2
\$\begingroup\$

Multisim plot of IsubL over RsubLThe constant current network and problem statement with values

I am stuck trying to determine the applicable formula for the load current. I have used \$I_b = \frac{Vin-Vbe}{(Beta+1)*Re}\$ and then \$I_L= (Beta)*I_b \$. The problem is I can't find a proper relationship for why the current drops or increases as R_L changes. Can someone please just point me in the right direction. I want to solve it myself, just need to know what I'm missing.

\$\endgroup\$
  • \$\begingroup\$ Do you know what saturation is? And for what Rc value your BJT's will start to enter into saturation region? \$\endgroup\$ – G36 May 29 at 19:33
  • \$\begingroup\$ When they write "assume \$\beta\$ is a high value" do they want you to assume it is \$\beta=\infty\$? \$\endgroup\$ – jonk May 29 at 19:53
  • \$\begingroup\$ I know the lab book says 100mV but doesn't say the resistor value. And this is where our text does a horrible job. It asks what the highest value of R_L can be before the basic equation is no longer valid, and I assume this is where the curve in the plot is non-linear. I just don't have any clue how to get there mathematically \$\endgroup\$ – WARmachin3 May 29 at 19:54
  • \$\begingroup\$ The \Beta value given is 200 so, this lab book is seriously vague on some topics \$\endgroup\$ – WARmachin3 May 29 at 19:55
  • \$\begingroup\$ @WARmachin3 Basically, this is just an emitter follower circuit. The emitter voltage will follow the base voltage (\$V_\text{IN}\$), less a \$V_\text{BE}\$ drop. That voltage, combined with \$R_\text{E}\$, determines the emitter current. This emitter current, less any base current required for recombination, is the collector current. That collector current causes a voltage drop on \$R_\text{C}\$. You have to subtract that voltage drop from the supply voltage rail. It stays in active mode until the collector voltage reaches the base voltage, \$V_\text{IN}\$. (Or less, if your text allows.) \$\endgroup\$ – jonk May 29 at 19:58
0
\$\begingroup\$

In active mode (not saturated):

\$\begin{align*} I_\text{L}&=I_\text{E}-\frac{I_\text{E}}{\beta+1}\\\\ I_\text{E}&=\frac{V_\text{IN}-V_\text{BE}}{R_\text{E}}\\\\ V_\text{C}&=V_\text{CC}-I_\text{L}\cdot R_\text{L} \end{align*}\$

Saturation just barely begins to happen right at the point where \$V_\text{C}=V_\text{IN}\$, which is right at the point where the BC junction of the BJT just begins to become forward-biased. But it's not really very noticeable at that point.

If you were designing an amplifier stage where you wanted the BJT to always be in active mode and never saturated at all, then you'd definitely want to maintain \$V_\text{C}\ge \left[V_\text{B}=V_\text{IN}\right]\$. But if you wanted to actually readily see some significant change in the "linear" declining collector voltage behavior, this doesn't really start to happen enough to be very visible until \$V_\text{C}\$ nears about \$500\:\text{mV}\$ below the base voltage (in your case, \$V_\text{IN}\$.) So you use what's appropriate to the situation at hand.

Let's call the variable defining saturation as \$V_{\text{CE}_\text{SAT}}\$. If you set \$V_{\text{CE}_\text{SAT}}=V_\text{BE}\$, then you are using a definition where saturation is defined to occur when it is just barely starting to happen. On the other hand, if you set \$V_{\text{CE}_\text{SAT}}=V_\text{BE}-500\:\text{mV}=200\:\text{mV}\$ (in your case, let's say), then you are using a definition where saturation is defined to occur when it moving into deep saturation. This latter definition is where you really see the cornering voltage at the collector.

Either way, the general equation for when saturation begins is:

$$V_\text{E}=V_\text{IN}-V_\text{BE}=\left(V_\text{CC}-V_{\text{CE}_\text{SAT}}\right)\cdot\frac{R_\text{E}}{R_\text{E}+\frac{\beta}{\beta+1}\,R_\text{L}}$$

(Solve for \$V_\text{IN}\$, of course, if you want that value.)

It may not be immediately obvious to you. But this is the same equation you'd get from the following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

In effect, this as a voltage divider between \$R_\text{E}\$ and \$R_\text{L}\$. The very slight adjustments involve reducing \$V_\text{CC}\$ by whatever value you pick for \$V_{\text{CE}_\text{SAT}}\$ and by a very minor adjustment in the value of \$R_\text{L}\$ to account for the fact that only \$\frac{\beta}{\beta+1}\$ of the emitter current becomes a collector current. Otherwise, it's really simple.

\$\endgroup\$
0
\$\begingroup\$

You have the variation in the large signal model with Vbe (and beta) as the transistor heats, but there is more to it than that.

I think you should use the hybrid-pi model for the transistor. With a "stiff" voltage source on the base, the output resistance of your current sink is ro, which is related to the Early Voltage Va that you will find in a transistor SPICE model.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.