# Differential amp operation

Discussion in 'General Electronics Chat' started by gerwiz, Feb 14, 2011.

1. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Hi Guys,

I've read lots of explanations in various books and online, but not certain I completely understand why, with a differential amp, if you hold the voltage constant at the non-inverting input, and raise the voltage at the inverting input, the output voltage is amplified at the non-inverting transistor's collector?

Is it because when the inverting input voltage is raised, this raises the voltage at the emitter, which in turn effectively lowers the bias of the non-inverting transistor's base, thus reducing the base current, thus reducing the collector current?

If someone can enlighten me I'd be most grateful

2. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Welcome to AAC!

Think in terms of the old fashion balance scales. There is only so much current going through the two transistors (throttled at their emitters), if the transistors are balanced both get equal current, but if one is turned on a bit harder guess who gets the current?

3. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Thanks Bill,
Yes, I understand that if both transistors are matched with equal inputs, then they will both have the same current, but if one transistor drives harder, and takes more current, what causes the other to take less?

I guess I'm trying to understand why (if using just a resistor in the tail) is the tail current constant?

Does my question make sense?

4. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Because they are feed with a constant current source (in theory). The total current is a hard value, not a variable.

If they are feed with a constant current as they are supposed to, this will make the collector resistor extremely stable, no matter what values the base inputs are at.

Look up current mirrors, then compare them with the op amp schematics you see.

Volume 3, Chapter 4 - Current mirrors

5. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Yes, I understand how it all works if the tail resistor is sinking a constant current - but the gap in my understanding is what makes this circuit a constant current source?

One book explained that a rise in voltage at one input causes a rise in voltage at its emitter, which is "felt" by the emitter in the other transistor and thus causes less current to flow in its collector.

I guess I'm trying to understand how the second transistor is responding to the change in input voltage - hence my original supposition about base bias.

Is this how it works?

6. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Neither one of the differential transistors is a constant current source, we are talking the part below the emitters, the third and forth transistors.

Why not post a schematic and we can parse it.

7. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Hi Bill,

Here's the circuit I've been looking at (attached)

File size:
6.6 KB
Views:
24
8. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
OK, that helps. You have to understand that schematic is a gross oversimplification. If the base of Q2 is not grounded then the output is going to be all over the place. As is, by grounding Q2 you help set a reference for the current.

However, most of what I've said still applies. The current at the emitters may not be exact, but it is approximately the same throughout. A good design will use a constant current source.

It is fairly easy to do the math. Look at -V, you know the emitter of Q2 is around -0.6 (base emitter drop), so you can calculate the current. If Q1 is also grounded half the current goes one way and half goes the other, but the emitter resistor sets the current and it is extremely predictable.

Now move the base of Q1 up +0.1V. Q2 will still try to have the same voltage, but Q1 now has much more current through it's base emitter, turning it on (probably saturating it). It will take almost all the current, as Q2 will turn off since it is getting almost no current through its base emitter. Remember the scales analogy?

Move the base of Q1 down -.1V below ground. The situation is now reversed, Q2 is getting much more current through the base emitter, and Q1 is turned off. The scales have gone the other way.

You have to remember the 0.6 to 0.7V drop on the base emitter is important. It has to be met for the the transistor to turn on. Go below it and the transistor turns off. The common emitters provide a form of feedback that amplifies this.

The high input impedance of a differential amp is simple enough, look at common collector biasing of the transistor.

The problem is easy to illustrate though, just go through the 3 scenarios I named.

9. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
OK - so in the situation where both transistors are in active mode, what happens? Does the emitter resistor still set the overall current being sourced?

10. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
That is the linear area, the spot where the differential amplifier is a usable circuit. At this point very little change will swing the balance with very little difference, which is the point.

I used the scenario's earlier to show the extreme examples, but the balance (I used that analogy for a reason) is very sensitive. I assume you have studied op amps and negative feedback. If you haven't yet you will soon.

Anyhow, try using a tablet and draw the circuit with the voltages. It is easy to do.

11. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Thanks Bill. Yes, I have studied op amps. I also appreciate why the diff amp needs to operate at is Q point in order to prevent it going into saturation or cut-off, and I'm possibly getting confused, but just wanted to understand why the current gets balanced between the 2 transistors.

I have heard one explanation which uses the analogy of the transistors acting like variable resistors, and this makes some sense - ie. if these resistors are in parallel then current flow is going to be balanced between them depending on their resistance ratio. But this doesn't seem to explain what actually happens inside the transistors?

So when a slight increase in voltage is applied at the input of Q1, increasing the current through Q1, which presumably raises the voltage slightly on Q1's emitter, what happens at Q2 to cause it to reduce it's current flow?

I think I'm missing something obvious here - but not sure what. Sorry if I seem to be repeating myself, and I do appreciate what you've said so far - it has been very helpful.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
For your posted circuit one might reasonably assume for the sake of understanding that transistors Q1 & Q2 are perfectly matched.

After some algebraic manipulations one can then show, for the small signal case that

Δib2=-[RE/(RE+re)]*Δib1

where
Δ is a change
ib1 is Q1 base current
ib2 is Q2 base current
re is the dynamic emitter resistance
RE is the added common emitter leg resistor

If RE>>re then

Δib2=-Δib1

In other words the feedback arrangement afforded with RE in place, ensures any change in either Q1 or Q2 base current is exactly matched by a change of equal magnitude but of opposite sign in the other transistor's base current.

A little thought on that score leads one to the conclusion that the total effective current flow in RE is the same over the linear operating range - with perfect matching of Q1 & Q2 still assumed.

If RE is replaced by a current source - with notionally infinite impedance - then it is easy to satisfy the requirement RE>>re.

Last edited: Feb 14, 2011
13. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Thanks. OK I see what the maths tells me about the relationship between ib1 and ib2, but in the small signal case, how does the emitter feedback work in terms of a rise of voltage and current at Q1's emitter? Does Q2 drive less current because the voltage at the common emitter has risen?

I'm sorry if this sounds like a very basic question, but just trying to clarify the operation in my head!

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
In practice, yes that's what will happen. The rise only needs to be relatively small to produce the change. For instance, a modest change of ~100mV in the effective VBE for Q2 might produce a full swing in the output voltage at the collector.

gerwiz likes this.
15. ### gerwiz Thread Starter New Member

Feb 11, 2011
8
0
Great - all is now clear - many thanks.