Help finding this node voltage

[Boltzman Oscillation]

Homework Statement

All the BJTs are the same, prove Vref = 2Vbe

Relevant Equations

KVL, ohms law, BJT properties

I am given that all the BJTs are the same.

First of all, when finding node voltages like Vref should I make the node voltage a voltage source first and then do circuit analysis? If so then I would use KVL but that does not give me Vref = 2Vbe.

I do know that Vref = Vb1 = Vb2. What am I doing wrong here? How would you find Vref?

 

[Joshy]

What is KVL giving you? I apologize if I'm being silly, but at a first glance I would say $2V_{be}$ too.

edit:

I was thinking about your test source. I'm not confident with that approach. Couldn't the test source have a nonzero current affect your results? You can use KVL just as it is.

 

[Boltzman Oscillation]

What is KVL giving you? I apologize if I'm being silly, but at a first glance I would say $2V_{be}$ too.

edit:

I was thinking about your test source. I'm not confident with that approach. Couldn't the test source have a nonzero current affect your results? You can use KVL just as it is.

Yeah I can't use KVL around the entire circuit unless I take into consideration the current source. I am almost sure now that the node voltage was found by adding all of the voltage drops between Vref and ground. That would be the Vbe1 and Vbe3. Since they are the same then they add up to 2Vbe.
Nodes are still confusing to me but hopefully this cleared things up for myself. Thanks for the help.

 

[Joshy]

If you don't mind me asking: What's confusing or makes you feel uneasy about nodes?

It use to confuse me too. I think it's totally okay, but it's good to remove this confusion as early as possible.

 

[Boltzman Oscillation]

If you don't mind me asking: What's confusing or makes you feel uneasy about nodes?

It use to confuse me too. I think it's totally okay, but it's good to remove this confusion as early as possible.

I don't quite understand how to find a voltage of a node. I think maybe I should think as a node voltage as two points instead of one. I should mentally connect the positive end if a voltmeter to the node and the negative end of the voltmeter to what my reference is, in this case it is ground. Thus my node voltage can be found using KVL if I just take the voltmeter connections and the circuit it makes as its own closed circuit. For example, here I would connect the positive end of the voltmeter to Vref, negative end to ground and then I would do KVL going counterclockwise through the voltmeter, the Vbe1 and Vbe3.

 

[Boltzman Oscillation]

I don't quite understand how to find a voltage of a node. I think maybe I should think as a node voltage as two points instead of one. I should mentally connect the positive end if a voltmeter to the node and the negative end of the voltmeter to what my reference is, in this case it is ground. Thus my node voltage can be found using KVL if I just take the voltmeter connections and the circuit it makes as its own closed circuit. For example, here I would connect the positive end of the voltmeter to Vref, negative end to ground and then I would do KVL going counterclockwise through the voltmeter, the Vbe1 and Vbe3.

Alternative I could go around Vbe2 and Vbe4 instead and get the same answer.

 

[Joshy]

I don't think I could have said it better. You're right: The voltage is not a measurement of something at one point; it's a measurement with respect to a reference (ground), and so you need two points. The voltmeter was a perfect example because I'm sure you've swapped the red (+) and black (-) probes before and got the negative answer of what you were expecting, but it didn't break your circuit; you only moved your reference and so you were looking up instead of down.

Something that helped me clear this up was trying ideas on circuits I was more comfortable with. That would be a voltage divider for me. What I did is I "broke" the voltage divider by moving the reference (ground) to where I normally like to see $V_{out}$. What I saw is the voltage across each element did not change. To solve this "broken" voltage divider I thought of $V_{out}$ as the voltage across $R_2$ and you'll get the exact same answer even though the voltage at my preferred node is now $0$. The node above $V_{in}$ with this configuration is also no longer $V_{in}$ because $0$ was no longer right beneath it (however the voltage across $V_{in}$ is still $V_{in}$).