# Voltage across a transistor

See attachment.

## Homework Equations

Voltage division equation.
Ohm's law.

## The Attempt at a Solution

From a) I know that Rx = 86k ohms.
Part b) and c) befuddle me. I think it is relevant to the transistor being on and off.

In b) I would think that because Rx is so high, Vbe would be too low to activate the transistor, and therefore 10V would flow through the 10kohm and the 866k eq. resistor, both losing 10V along the way.

In c) I would think that Rx is 8600. Therefore Vbe = 4.1V. I don't know where to go from there or even if that is correct.

#### Attachments

• transistor.png
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NascentOxygen
Staff Emeritus
From a) I know that Rx = 86k ohms.
Part b) and c) befuddle me. I think it is relevant to the transistor being on and off.

In b) I would think that because Rx is so high, Vbe would be too low to activate the transistor, and therefore 10V would flow through the 10kohm and the 866k eq. resistor, both losing 10V along the way.
Martinet, voltage doesn't flow anywhere. We say current flows (or if you wish to be pedantic, charge flows) but voltage only exists between two points or across elements.

If you are always careful to remember this in your attempts to explain a circuit's working, it will save you from many pitfalls.

Now, if you are picturing current flowing through the 10k resistor, where will it then go to?

Martinet, voltage doesn't flow anywhere. We say current flows (or if you wish to be pedantic, charge flows) but voltage only exists between two points or across elements.

If you are always careful to remember this in your attempts to explain a circuit's working, it will save you from many pitfalls.

Now, if you are picturing current flowing through the 10k resistor, where will it then go to?

That is a good question..
Through whatever device Vout is being measured across and to earth? I am right in saying it can't flow through the transistor aren't I?

NascentOxygen
Staff Emeritus
That is a good question..
Through whatever device Vout is being measured across and to earth?
Vout is shown being measured by an ideal volt measurer thing. It gives a perfect reading yet draws no current.

And what does this mean in relation to my problem? I am completely lost with this one..

NascentOxygen
Staff Emeritus
And what does this mean in relation to my problem?
It means you might need to give it more thought. If you "guess" that current might be flowing in some element, you must justify this by also being prepared to account for where that current comes from, exactly, and where it then goes to.

So does current only flow through the left side of the circuit? As that is the only path along which current can flow in light of infinite resistance from the inactive transistor and the wire that goes to nowhere.

Edit. Potential brainwave. For partis and b do you just work out the voltage as divided by the left voltage divider and thus determine whether or not the transistor is active and therefore it's resistance, being negligible or infinite, and apply those resistances to the voltage divider on the right to get b) 10V and c) 0V?

But this then begs the question as to where the current flows to from the upper terminal. The potential stored within it is 10V and resistance is negligible. Applying I = V / R to this would give a very large current butheir rent implies charge is moving and it can't be. Is it possible for electrons in the terminal to just sit there with 10V of potential stored in them?

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Martinet,

OK, let's get your head screwed on correctly. The problem says that the base current is negligible in comparison with the current existing in the Rx/R2 combination, so we will ignore it. So we have a voltage divider consisting of R2,Rx, and 10 volts. The part A problem tells you what the base voltage is to make the transistor conduct. Can you figure out what the Rx resistance should be to make it happen? Just work on part A first.

Ratch

I already have done part A. That was the simple part it's just solving a voltage div equation for Vout = 0.65V.

b) and c) were the problem, but I think I understand and my possible understanding lines up with the answers given. Are you able to tell me if what I said in my last post is correct?

Martinet,

You said a lot of things in your last post, many of which I think are irrelevant or I don't agree with. Which particular statement did you want me to verify?

Ratch

The second paragraph and the answers to b) and c).

NascentOxygen
Staff Emeritus
So does current only flow through the left side of the circuit? As that is the only path along which current can flow in light of infinite resistance from the inactive transistor and the wire that goes to nowhere.

Edit. Potential brainwave. For partis and b do you just work out the voltage as divided by the left voltage divider and thus determine whether or not the transistor is active and therefore it's resistance, being negligible or infinite, and apply those resistances to the voltage divider on the right to get b) 10V and c) 0V?
That sounds right. (Just be wary of saying a resistance is infinite (or zero); some people take issue with such extremes, their being unrealistic. But I know what you mean.)
But this then begs the question as to where the current flows to from the upper terminal. The potential stored within it is 10V and resistance is negligible. Applying I = V / R to this would give a very large current butheir rent implies charge is moving and it can't be. Is it possible for electrons in the terminal to just sit there with 10V of potential stored in them?
Probably, if I understand you correctly. This "very large current" you theorised, would its path also happen to take it through Rc? https://www.physicsforums.com/images/icons/icon6.gif [Broken]

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Yes. Given that potential is lost, however small an amount, across Rc, current would have to flow through it. This current cannot then flow through the transistor to lose its potential at earth..

(I now see what you're getting at. Okay, The current is actually very small.)

But still, there is current and it has nowhere to go. Am I wrong? But then if I think of the transistor as a very high-value resistor, there *would* still be current flowing through it, equal to this very small amount and it would then reach earth and lose its potential. I think. The question is am I allowed to consider a transistor in this case as a resistor? It does have some current flowing through it, right? Thermal leakage or something?

Also. Apologies for the terrible grammar in my earlier post. I wrote it on my iPad and clearly the autocorrect/lack of haptic feedback from the keyboard did some damage.

NascentOxygen
Staff Emeritus
Yes. Given that potential is lost, however small an amount, across Rc, current would have to flow through it. This current cannot then flow through the transistor to lose its potential at earth..

(I now see what you're getting at. Okay, The current is actually very small.)

But still, there is current and it has nowhere to go. Am I wrong?
With nowhere to go, there can be no current. So with no current through Rc there is no voltage drop across Rc. So with +10V on one end, no voltage drop across Rc, there must be +10V on its other end also.
am I allowed to consider a transistor in this case as a resistor? It does have some current flowing through it, right? Thermal leakage or something?
A tiny current. Negligible in the present context.

So basically, given the tiny current that comes about as a result of the massive resistance of the transistor, and given the tiny voltage drop across Rc that comes with this current, I can essentially look at that path through Rc to Vout as an extension of the cell.

Martinet,

So basically, given the tiny current that comes about as a result of the massive resistance of the transistor, and given the tiny voltage drop across Rc that comes with this current, I can essentially look at that path through Rc to Vout as an extension of the cell.

The problem says that the base current of the transistor is negligible, so don't worry about the resistance of the transisitor and Rc when you calculate Rx.

When LDR is 10Rx, the transistor will turn off regardless of the resistance of the transisitor and Rc.

When LDR is Rx/10, the transistor will saturate regardless of the resistance of the transisitor and Rc.

So, don't get hung up on the resistance of the transisitor and Rc. They don't affect any parts of the problem.

Ratch

Ratch, your addressing questions to which I already know the answer. What I wanted to know was the voltage across the transistor when the LDR as at the resistances given, and why it is so. I now know that because the transistor is off 10V of potential is stored in the wire, and that 10V of potential are expended through the LDR and 6kR, meaning the voltage across the resistor is 10V.

Martinet,