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This problem is from Transistor circuits and applications by Cowles:

enter image description here

Attempt at solution

The circuit diagram seems incomplete to me, because they assert a +12 V potential at the emitter but the circuit has no gnd reference. The inclusion of the switch also seems weird to me, as its purpose isn't relevant to this practice problem. In any case, I redraw the schematic of how I think it should look: -

schematic

simulate this circuit – Schematic created using CircuitLab

For the PNP transistor to stay in the active region (which I guess we want?) we require $$V_\text{BE} = -0.7 \: \text{V} \Rightarrow V_\text{B} = 11.3 \: \text{V} $$

The current through \$R_A\$ is \$I_B = \frac{V_B}{R_A} = \frac{11.3 \: \text{V}}{R_A} \$ and the current through the 5 ohm resistor is \$I = \beta I_B = \beta\frac{11.3 \: \text{V}}{R_A} \$.

And here I'm stuck. I don't know how to answer the question "what is the largest resistance \$R_A\$ that you should specify. It doesn't seem to me that there are any restrictions on what values \$R_A\$ can take. Is the problem missing some vital information?

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  • \$\begingroup\$ when the switch closes, Ra is effectively in series with your 5ohms load. I like to believe that you need to choose Ra such that your transistor operates in the saturation region. People that are more experienced than me can help you out. \$\endgroup\$
    – DRF
    Commented Jul 9 at 19:15

4 Answers 4

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The ground is the line at the bottom common to the switch and heater resistance, they just didn't mark it as such.

The switch is relevant, it is the thermostat switch as mentioned in the problem. It switches on when the temperature gets below a set point, providing a path for base current that will turn the transistor on allowing current through the heater resistance.

You don't want the transistor to be in the active region, you want it in either saturation or cutoff. A thermostat is a binary device, on or off. When on, you want saturation, when off you want cutoff.

As a rule of thumb, the base current for saturation is generally taken as the collector current divided by 10, so you need to figure out how much current the heater will take and figure the base current from that. Your instructor or the book (look back through the material leading up to this question) may have set a different rule for saturation current, if so you should follow that, but historically people have used \$I_{B_{SAT}} = I_{C_{SAT}}/10\$. This is to ensure that the transistor will be fully saturated without having to worry too much about the transistor's beta, as long as it's more than 10 it's should be okay, and things like temperature changes or replacing the transistor won't affect the circuit functioning properly.

Once you've got the necessary base current, it's just a matter of calculating the resistor using Ohm's Law. The base voltage will be \$12 V - V_{BE} \approx 11.3 V\$, divide that voltage by the base current and you've got it.

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    \$\begingroup\$ Addendum: The ground is best assumed to be the line at the bottom, yes, but this is not a general convention--I would call this a mistake in the diagram. You shouldn't draw diagrams with only one potential marked unless you really do mean that the entire circuit is floating at that potential. \$\endgroup\$
    – Hearth
    Commented Jul 9 at 19:41
  • \$\begingroup\$ @Hearth Yes, I find a lot of books just left things out like this. Some of the schematics in old military training manuals will drive you nuts. \$\endgroup\$
    – GodJihyo
    Commented Jul 9 at 19:44
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Is the problem missing some vital information?

Yes, very much, as you note yourself. It is missing:

  • Ground connection. This is reasonably assumed to be the horizontal at the bottom.
  • Transistor hFE specification (current gain in this common-emitter mode), to give you β in your equations.

Without knowing these, the question is impossible, as you surmised. You can make presumptions but that never feels like a good road to go down with a formal textbook.

However:

  • The ground connection is a pretty safe bet.

  • The transistor hFE may be stated earlier in the book or that section of it. For this 1968 book, one can guess at the typical textbook value of 10 but it will be a guess.

(There may be a book of answers for your book but I couldn't find it in a quick internet search.)

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The inclusion of the switch also seems weird to me, as its purpose isn't relevant to this practice problem.

Re-reading the problem statement: The 5Ω heater in Fig. P-1.6 is to be turned on by closing the low current thermostat switch S.

So that's what the switch is. It is the thermostat that senses the temperature of whatever the heater is heating, and turns the heater on and off indirectly via the transistor to maintain the desired temperature.

The words low current ... switch S mean that the switch does not have enough current carrying capacity to control the heater directly.

A 5Ω heater driven from 12V draws \$12{\rm\,V}/5\,\Omega=2.4{\rm\,A}\$. So we know that the thermostat S cannot carry 2.4A. It'd be reasonable to assume it can carry an order of magnitude smaller current, so 240mA max.

Thus, the transistor must act as a saturating switch. To ensure saturation, it is reasonable to drive 1/10 of the collector current into the base, as suggested in another answer.

We assume that there's no voltage drop across the thermostat S. So, RA needs to draw 240mA from \$12{\rm\,V} - V_{BE} = 11.3{\rm\,V}\$. From Ohm's law you can figure out what value RA needs to be.

The current in the thermostat will be at most what we designed it to be - 240mA. Most transistors will have a beta much higher than 10, so the thermostat current will be lower than 240mA in practice. More precisely, it will be the heater current divided by the transistor beta at the operating point of the transistor. The operating point is saturation with 2.4A of collector current.


As an aside, the approach above will draw unnecessarily high base current from the transistor, just to ensure saturation in all conditions. Instead, we can control the base current to be just enough to saturate the transistor and no more.

This can be done with an op-amp:

schematic

simulate this circuit – Schematic created using CircuitLab

R1-VR1-R2 establish a reference collector voltage for the saturated condition of Q1. R3-R4 divide the collector voltage by half to provide feedback within the op-amp's input voltage range. C1 stabilizes the op-amp. R5-R6 set the operating point for the op-amp's output voltage and current. This ensures that the op-amp's output voltage is close to the center of the range, and that the op-amp's output only needs to carry a fraction of the base current.

This can also be accomplished using a discrete differential amplifier to control the output transistor:

schematic

simulate this circuit

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The schematic does not explicitly specify the potential of the lower node, and you are correct to question this. You are also probably correct that the author intended this to be 0V.

For the PNP transistor to stay in the active region (which I guess we want?)

No, we need the transistor to be saturated, or cut-off. That's because we require either 0V across the load, or the full 12V. In active mode, collector potential will be somewhere between the extremes of 0V and +12V, which is not the required behaviour. I can assume this, because control is via a switch, and not some arbitrary analogue signal, implying full on/off behaviour.

You are not told (in the snippet you provided) what the current gain \$\beta\$ of the transistor is. Maybe it's written elsewhere. Without that information, you can't know what base current is required to obtain the requisite amount of collector current.

I'll assume \$\beta=100\$.

You know the two required conditions of the transistor, fully off (cut-off) and fully on (saturated). In cut-off, collector current \$I_C=0\$, and the load will have 0V across it. For that condition, base current must also be \$I_B=0\$, which occurs when switch S is open. No further analysis required.

When the transistor is saturated, \$V_{CE}\$, the difference in potential between collector and emitter, is close to zero. I say close, because that's never true, \$V_{CE}\$ will more likely be 0.2V or so, or even more depending on the model.

Just for pedantry, being PNP, this transistor's datasheet will show a negative value for the parameter \$V_{CE}\$. Technically:

$$ V_{CE} = V_C - V_E $$

Since for a PNP transistor in normal use \$V_E\$ is always more positive than \$V_C\$, this will yield a negative value, but in a simple circuit like this, it's easier to say that in saturation, since the supply is 12V, and the transistor takes 0.2V of that, the load must have the remaining 11.8V across it.

It's not clear whether the question expects you to consider a non-zero \$V_{CE}\$ in your answer, but I'm going to ignore it here, and assume \$V_{CE}=0\$, and the full 12V appears across the load. This reveals collector current:

$$ I_C = \frac{12V}{5\Omega} = 2.4A $$

Then we can find the corresponding base current:

$$ I_B = \frac{I_C}{\beta} = \frac{2.4A}{100} = 24mA $$

As you correctly pointed out, the base-emitter junction of the transistor will develop about 0.7V. Strictly speaking \$V_{BE}=V_B-V_E=(+11.3)-(+12)=-0.7V\$, but again at a glance, we know that the remaining 11.3V must exist across \$R_A\$. Knowing the required base current \$I_B=24mA\$:

$$ R_A = \frac{11.3V}{24mA} = 470\Omega $$

If \$R_A\$ were larger than that, base current would fall to a level insufficient to saturate the transistor. With a smaller \$R_A\$, base current would increase, and the transistor would remain saturated.

Therefore this value for \$R_A\$ is a maximum.

In real life, there are some caveats to consider. It's common practice to aim for a base current at least twice the calculated value (\$R_A=\frac{11.3V}{48mA}=240\Omega\$), or up to ten times that (depending on the application and the load) for a couple of reasons:

  • \$\beta\$ is very approximate. It could be significantly lower or higher, varying from transistor to transistor, so you must design for worst-case conditions. Perhaps you'll get unlucky and find a device with \$\beta=50\$.

  • \$\beta\$ drops off dramatically as the transistor approaches saturation.

Since in practice we aim for base current significantly higher than the theoretical minimum for saturation, the small inconvenience \$V_{CE}\ne 0\$ is really insignificant.

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