Worksheet link: https://www.allaboutcircuits.com/worksheets/dc-branch-current-analysis/

First of all, I don't get why transistor is treated as a current source. Isn't it just a switch that lets current through depending if gate is on or off? Additionally, the question calls out two sources (two batteries) but if transistor is a source then shouldn't the text in the question talk about three sources?

If I assume that current flowing down from that junction is 5mA then it makes sense to me why we would define current from 7.2V battery as \( 5\text{mA} - I \).
I've redrawn it into (hopefully) an equivalent circuit:



Still, I don't know how to take it from there.
When I calculate a voltage drop in 6v source(+) -> 5mA "source" -> 6v source(-) loop, 6V is being dropped so it'd seem that 6 mA has to be dropped: \[ I = \frac{6}{1000} (\frac{\text{V}}{\Omega}) = 6 \text{mA} \] but there's a 5mA source on this path and I don't know what to make of it.

I suppose I'm missing something obvious again. All suggestions welcome.

- LD

 

A transistor is a device that has certain relationships between the voltages and currents at its three pins. Depending on the constraints we place on what some of those voltages and currents are, we can approximate the resulting voltages and currents elsewhere in a number of useful ways.

If we constrain the base-emitter junction so that it is either well below the threshold voltage (of about 600 mV for a typical BJT) or that it has sufficient current through the base, we can approximate it as acting like a switch between the collector and emitter.

But if we force it to operate between those two extremes, so that we supply it with a base current that is sufficient to get it out of cutoff but not enough to get it into saturation, then we can place it in the active (or linear) mode of operation in which, to an approximation that is good enough for many applications, as a current-controlled current source.

In the circuit you have shown, there ARE three sources. The 6 V source, the 7.2 V source, and the 5 mA source.

You calculation is showing one of the classic mistakes people make when using Ohm's Law.

You are grabbing a voltage from one place (the 6 V) and the resistance from someplace else (the 1000 Ω) and throwing them at Ohm's Law to come up with a current -- but that current is meaningless because you have abused Ohm's Law.

Ohm's Law applies to very specific voltages and currents. It is the resistance of a particular resistor, the voltage across THAT resistor, and the current through THAT resistor. The 6 V is the voltage across the 6 V battery, it is NOT the voltage across the resistor.

To use KVL around the loop, you need to sum up the voltage drops around the loop. For loops involving the current source, that means including the voltage across the current source. If it is an unknown quantity, then simply treat it as an unknown quantity that must be solved for.

A simple and explicit way of doing this is to assign a voltage label to each node and then express the voltage



KVL around the left-hand loop is then:

\(
V_{AB} \; + \; V_{BD} \; + \; + V_{DA} = 0
\)

We know that V_DA = -6 V.

We can use Ohm's Law to express V_BD in terms of the current through the 1 kΩ resistor, but first we have to define that current (and it's reference direction).


So

V_AB = I_1 * 1 kΩ

Finally, V_BD is simply the unknown voltage across the current source. But this same voltage will appear in the KVL equation for the other loop, which gives us the ability to eliminate it from the equations (or, if we want to, to solve for it).

 

Thanks WBahn! I got it now.

Posting solution for completeness:
Left and right loops according to KVL:
\[ V_{\text{AB}} + V_{\text{BD}} + V_{\text{DA}} = 0\]
\[ V_{\text{CB}} + V_{\text{BD}} + V_{\text{DC}} = 0\]
If we apply Ohm's law to \(V_{\text{AB}}, V_{\text{CB}} \) and known voltages:
\[ R_1 I_1 + V_{\text{BD}} - 6 = 0\]
\[ R_2 I_2 + V_{\text{BD}} - 7.2 = 0\]
\( I_2 \) can be defined in terms of \( I_1 \). If we do that and apply known resistances then we can solve this system for \( I_1 \).

I got correct result with that approach but I've now noticed that it's an overkill. Since both L and R loops equal to 0, they're equal.
\[ V_{\text{AB}} + V_{\text{BD}} + V_{\text{DA}} = V_{\text{CB}} + V_{\text{BD}} + V_{\text{DC}} \]
Most of these quantities are known and \( V_{\text{BD}} \) is eliminated:
\[ V_{\text{AB}} -6 = V_{\text{CB}} -7.2 \]
We can easily solve it with Ohm's law.

 

Thanks WBahn! I got it now.

Posting solution for completeness:
Left and right loops according to KVL:
\[ V_{\text{AB}} + V_{\text{BD}} + V_{\text{DA}} = 0\]
\[ V_{\text{CB}} + V_{\text{BD}} + V_{\text{DC}} = 0\]
If we apply Ohm's law to \(V_{\text{AB}}, V_{\text{CB}} \) and known voltages:
\[ R_1 I_1 + V_{\text{BD}} - 6 = 0\]
\[ R_2 I_2 + V_{\text{BD}} - 7.2 = 0\]
\( I_2 \) can be defined in terms of \( I_1 \). If we do that and apply known resistances then we can solve this system for \( I_1 \).

I got correct result with that approach but I've now noticed that it's an overkill. Since both L and R loops equal to 0, they're equal.
\[ V_{\text{AB}} + V_{\text{BD}} + V_{\text{DA}} = V_{\text{CB}} + V_{\text{BD}} + V_{\text{DC}} \]
Most of these quantities are known and \( V_{\text{BD}} \) is eliminated:
\[ V_{\text{AB}} -6 = V_{\text{CB}} -7.2 \]
We can easily solve it with Ohm's law.

Before I comment on your last part, let me strongly recommend that you start tracking your units properly. This will pay off huge dividends down the road.

For instance, you should have:

\[ R_2 I_2 + V_{\text{BD}} - 7.2 \; V= 0\]

If you are tracking your units, then when you see this equation you can quickly verify that it is dimensionally sound. The first term is the product of a resistance and a current, which yields a voltage. The second term is a symbolic voltage, and the third term is an explicit voltage. So these can be added/subtracted with no problem.

You will make mistakes in your work -- we all do. If you track your units (and actually work with them as mathematical factors when manipulating expressions), most (not all) mistakes you make will mess up the units, allowing you to catch the mistake almost immediately and usually at a point where identifying and correcting the mistake is very easy and quick.

Now, it's certainly true that you can set the two loop equations equal to each other, but doing so simply results in the loop equation for the perimeter of the circuit. By itself, that's not enough to solve the problem because you have two unknown currents in the two resistors. So you still need to apply KCL at the top of the current source to take that into account.

If I were given this circuit to analyze, I would note that the voltages at A and C are fixed (relative to the bottom node) and thus I can write KCL at the current source directly in terms of Ohm's Law for the two resistors. Normally, I like to work symbolically and plug in actual values at the end, but I'll work with the values given directly in this instance.

The first equation is one I would likely not write down unless I am trying to be nice to whomever will be looking at my work (such as a grader or a customer). It's included here for the same reason -- to make where I am starting crystal clear with minimal effort/guessing/reverse-engineering on the reader's part.

\(
\frac{V_A \; - \; V_B}{1 \; k\Omega} \; + \; \frac{V_C \; - \; V_B}{1 \;k\Omega} \; = \; 5\; mA \\
\left( 6\;V \; - \; V_B \right) \; + \; \left( 7.2\;V \; - V_B \right) \; = \; 5\; mA \cdot 1 \;k\Omega \; = \; 5 \; V \\
-2 V_B \; = \; 5 \; V \; - 6\;V \; - \; 7.2\;V \; = \; -8.2\;V \\
V_B \; = \; 4.1\;V
\)

Now it is a simple application of Ohm's Law for each resistor to find the currents in each, and confirm that they add up to 5 mA.

 

Worksheet link: https://www.allaboutcircuits.com/worksheets/dc-branch-current-analysis/

First of all, I don't get why transistor is treated as a current source. Isn't it just a switch that lets current through depending if gate is on or off? Additionally, the question calls out two sources (two batteries) but if transistor is a source then shouldn't the text in the question talk about three sources?

If I assume that current flowing down from that junction is 5mA then it makes sense to me why we would define current from 7.2V battery as \( 5\text{mA} - I \).
I've redrawn it into (hopefully) an equivalent circuit:

View attachment 345630

Still, I don't know how to take it from there.
When I calculate a voltage drop in 6v source(+) -> 5mA "source" -> 6v source(-) loop, 6V is being dropped so it'd seem that 6 mA has to be dropped: \[ I = \frac{6}{1000} (\frac{\text{V}}{\Omega}) = 6 \text{mA} \] but there's a 5mA source on this path and I don't know what to make of it.

I suppose I'm missing something obvious again. All suggestions welcome.

- LD

Hi there,

In electronics there are a lot of a lot of a lot of analogous ways of looking at circuits and circuit components. The one you seem to be running into is the current controlled current source (CCCS) model for a bipolar transistor. It's just one of many. It assumes that a current through the base emitter controls a current through the collector emitter. That current through the collector emitter is where the current source comes in for the model.

That can be used for a lot of things including when we look at the transistor as a switch. A small current in the base produces a large current in the collector and so as we vary the base current from a small value to a large value the collector current goes from a small value to a larger value and that makes it look like a switch.
This might seem odd at first because we usually think of an actual switch as something that is either 'on' or 'off', so it either conducts or does not conduct. The analogy with the CCCS model is that since the small base current causes either a very small current in the collector or a very large current, it starts to look a lot like a switch which also conducts a lot when it is 'on' and not so much when it is 'off'. It is a way to make the math a little more manageable in some cases.

It also depends a lot on the application. For some applications it is more useful to view it as an actual switch that is fully on or fully off. In that case though we have to consider a change in topology which can be interesting but sometimes unnecessarily complicated. Using a model like the CCCS model we can get away without doing that and then the math is a little more straightforward.

The other simpler model commonly used is the voltage controlled current source (VCCS) where we imagine the collector current as being controlled by a voltage across the base and emitter. As the BE voltage increases, the collector current increases. This can also be viewed as a switch when we restrict the BE voltage to just two levels, and that gives rise to just two levels of collector current either high or very low, which also makes it again look like a switching action.

There is still yet another simpler model that views the transistor as a charge-controlled device. This is not as often used through except in some situations where charge is a useful quantity to use to calculate something about the transistor that helps understand the circuit operation.

 

Hi guys,

Thanks for all the suggestions and insights on the transistor. WBahn is right that my calculations are sloppy and I should be more methodical. I just want to point out to potential future readers that despite that, the calculations are correct. In my (even sloppier) notebook I've carried both of the approaches to the correct answer of 1.9mA.

Cheers!
- LD

 

Hi guys,

Thanks for all the suggestions and insights on the transistor. WBahn is right that my calculations are sloppy and I should be more methodical. I just want to point out to potential future readers that despite that, the calculations are correct. In my (even sloppier) notebook I've carried both of the approaches to the correct answer of 1.9mA.

Cheers!
- LD

It's good to hear you are making progress