Understanding loops and voltages in transistor circuits.

Update Svc

Joined Jan 10, 2018
10
We have been given the question below to identify the missing variables.

I am a bit new to transistors, so far in class we covered a few simplified circuits,
I am finding it difficult to identify the loops so I can use KVL to identify currents.
Plus so far I have never done anything that has a "voltage terminals" instead of my usual comfort zone of Cells.
I would like an explanation of the loops: how they are connected , then I think I will be good.

I had the above circuit in mind is it right?
I got the base current IB as 0.000036994A but my intiution is wrong
I used an KVL equation : 20V-IB*47000-4700(1IB+100IB)-0.7=0

WBahn

Joined Mar 31, 2012
26,398
Your circuit is almost correct. In fact, it IS correct except for your choice of ground reference.

The bottom of R3 is NOT tied to ground. It is tied to -Vee which is shown as being -20 V. The left of R1 is tied to ground and not +20V.

Replace your BAT1 with two 20 V batteries connected in series and make the junction between them your ground reference. Do you see that the top of the top battery is at +20 V relative to your ground and the bottom of your bottom battery is at -20 V relative to your ground? Do you see how to connect the three nodes of this battery stack to the remainder of the circuit to get what is shown in the problem?

WBahn

Joined Mar 31, 2012
26,398
I got the base current IB as 0.000036994A but my intiution is wrong
I used an KVL equation : 20V-IB*47000-4700(1IB+100IB)-0.7=0
Why do you think your intuition is wrong?

I would recommend using units properly and putting the terms in the same order as you encounter them as you go around the loop when you set up your work. This greatly reduces the chance of making a mistake and makes it a lot easier to track it down if you do. So I would have put it as:

20 V - IB·47kΩ - 0.7 V - 4.7kΩ·(1·IB + 100·IB) = 0

I'd also recommend working in engineering units (i.e., scientific notation except your exponents are multiples of 3 and expressed using SI prefixes) so you and is

IB = 37.0 μA

As a general rule, final answers should be expressed to 3 sig figs. You are actually seldom justified in reporting them to that since even 1% tolerance components only justifies two sig figs. Intermediate results that are going to be used in subsequent calculations should generally carry two addition sig figs to prevent cumulative round off error.

Update Svc

Joined Jan 10, 2018
10

Will these be the equivalent cell circuit

WBahn

Joined Mar 31, 2012
26,398
You're getting colder. Now the bottom of R3 is at +20 V, to top of R2 is at +20 V, and the left of R1 is at +40 V.

Stack two 20 V batteries on top of each other just like you (usually) put batteries in a flashlight -- the negative terminal of the top one connecting to the positive terminal of the bottom one.

Update Svc

Joined Jan 10, 2018
10
Why do you think your intuition is wrong?

I would recommend using units properly and putting the terms in the same order as you encounter them as you go around the loop when you set up your work. This greatly reduces the chance of making a mistake and makes it a lot easier to track it down if you do. So I would have put it as:

20 V - IB·47kΩ - 0.7 V - 4.7kΩ·(1·IB + 100·IB) = 0

I'd also recommend working in engineering units (i.e., scientific notation except your exponents are multiples of 3 and expressed using SI prefixes) so you and is

IB = 37.0 μA

As a general rule, final answers should be expressed to 3 sig figs. You are actually seldom justified in reporting them to that since even 1% tolerance components only justifies two sig figs. Intermediate results that are going to be used in subsequent calculations should generally carry two addition sig figs to prevent cumulative round off error.
Thanks,
I thought I was wrong since my friends got different answers,
am about to tell them whose right.

Update Svc

Joined Jan 10, 2018
10
You're getting colder. Now the bottom of R3 is at +20 V, to top of R2 is at +20 V, and the left of R1 is at +40 V.

Stack two 20 V batteries on top of each other just like you (usually) put batteries in a flashlight -- the negative terminal of the top one connecting to the positive terminal of the bottom one.
So I just flip Battery 4?
I use circuit labs to create the circuit, you ca edit the schematic here:https://www.circuitlab.com/editor/#?id=asb77yj75dmn
Will help without going thru numerous revisions.

whitehaired novice

Joined Jul 15, 2017
289
No. You have added a new battery which you call BAT2--take it out and ground the end or R1.

WBahn

Joined Mar 31, 2012
26,398
So I just flip Battery 4?
Strictly speaking, yes.

But notice that the original circuit has the left end of R1 tied directly to ground. So in your schematic, tie it directly to ground -- run a wire from the left of R1 to the node you have stuck your ground symbol on.

If you just flip Battery 4, the left end of R1 will be at 0 V, but only by going down to -20 V and then back up to 0 V. No need for that.

I use circuit labs to create the circuit, you ca edit the schematic here:https://www.circuitlab.com/editor/#?id=asb77yj75dmn
Will help without going thru numerous revisions.
The numerous revisions is how YOU will learn what you need to learn in order for YOU to know how to do it -- and THAT's the whole point, isn't it?

WBahn

Joined Mar 31, 2012
26,398
Thanks,
I thought I was wrong since my friends got different answers,
am about to tell them whose right.
On of the beautiful things about most engineering problems (even in the real world) is that the correctness of an answer can be determined from the answer itself.

So you just need to slug through and get your answers and then check to see whether those answers are consistent with the original problem. If they are, then you are correct. If not, then you know you are wrong and can start figuring out where you went wrong.

Update Svc

Joined Jan 10, 2018
10
Strictly speaking, yes.

But notice that the original circuit has the left end of R1 tied directly to ground. So in your schematic, tie it directly to ground -- run a wire from the left of R1 to the node you have stuck your ground symbol on.

If you just flip Battery 4, the left end of R1 will be at 0 V, but only by going down to -20 V and then back up to 0 V. No need for that.

The numerous revisions is how YOU will learn what you need to learn in order for YOU to know how to do it -- and THAT's the whole point, isn't it?
Based on your explanation this is what I got
.

WBahn

Joined Mar 31, 2012
26,398
Closer.

Look at the original schematic in the problem. Does the left of R1 connect to the same node as the bottom of R3?

No. But you have then connected together. In fact, the left side of R1 is now tied to two different nodes (making them a single, combined node). It is tied directly via draw wire to the negative side of BAT 4. Then it is also tied directly, via the ground system, to the junction of BAT3 and BAT4.

Don't make things more complicated than they are.

ALL the ground symbol is is a shortcut way to draw a wire and to declare the voltage on one particular node. Draw wires connecting every ground symbol on the schematic together and call the voltage on that node of wires "zero volts". That's all there is to it.

So, as shown, you have a wire that leaves the top of BAT4, goes to the ground symbol, goes through the wire connecting it to the other ground symbol, and goes back to the bottom of BAT4. Hence BAT4 is shorted out.

MrAl

Joined Jun 17, 2014
8,984
We have been given the question below to identify the missing variables.

I am a bit new to transistors, so far in class we covered a few simplified circuits,
I am finding it difficult to identify the loops so I can use KVL to identify currents.
Plus so far I have never done anything that has a "voltage terminals" instead of my usual comfort zone of Cells.
I would like an explanation of the loops: how they are connected , then I think I will be good.

I had the above circuit in mind is it right?
I got the base current IB as 0.000036994A but my intiution is wrong
I used an KVL equation : 20V-IB*47000-4700(1IB+100IB)-0.7=0
Hi,

A different view is that your second schematic is correct but you shifted the ground. UP by +20v. All your results will therefore come out 20 volts higher than they really are in the original circuit, so you'd have to subtract 20v from every voltage result to get the right results for the orgiinal circuit.

Update Svc

Joined Jan 10, 2018
10
Hi,

A different view is that your second schematic is correct but you shifted the ground. UP by +20v. All your results will therefore come out 20 volts higher than they really are in the original circuit, so you'd have to subtract 20v from every voltage result to get the right results for the orgiinal circuit.
So you from second circuit I should flip battery two and battery four?

Update Svc

Joined Jan 10, 2018
10
Closer.

Look at the original schematic in the problem. Does the left of R1 connect to the same node as the bottom of R3?

No. But you have then connected together. In fact, the left side of R1 is now tied to two different nodes (making them a single, combined node). It is tied directly via draw wire to the negative side of BAT 4. Then it is also tied directly, via the ground system, to the junction of BAT3 and BAT4.

Don't make things more complicated than they are.

ALL the ground symbol is is a shortcut way to draw a wire and to declare the voltage on one particular node. Draw wires connecting every ground symbol on the schematic together and call the voltage on that node of wires "zero volts". That's all there is to it.

So, as shown, you have a wire that leaves the top of BAT4, goes to the ground symbol, goes through the wire connecting it to the other ground symbol, and goes back to the bottom of BAT4. Hence BAT4 is shorted out.
This is what I got

WBahn

Joined Mar 31, 2012
26,398
Yep.

A more conventional way to draw it (assuming you don't want to use node labels as done in the original problem) is to put the stacked batteries on the far left. Then you can come straight across with a wire from the junction to the left side of R1 and attack the ground symbol about midway along it. This also frees up the right side of the drawing for the circuitry that will invariably follow since that transistor stage is their for the purpose of driving something.

Update Svc

Joined Jan 10, 2018
10
Yep.

A more conventional way to draw it (assuming you don't want to use node labels as done in the original problem) is to put the stacked batteries on the far left. Then you can come straight across with a wire from the junction to the left side of R1 and attack the ground symbol about midway along it. This also frees up the right side of the drawing for the circuitry that will invariably follow since that transistor stage is their for the purpose of driving something.
I am still find it hard to understand how the transistor will be active, with regard to where the base current will come from.
Can you explain a bit of electron flow based on the schematic above.

shteii01

Joined Feb 19, 2010
4,644
I am still find it hard to understand how the transistor will be active, with regard to where the base current will come from.
Can you explain a bit of electron flow based on the schematic above.
As long as you have 0.7V drop from B to E, the transistor is On. That is all you care about. With B junction at 0V and E junction at close to -20V, you have that drop, so the transistor is On.

And yes, technically, the current will flow from ground, through R1, into B junction, through the "diode", and come out of the E junction.

MrAl

Joined Jun 17, 2014
8,984
So you from second circuit I should flip battery two and battery four?
Hello again,

Here is a pic of all three drawings in one.

Drawing number 3 matches the original (drawing 1) but drawing 2 is easier to work with.

To see the base current it should not be hard. In drawing 2 just think of the current flowing from the left battery through the base resistor and into the base, then through the base emitter than through the emitter resistor, then back to the other terminal of the battery on the left. If you assume a base emitter voltage of 0.7v then you can easily calculate the base current if you know the Beta, which you do. So one of your equations is right for drawing 2.

Note that you have two enclosed spaces graphically, one on the left and one on the right. Those are your two current loops.

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Update Svc

Joined Jan 10, 2018
10
Hello again,

Here is a pic of all three drawings in one.

Drawing number 3 matches the original (drawing 1) but drawing 2 is easier to work with.

To see the base current it should not be hard. In drawing 2 just think of the current flowing from the left battery through the base resistor and into the base, then through the base emitter than through the emitter resistor, then back to the other terminal of the battery on the left. If you assume a base emitter voltage of 0.7v then you can easily calculate the base current if you know the Beta, which you do. So one of your equations is right for drawing 2.

Note that you have two enclosed spaces graphically, one on the left and one on the right. Those are your two current loops.
Thanks, could you explain what you mean by one of my equations is right.