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I am studying Transistors from the book - "Arts of Electronics" by Paul Horowitz and Winfield Hill (second edition).

In the text, there is the following circuit (under title - Transistor Switch) on page 63:

schematic

simulate this circuit – Schematic created using CircuitLab

I understand that the transistor would work in the saturation state, because otherwise the collector voltage would be -84V, i.e. less than the emitter voltage (which in this case in zero).

However the text mentions,

Overdriving the base (we used 9.4mA when 1.0mA would have barely sufficed) makes the circuit conservative.

Incidentally, in a real circuit you would probably put a resistor from base to ground (perhaps 10k in this case) to make sure the base is at ground with the switch open.

My questions: 1) What does the first line (from the text) mean? What is the meaning of a circuit becoming conservative?

2) Why would we have to place a resistor from base to ground? How do we decide the base voltage in absence of current i.e. switch is open? Because in that case collector voltage would be 10V and emitter voltage would be 0V, how do we determine base voltage?

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  • \$\begingroup\$ How did you get a collector voltage of -84 volts? \$\endgroup\$ – Peter Bennett Apr 4 '18 at 22:05
  • \$\begingroup\$ @PeterBennett, Ic = (beta)Ib. Therefore, Ic = 940 mA. R_lamp = 100 ohms, hence voltage drop across lamp = 940 mA * 100 ohms = 94 V. Hence if the positive side is 10 V, then the negative terminal must be -84 V. (Sorry for the late reply :) \$\endgroup\$ – Mohammed Arshaan Apr 8 '18 at 23:42
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A circuit is conservatively calculated if you could drop in another transistor type or load without any change in the operating mode.

For example, if your load was a bit higher and your transistor had a bit smaller beta, you could still use the same base current if the original calculated values had been conservative – not bleeding edge.


Any transistor with an open base is susceptible to stray currents. Touch the base terminal at an EMC pestered place and you are a nice antenna for 50/60Hz currents. You don't even have to touch darlington types, coming near the base terminal is sufficient. Placing a resistor or capacitor to GND shorts those stray currents.

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  • \$\begingroup\$ What did you mean by - bleeding edge? \$\endgroup\$ – Mohammed Arshaan Apr 1 '18 at 0:15
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    \$\begingroup\$ It's the opposite, calculating everthing to corner cases thin as a razor so you can hurt yourself easily. \$\endgroup\$ – Janka Apr 1 '18 at 0:21
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Janka's answer is a good one, but I wanted to focus in on the transistor current gain thing. That entire chapter assumes a typical beta of 100. But the following chart from a common 3904's datasheet shows quite a large (normalised) variation of beta depending on temperature and collector current:-

beta

In essence it can change. And if you thought that you might just measure it with a DMM on hFE setting, consider that it changes so that the Ic your meter uses might not be anywhere near your circuit's operating Ic. So you take a conservative estimate for Ic and therefore Ib. Just to make sure it all works properly.

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To more directly answer question #2, as the circuit is drawn, you can't actually know what the voltage at the base is when the switch is open. As the book mentioned, in an actual design, unless you want your circuit to behave erratically, you would attach a resistor from the base to ground ensuring that when the switch is open, you know that the transistor would be 'off'. The resistor would be sized to be 'large' enough that it's not going to have an adverse effect on the circuits normal operation, but 'small' enough that you can be sure that the base is grounded when the switch is open. (1k - 10k are pretty standard pull-up/pull-down resistor values)

https://en.wikipedia.org/wiki/Pull-up_resistor

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