# Calculating voltage, current and magnification for BJT circuit

#### beggi9

Joined Mar 3, 2016
42
a) For the following circuit (See attachment) determine VE,VC and IC when Vin=0
b) Determine VE,VC and IC when Vin=2V

I'm not sure how to calculate this, mostly because I don't know how I can find a voltage in just a certain point in the circuit.
Would it be the same as calculating VBE and VCE?
Should I use Kirchoff's rule?

All help well appreciated!

#### Attachments

• 30.4 KB Views: 19

#### MrAl

Joined Jun 17, 2014
7,891
Hi,

What approximations are you allowed to use, such as base emitter voltage equal to 0.7v ?

#### Jony130

Joined Feb 17, 2009
5,212
You can use any circuit network analysis techniques you know or like. As for Ve and Vc. Ve is a voltage at emitter, to be more precise Ve is the voltage difference between emitter and ground. And Vc is a voltage difference between Vc and ground.

#### WBahn

Joined Mar 31, 2012
26,398
a) For the following circuit (See attachment) determine VE,VC and IC when Vin=0
b) Determine VE,VC and IC when Vin=2V

I'm not sure how to calculate this, mostly because I don't know how I can find a voltage in just a certain point in the circuit.
Would it be the same as calculating VBE and VCE?
Should I use Kirchoff's rule?

All help well appreciated!
A bit about subscript notation might help.

When we use a single subscript, such as Vc or Ve, we are generally referring to the voltage at a particular node relative to whichever node we have declared to be the 0 V reference (a.k.a. the "common" or the "ground", depending on context).

When we use a double subscript, such as Vce, we are talking about the voltage at the first node relative to the second node -- in other words the voltage difference between the two. Mathematically, we have

Vce = Vc - Ve

When both subscripts are the same, such as Vee, this would always be zero (Ve - Ve), at least in circuit that don't involve nonconservative electric fields. So, by convention, we use this to indicate a power supply voltage. In this case it is the power supply associated (not necessarily directly connected to) the emitter.

It's not that simple (things seldom are), but that will get you 90+% of the way there.

#### beggi9

Joined Mar 3, 2016
42
Hi,

What approximations are you allowed to use, such as base emitter voltage equal to 0.7v ?
Yes I'm allowed to assume base emitter is equal to 0.7V.

#### beggi9

Joined Mar 3, 2016
42
A bit about subscript notation might help.

When we use a single subscript, such as Vc or Ve, we are generally referring to the voltage at a particular node relative to whichever node we have declared to be the 0 V reference (a.k.a. the "common" or the "ground", depending on context).

When we use a double subscript, such as Vce, we are talking about the voltage at the first node relative to the second node -- in other words the voltage difference between the two. Mathematically, we have

Vce = Vc - Ve

When both subscripts are the same, such as Vee, this would always be zero (Ve - Ve), at least in circuit that don't involve nonconservative electric fields. So, by convention, we use this to indicate a power supply voltage. In this case it is the power supply associated (not necessarily directly connected to) the emitter.

It's not that simple (things seldom are), but that will get you 90+% of the way there.
Thank you very much for the detailed information! But okay if it's the difference between VE and the ground I should be able to use Kirchoff's law to isolate the currents IB and IC and then use those to find VE and VC right?
I got: 5-0.7=3900*(IB+IC) and 20-0.7=6800IC+3900*(IC+IB)
Only problem is that from that I get IB as a negative and that can't be right since there is a diode that only let's positive current through or am I mistaken?

#### WBahn

Joined Mar 31, 2012
26,398
Since you don't have a beta for the transistor, you don't have any way to determine a value of Ib.

But you CAN determine a value for Ie by only looking at the left-hand loop.

Focus on that for now.

#### Jony130

Joined Feb 17, 2009
5,212
As yourself a question what is the voltage at the emitter (VE) for Vin = 0V and Vin = 2V next what is the voltage at the bottom of a R1 resistor?
The "top" voltage is VE and the bottom one is ?? Then use Ohm's law to find current that will flow through R1 (Emitter current).

#### beggi9

Joined Mar 3, 2016
42
Since you don't have a beta for the transistor, you don't have any way to determine a value of Ib.

But you CAN determine a value for Ie by only looking at the left-hand loop.

Focus on that for now.
Ah okay I see. But okay if I look only at the left loop I should have:
5-0.7=3900*IE so IE=1.10256*10^-3A right?
So VE = 3900*(1.0256*10^-3)+5=9.2999V
And from there I would find IC and VC with the right loop?

#### WBahn

Joined Mar 31, 2012
26,398
Ah okay I see. But okay if I look only at the left loop I should have:
5-0.7=3900*IE so IE=1.10256*10^-3A right?
So VE = 3900*(1.0256*10^-3)+5=9.2999V
And from there I would find IC and VC with the right loop?
Not quite. As you go around the left loop starting at Vb, the first thing you need is the voltage on Vb, which is NOT 5 V. What is it?.

Now the next thing for me to harp on is.... units (the pause for effect is to allow long-time readers -- members for more than a week -- time to shake and/or smack there heads).

Even physical quantity (nearly) has two parts -- a magnitude and a unit -- that are multiplied together. They should be treated that way and, if you get in that habit, then most mistakes you make will mess up the units letting you spot the mistake right away and fix it before spending pages on a solution that was guaranteed to be wrong hours ago or, worse, not catching the mistake at all.

So here you have:

5 V - 0.7 V = (3.9 kΩ)·Ie (Note that I'm keeping your mistake in place)

Ie = (5 V - 0.7 V) / (3.9 kΩ) = 1.1026 mA

As for the next step:

So VE = 3900*(1.0256*10^-3)+5=9.2999V
Ask your self if this makes any sense. In the prior step you assumed (wrongly) that Vb = 5 V. If that is correct, does it make sense for the voltage at the emitter to be 9.3 V?

You have two ways to get Ve at this point. The first, and simplest, is to note that if you know the base voltage and you know that the transistor is on, then you know that Vbe = 0.7 V.

Remember that Vbe is a shorthand for Vb - Ve, thus

Vbe = Vb - Ve
Ve = Vb - Vbe

The other way, which is what you tried to do, is to walk from a node of known voltage to the node voltage you are interested in (and, the method above is the same thing, really).

Vee is a supply voltage relative to ground, so

Vee - 0V = -5V

where Node 'ee' is the node at the junction of the negative supply and R1.

The voltage across R1, V_R1, is the voltage on the top minus the voltage on the bottom (of the resistor), or

V_R1 = Ve - Vee.

By Ohm's Law, this is also

V_R1 = Ie·R1

where Ie is the current flowing through the resistor going downward (i.e., from the top to the bottom in order to be consistent with how we defined V_R1).

So you want Ve:

Ve = V_R1 + Vee = Ie·R1 + Vee

NOW plug in your numbers (after correcting Ie).

#### beggi9

Joined Mar 3, 2016
42
Not quite. As you go around the left loop starting at Vb, the first thing you need is the voltage on Vb, which is NOT 5 V. What is it?.

Now the next thing for me to harp on is.... units (the pause for effect is to allow long-time readers -- members for more than a week -- time to shake and/or smack there heads).

Even physical quantity (nearly) has two parts -- a magnitude and a unit -- that are multiplied together. They should be treated that way and, if you get in that habit, then most mistakes you make will mess up the units letting you spot the mistake right away and fix it before spending pages on a solution that was guaranteed to be wrong hours ago or, worse, not catching the mistake at all.

So here you have:

5 V - 0.7 V = (3.9 kΩ)·Ie (Note that I'm keeping your mistake in place)

Ie = (5 V - 0.7 V) / (3.9 kΩ) = 1.1026 mA

As for the next step:

Ask your self if this makes any sense. In the prior step you assumed (wrongly) that Vb = 5 V. If that is correct, does it make sense for the voltage at the emitter to be 9.3 V?

You have two ways to get Ve at this point. The first, and simplest, is to note that if you know the base voltage and you know that the transistor is on, then you know that Vbe = 0.7 V.

Remember that Vbe is a shorthand for Vb - Ve, thus

Vbe = Vb - Ve
Ve = Vb - Vbe

The other way, which is what you tried to do, is to walk from a node of known voltage to the node voltage you are interested in (and, the method above is the same thing, really).

Vee is a supply voltage relative to ground, so

Vee - 0V = -5V

where Node 'ee' is the node at the junction of the negative supply and R1.

The voltage across R1, V_R1, is the voltage on the top minus the voltage on the bottom (of the resistor), or

V_R1 = Ve - Vee.

By Ohm's Law, this is also

V_R1 = Ie·R1

where Ie is the current flowing through the resistor going downward (i.e., from the top to the bottom in order to be consistent with how we defined V_R1).

So you want Ve:

Ve = V_R1 + Vee = Ie·R1 + Vee

NOW plug in your numbers (after correcting Ie).
Thanks a lot (and sorry about the units, won't happen again) so it would be:
Ve=Ie*R1-5V or 0.7V=0V-Ve so Ve=0.7V
I can then assume Ie=Ic so I have:
Ie=(Vbe-Vee)/R1 = 1.46154mA
Vc = 15V-(1.46154*10^-3)A*6800Ω = 5.06153V
Am I on the right path now or is there still something I'm overlooking?

Last edited:

#### WBahn

Joined Mar 31, 2012
26,398
Thanks a lot (and sorry about the units, won't happen again) so it would be:
Ve=Ie*R1-5V or 0.7V=0V-Ve so Ve=0.7V
Remember, ask if your answer makes sense. If the voltage at the base of the transistor is 0V (I assume we are doing part (a), but you don't indicate this), and if we drop 0.7 V across the base-emitter junction, then does it make sense for the voltage at the emitter to be 0.7 V?

I can then assume Ie=Ic so I have:
Ie=(Vbe-Vee)/R1 = 1.46154mA
You are claiming that the voltage across R1 is (Vbe - Vee). What is the basis for such a claim?

#### beggi9

Joined Mar 3, 2016
42
Remember, ask if your answer makes sense. If the voltage at the base of the transistor is 0V (I assume we are doing part (a), but you don't indicate this), and if we drop 0.7 V across the base-emitter junction, then does it make sense for the voltage at the emitter to be 0.7 V?

You are claiming that the voltage across R1 is (Vbe - Vee). What is the basis for such a claim?
Well no technically it should be -0.7V I think.
The basis for Ie=(Vbe-Vee)/R1 would be that I though Ve = 0.7V (-0.7V if I went to the left and +0.7V if I went down) so using
V_R1 = Ve - Vee it would become V_R1=Vb - Vee. ( I use IC as positive going into the collector. so I go from + to - for positive value.)
Is that a wrong conclusion?

#### Jony130

Joined Feb 17, 2009
5,212
Voltage at emitter with respect to GND is equal to VE = -0.7V, the voltage at the bottom of a R1 resistor is -5V. Therefore voltage across R1 resistor is equal to -0.7V - (-5V) = 4.3V

#### beggi9

Joined Mar 3, 2016
42
Voltage at emitter with respect to GND is equal to VE = -0.7V, the voltage at the bottom of a R1 resistor is -5V. Therefore voltage across R1 resistor is equal to -0.7V - (-5V) = 4.3V
Of course that makes more sense, thank you.
So it should be: Ve=-0.7V and from there:
Ie=4.3V/3900Ω= 1.10256*10mA (=Ic)
Vc = 15-Ic*6800=7.5V right?

And as for (b) I would simply do the same except now Ve=1.3V (because Vin=2V and Vbe=0.7V)

#### Jony130

Joined Feb 17, 2009
5,212
1.10256*10mA
??
Simply if we have a voltage in volts and resistance in kohms the result will be in milliamperes
4.3V/3.9kΩ = 4.3/3.9 = 1.1025 = 1.1025mA
5V/0.2kΩ = 5/0.2 = 25 = 25mA
Vc = 15-Ic*6800=7.5V right?
And as for (b) I would simply do the same except now Ve=1.3V (because Vin=2V and Vbe=0.7V)
Looks good.

#### WBahn

Joined Mar 31, 2012
26,398
Well no technically it should be -0.7V I think.
The basis for Ie=(Vbe-Vee)/R1 would be that I though Ve = 0.7V (-0.7V if I went to the left and +0.7V if I went down) so using
V_R1 = Ve - Vee it would become V_R1=Vb - Vee. ( I use IC as positive going into the collector. so I go from + to - for positive value.)
Is that a wrong conclusion?
There is no "technically" about it. Ve is -0.7 V. Period. It is not, in any way, shape, or form, +0.7 V.

Remember, Ve is shorthand for the voltage at node E relative to the common (0 V) reference. Because the base is fixed at 0 V, and since Ve is 0.7 V LOWER than the base, it is also 0.7 V lower the common, so it is -0.7 V. Negative signs are not optional and they are not discretionary. Try jump starting a car by hooking up the batteries without regard to the minus sign and see what happens (hint -- I do NOT recommend actually carrying out the experiment).

How does

V_R1 = Ve - Vee

become

V_R1 = Vb - Vee ?

This could only be true if Vb = Ve, but since Vbe = 0.7 V, we know that Vb = Ve + 0.7 V, so they are NOT the same.

#### beggi9

Joined Mar 3, 2016
42
??
Simply if we have a voltage in volts and resistance in kohms the result will be in milliamperes
4.3V/3.9kΩ = 4.3/3.9 = 1.1025 = 1.1025mA
5V/0.2kΩ = 5/0.2 = 25 = 25mA

Looks good.
Yeah sorry that "*10" shouldn't be there. Thanks for your help

#### WBahn

Joined Mar 31, 2012
26,398
Of course that makes more sense, thank you.
So it should be: Ve=-0.7V and from there:
Ie=4.3V/3900Ω= 1.10256*10mA (=Ic)
Vc = 15-Ic*6800=7.5V right?

And as for (b) I would simply do the same except now Ve=1.3V (because Vin=2V and Vbe=0.7V)
Ie = 4.3 V / 3900 Ω = 1.103 mA (don't know where the *10 is coming from).

UNITS!!!!

Vc = 15 V - (Ic * 6800 Ω) = 7.51 V

A note on sig figs: As a rule of thumb, results should generally be given with three sig figs (four if the leading digit is a 1) and intermediate results should usually have one more, particularly if it isn't being carried forward in a calculator's memory.