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OK, calling me an amateur is being polite. But I am trying to learn more about how these circuits work.

One example I am learning from is the ZX Spectrum ULA book. There is a oscillator circuit on page 87 that is really confusing me. I can post a picture if anyone wants but it's mostly irrelevant for my question.

Anyway, that circuit has a transistor where the base and emitter are tied to ground. But, the collector is tied to the base of another transistor.

Everywhere I read about transistors, the water analogy is used. But I then read this reference (https://electronics.stackexchange.com/a/61787/32770) that used a natural gas analogy which really made sense to me. Basically, the emitter is a pipe that goes deep underground to highly pressurized natural gas. The base is a valve and the collector is a pipe that just goes into the air. By turning the valve (base) on, the gas (current) is allowed to flow upwards into the air. Which is opposite of the conventional current flow. This was an ah ha! moment for me.

But, looking back at the circuit, it would seem that a transistor's collector tied to the base of another transistor would be useless because the valve (original base) would always be "on". So why have it? Why not just tie ground to the base of the SECOND transistor?

I hope my question makes sense. I can post the actual schematic if people need it.

Thanks

EDIT

Here is the diagram. Hopefully it will explain my question better.

Take a look at Q5 and Q6

enter image description here

EDIT 2

It's embarrassing but it appears that my ancient eyeballs could not see the original connections of the transistors and that the base's are connecting to other items as well.

Thanks to @Ignacio Vazquez-Abrams for pointing that out. If this is true, would you mind answering it? I want to give credit where it's deserved.

BTW, thanks for rasterizing that photo!

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  • \$\begingroup\$ The configuration you describe doesn't seem normal. In this case I think the relevant part of the circuit diagram would be very useful to anyone trying to answer your question. \$\endgroup\$ – brhans Jul 15 '15 at 13:25
  • \$\begingroup\$ I was afraid of that. I can't get the image at the moment but will try to post it later on today. I was curious if this was a "normal" practice. \$\endgroup\$ – cbmeeks Jul 15 '15 at 13:30
  • \$\begingroup\$ Is it possible you are mistaking the emitter for the collector? Sometimes transistors are used with the base connected to the collector to create a diode connection. \$\endgroup\$ – Null Jul 15 '15 at 13:39
  • \$\begingroup\$ It's entirely possible with me. lol. I uploaded a picture of the circuit for reference. I believe there are two examples here. Q6 and Q5. \$\endgroup\$ – cbmeeks Jul 15 '15 at 13:40
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    \$\begingroup\$ That transistor is connected to more than just the base of the other transistor. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 15 '15 at 13:41
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Oops. I got the order wrong in my original explanation, and have edited to reflect this. My apologies.

In your picture, Q5 and Q6 are being used as zener diodes. Current flows through the CB junction, which behaves as a reverse-biased diode. Since an NPN transistor has two PN junctions back-to-back, with the base grounded it looks like

schematic

simulate this circuit – Schematic created using CircuitLab

Note that you ordinarily avoid exceeding the voltages where a transistor junction will act like a zener, but that doesn't mean you can't do it when you want to.

The other configuration you show, consisting of 2 transistors with common bases and the emitter of one tied to the base, is widely used in IC design, and is called a current mirror https://en.wikipedia.org/wiki/Current_mirror.

In effect, the first transistor acts as a diode whose voltage (the base-emitter voltage) is controlled by the current forced through it by the collector resistor. If the two transistors are perfectly matched and thermally maintained at the same temperature, a pretty good description of adjacent transistors on an IC die, the current through the second transistor will be forced to equal the current in the first, since both have identical Vbe characteristics.

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    \$\begingroup\$ "Q5 and Q6 are being used as zener diodes": I think this is the thing that wants explaining, because it's very unusual. The current mirrors are Q7/Q8 and Q9/Q10 - the base of Q5 is not connected to the base of Q1. \$\endgroup\$ – pjc50 Jul 15 '15 at 14:03
  • \$\begingroup\$ Great, thanks for forcing me to have to read more material. LOL. Seriously, if they are being used as diodes, is this because you can't actually put a "real" diode inside an IC? In the same regards of not being able to put capacitors in an IC? \$\endgroup\$ – cbmeeks Jul 15 '15 at 14:03
  • \$\begingroup\$ @pjc50 - yes, my bad, I got carried away with the interesting stuff. Please see the edited version. \$\endgroup\$ – WhatRoughBeast Jul 15 '15 at 14:12
  • \$\begingroup\$ @WhatRoughBeast thanks for the correction. To me, this is more interesting because I have a digital design background and this is the forbidden secrets of analogue IC design :) Can we infer what this Zener's value is (6V?) and what its function is (clamping the oscillator? temperature compensation?) \$\endgroup\$ – pjc50 Jul 15 '15 at 14:16
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    \$\begingroup\$ @cbmeeks - Oh, no, it's perfectly possible to put diodes in an IC. As a matter of fact, every BJT in an IC is composed of at least two diodes (or at least 2 PN junctions). It's not useful to put a big honking power diode in an IC, but that's just because it takes up a lot of expensive real estate. And capacitors are built into ICs on a grand scale - the storage elements in DRAMs are nothing but little capacitors. As with diodes, big capacitors are possible, but the economics of IC design make it a bad idea. Please see my edit, if you haven't already. \$\endgroup\$ – WhatRoughBeast Jul 15 '15 at 14:17

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