduct more even when the input voltage goes up.

This isn't a complete answer, but it is at least something to chew on.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.