# Setting optimum base current for saturation and power dissipation of a BJT

Regarding the following circuit example, I want to find the optimum current for saturation of Q1. By saturation I mean the safe one. For simplicity I didn't bias the transistor and I'm assuming beta is very very stable.

I sweep Vb from zero to 5V and obtain the following plots:

[left-click to enlarge the plots]

Green plot is the power for the Q1 with respect to Vb.

So by using a cursor I find the minimum point(after cut-off) for the BJT power. This is where the saturation regime begins. At this point Ic is maximum and I find Ib is around 26uA.

So if this transistor needs to be in ON OFF mode for long time, does it make sense that I set Ib as 26uA to obtain the optimum saturation? Or should I do some overshooting as common practice? I'm wondering how would it be safe set in common practice.

• It is unclear to me what you're trying to achieve here. What is the intended usage mode of the transistor, as a switch? Why saturation? Overshooting relates to transient (time) behavior, you consider DC, not time. – Bimpelrekkie Sep 25 '17 at 12:04
• @Bimpelrekkie I want to learn a method/reasoning when I need to saturate a BJT. But as you see after Vbe more than 1.86V or Ib is more than 25uA, the BJT saturates. See the plots. But then what is the optimum value good value for saturation here? Should I set Ib such that BJT power minimum? I mean any base current between 26uA and 3mA base current causes saturation. But what is the best value? What is the common practice when setting it? – GNZ Sep 25 '17 at 12:11
• Your starting point is "wanting saturation", but then I ask why?. How do you think operating the BJT in saturation is a goal by itself? There are infinite ways to operate the transistor in saturation. You have to narrow down the conditions. But what is the best value? Then first define what Best is. Is a car always better than a bicycle? No, it depends what you want to achieve. Same here: it is unclear what you want to achieve. What do you want to do with the BJT? – Bimpelrekkie Sep 25 '17 at 12:16
• @Bimpelrekkie Basically by saturation I mean using it as a good switch. Good means in my case the switch wouldn't cheat you by sometimes going into linear region. And good also means by doing so it shouldn't dissipate so much power. Is it more clear now? There must be a common practice? Or no? – GNZ Sep 25 '17 at 12:23
• Making a good switch (low resistance) and then placing a 1 k resistance in series makes little sense. Why is the resistor in series with the emitter and not at the collector side. If you want to make a switchable 1k to ground then that is the way to do it. Like you have it, when current flows through the 1 k resistor Vbe will change and the transistor could come out of saturation. – Bimpelrekkie Sep 25 '17 at 12:51

I'll take your experimental question seriously for a moment. The circuit does have utility (though I suspect you just cobbled this up without a specific purpose in mind, as your explanation saying, "switch," just fails.)

You are supplying the base with a voltage source, as shown. So this BJT is being treated as an emitter follower. If you were to place some load (such as an LED) in the collector leg, as shown, then varying the voltage on the base allows you to set the collector current as $I_C\approx\frac{V_B-V_{BE}}{R_E}$. So this could potentially be a circuit that converts a known voltage source into a known current sink.

It is NOT a switch. At most, what happens at some point is that as the base voltage climbs up and the voltage drop of the collector load increases then $V_{CE}\lt V_{BE}$ and the BJT moves into varying degrees of saturation. However, when this starts to happen depends upon your circuit particulars.

You can easily solve the circuit equations to find the base voltage required, as:

$$V_B\approx \frac{V_{BE}+\left(V_{CC}-V_{CE}+\frac{R_C}{R_E}V_{BE}\right)\cdot\frac{\beta}{\beta+1}}{1+\frac{R_C}{R_E}\cdot\frac{\beta}{\beta+1}}\label{over}\tag{over-kill}$$

Treating $\frac{\beta}{\beta+1}\approx 1$, this means:

$$V_B\approx V_{BE}+\frac{V_{CC}-V_{CE}}{1+\frac{R_C}{R_E}}\label{good}\tag{good enough}$$

From either of the above, you can estimate saturation to start occurring when $V_{CE}=V_{BE}=700\:\textrm{mV}$, or with $V_{CC}=5\:\textrm{V}$ you have in your case:

$$V_B\approx 1.7\:\textrm{V}\tag{early sat.}$$

With relatively deep saturation at $V_{CE}=100\:\textrm{mV}$, then:

$$V_B\approx 1.84\:\textrm{V}\tag{deep sat.}$$

As you can see, this pretty much brackets the dip in BJT power. The reason is pretty obvious, as your chart shows. Right at the point where the BJT starts going into saturation is also exactly at the point where the base current starts to rise very rapidly. So now there is an ever-increasing base-emitter dissipation being added and this rapidly dominates.

All you learn from this is that optimal (lowest) BJT power dissipation takes place roughly when the BJT ceases to be an amplifier with a large $V_{CE}$ voltage across it (for some collector current) and starts entering into early saturation stages when $V_{CE}$ just starts to go below $V_{BE}$. (Technically, it also occurs when there is no base current, of course.) Pushing harder into saturation only dramatically increases the base current without usefully increasing the collector current and just increases dissipation.

Substantially before saturation, when the BJT is still acting with a viable $\beta$ and $V_{CE} \ge 2\:\textrm{V}$ (or so), the power dissipation follows the usual parabolic curve found between "circuit" (BJT here) and load ($R_E$ and $R_C$ here.) Maximum power transfer occurs when the power in $R_E$ and $R_C$ equals the power in the BJT. Since the collector current and emitter currents, in this case, are approximately equal then this happens when the voltage is split between the BJT and the two resistors, such that $V_{CE}\approx 2.5\:\textrm{V}$ in this case. If you plug that value into the $\ref{good}$ equation above, you'll find this happens when:

$$V_B\approx 1.3\:\textrm{V}\tag{peak BJT dissipation}$$

Which also matches up with your curve.

All this follows from some very basic ideas and a relatively simple circuit analysis.

• Why $V_T$ in the equation? – G36 Sep 25 '17 at 18:55
• @G36 I'll remove it (bad memory here.) Thanks! – jonk Sep 25 '17 at 19:44
• Thanks for the analysis. I think if you would write Vb explicitly would be better. I tried to write how Vb can be derived for Vce=100mV here.. Always Ve=Vb-Vbe. In the neighbourhood of saturation still assuming Ic=Ie. This yields: Vrc/Rc=Ve/Re. Since Vrc=5-0.1-Ve, and Ve=Vb-Vbe from these we can write: [5-0.1-(Vb-Vbe)]/Rc=(Vb-Vbe)/Re. Now assuming Vbe=0.7V, Vb=1.84V. – GNZ Sep 25 '17 at 22:06
• @161776 With $I_C\approx I_E$ while the transistor isn't yet saturated, it follows that $V_{CE}\approx V_{CC}-I_C\cdot\left(R_C+R_E\right)$ but also that $I_C\approx\frac{V_B-V_{BE}}{R_E}$. Solve those for $V_B$ and you will get the "good enough" equation. – jonk Sep 25 '17 at 22:31