# Analog foundations of digital electronics clear as mud

Discussion in 'General Electronics Chat' started by odenskrigare, Apr 19, 2010.

1. ### odenskrigare Thread Starter New Member

Apr 19, 2010
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I self-taught a fair amount of digital and a bit of analog circuits. The very computational part (e.g., put in two ones, get out a zero) which is what they have in digital logic texts is easy for me but at the more primitive level I am having a much harder time. See here for instance:

http://www.eng.utah.edu/~cs6710/handouts/AppendixB/appendixB.doc3.html

See the first diagram therein

"A high voltage at the base turns on the transistor. The output F is discharged to ground, getting close to 0 V but never quite reaching it (it reaches a voltage drop away from 0 V).

When a low voltage is placed on the base, the transistor is turned off. The output node F is charged up toward the power supply voltage through the pull-up/load resistor R1."

Could someone please clarify the bolded phrases for me? Furthermore, what is F exactly? Is it a wire that leads to ground? Is it just a point where you would attach one lead of a voltmeter or similar component, and connect the other to ground, thereby measuring the output?

Sep 7, 2009
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3. ### kingdano Member

Apr 14, 2010
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"A high voltage at the base turns on the transistor. The output F is discharged to ground, getting close to 0 V but never quite reaching it (it reaches a voltage drop away from 0 V).

When a low voltage is placed on the base, the transistor is turned off. The output node F is charged up toward the power supply voltage through the pull-up/load resistor R1."

1. To be clear - 'F' here is simply denoting the output signal - much like you would have on a logic gate. Your second guess is essentially correct, but you wouldn't need to connect a voltmeter necessarily - any device accepting the output voltage level could be connected.

2. What the first bolded statement is saying is this:

When the base of the transistor has a "high voltage" applied (high voltage could be any value which causes the junction between collector and emitter to conduct current - in its saturated state) it creates a path for current to flow from +Vcc to ground, through the resistor and transistor. With that junction closed - there is no path, and therefore the voltage there would read +Vcc.

3. Similarly, the second statement is the other case - where the base is exposed to a low voltage. In this case the voltage at point 'F' will quickly rise towards +Vcc, as the path for current to flow is closed.

Does that help?

(This is an inverter, by the way - if that helps. 5V input = 0.7V out (logic low) and 0V input = 5V output for example)

Last edited: Apr 19, 2010
4. ### SgtWookie Expert

Jul 17, 2007
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In the diagram, R1 is the collector resistor, and R2 the base current limiting resistor.

When current flows through R2 into the base of the transistor, the transistor begins to conduct current through the collector to the emitter. Up to a point (known as "saturation") increasing current through the base will increase the collector current. After saturation is reached, increasing the base current will have a negligible effect on collector current.

As current through the collector increases, current through R1 also increases, up to the point where nearly all of the supply voltage is dropped across R1. In effect, when the transistor is in saturation, it can be considered as resistor with a low value of resistance - you have a voltage divider with R1 on top, and a low-value resistor to ground.

When the current through the base of the transistor is stopped, the current flow through the collector is also stopped. Since the path to ground via the collector has been removed, the voltage divider still has the same resistance on the top (R1) and nearly infinite resistance on the bottom.

5. ### odenskrigare Thread Starter New Member

Apr 19, 2010
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Almost there. I need one more thing to clarify. If you measured voltage between R1 and the collector and ground while base voltage were not high enough to allow current flow, why would it read +Vcc? Wouldn't most of the voltage have been dropped across R1?

Maybe I'm just woefully confused

6. ### Markd77 Senior Member

Sep 7, 2009
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Without any current flowing through R1, both ends are at the same voltage +Vcc.

7. ### SgtWookie Expert

Jul 17, 2007
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If the collector is not sinking current, then it is effectively an open connection, or infinite resistance.

R1 is still connected to Vcc. Since there is nothing else in the circuit to sink current, the junction of R1 and collector will measure Vcc.

8. ### odenskrigare Thread Starter New Member

Apr 19, 2010
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Ahhh right so F isn't connected to ground at all ... silly me

Thanks for the help. I can now get a move on with (more interesting) digital logic poop with a somewhat better understanding of how it works inside

9. ### Saon Das New Member

Mar 27, 2010
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more e-books on analog electronics