Is this considered a common-base configuration?

Thread Starter

hp1729

Joined Nov 23, 2015
2,304
In basic circuit descriptions the common-base design have the base connected to ground. If we return the base to VCC through a resistor is this still considered a common base circuit? I can't remember ever seeing it demonstrated. This is the basic input design for a TTL gate. Or is it just the equivalent of the diode circuit?
 

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crutschow

Joined Mar 14, 2008
25,269
Q2 does perform the basic function of the two diodes so it's not really a common-base design, except that the transistor does conduct as a transistor when the input goes low to discharge any capacitance at the collector of Q2 and the base junction of Q1.
That's why the TTL circuits are faster then the DTL circuits they replaced.
So perhaps you could say its a mixed use. ;)
 

Thread Starter

hp1729

Joined Nov 23, 2015
2,304
Q2 does perform the basic function of the two diodes so it's not really a common-base design, except that the transistor does conduct as a transistor when the input goes low to discharge any capacitance at the collector of Q2 and the base junction of Q1.
That's why the TTL circuits are faster then the DTL circuits they replaced.
So perhaps you could say its a mixed use. ;)
When the input goes low Q2 doesn't really act as a transistor. It just conducts up through the resistor. When the input goes high that diode shuts off and the other diode conducts from base of Q1, through the diode and up through the same resistor. Or is my understanding wrong?
 

crutschow

Joined Mar 14, 2008
25,269
When the input goes low Q2 doesn't really act as a transistor. It just conducts up through the resistor. When the input goes high that diode shuts off and the other diode conducts from base of Q1, through the diode and up through the same resistor. Or is my understanding wrong?
It is incorrect.
When the input goes low, the base emitter is now forward biased due to the base resistor, allowing normal collector-emitter transistor action.
Since, at the instant this happens, the collector is at a higher voltage than the emitter, due to the charge on the stray capacitance at the collector node, the transistor action will discharge this capacitance until the collector and emitter voltages are essentially the same.
Of course, when the input goes back high, the base emitter current stops and it no longer acts as a transistor, with the base current now going through the base-collector junction.

This is shown in the below for a simulated TTL input.
I used a high value of stray capacitance to better show the action.
Note the fast fall-time at the capacitor due to the transistor action causing the high spike of current through the collector, but the slower risetime since then the capacitor charge current is being provided by the base resistor.

TTL Input.PNG
 

KL7AJ

Joined Nov 4, 2008
2,225
In basic circuit descriptions the common-base design have the base connected to ground. If we return the base to VCC through a resistor is this still considered a common base circuit? I can't remember ever seeing it demonstrated. This is the basic input design for a TTL gate. Or is it just the equivalent of the diode circuit?
That's really a lamely stated question. But yes, the first transistor is common base. I'd give an F to the question writer though. :)
 

RichardO

Joined May 4, 2013
2,271
Notice that if the input is pulled high enough, the transistor now has the emitter more positive than the collector. The transistor still has gain in this reversed condition but it is much lower than the normal gain.
 

Thread Starter

hp1729

Joined Nov 23, 2015
2,304
That's really a lamely stated question. But yes, the first transistor is common base. I'd give an F to the question writer though. :)
I'll take the F, of course. If I thought I understood the topic thoroughly I wouldn't pose the question.
 

crutschow

Joined Mar 14, 2008
25,269
Here's a interesting result with a more typical 3V TTL logic level input swing.
The output actually goes about 0.1V higher than the input due to charging from the base resistor.

TTL Input.PNG
 

Thread Starter

hp1729

Joined Nov 23, 2015
2,304
Notice that if the input is pulled high enough, the transistor now has the emitter more positive than the collector. The transistor still has gain in this reversed condition but it is much lower than the normal gain.
I just don't see the action of a transistor with the emitter and collector current being nearly equal. Either the emitter is conducting (low in) with low collector current, or the collector is conducting (high in) with low emitter current. So Q2 never actually acts as a transistor.
 

RichardO

Joined May 4, 2013
2,271
I just don't see the action of a transistor with the emitter and collector current being nearly equal. Either the emitter is conducting (low in) with low collector current, or the collector is conducting (high in) with low emitter current. So Q2 never actually acts as a transistor.
Good point. The transistor would only be backwards if the emitter voltage is above the power supply. This could be caused by ringing on the input, I suppose.
 

crutschow

Joined Mar 14, 2008
25,269
................
So Q2 never actually acts as a transistor.
As I discussed and showed in my simulation, it does momentarily act as a transistor when discharging the stray capacitance at the collector when the input goes low.
It's precisely this transistor actions that made TTL faster than the DTL circuits it replaced.
 
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