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Hello, I often see many differential amplifiers using constant current supplies on one side of their power supply rail. For example in the circuit below (not my design just grabbed it off a web page) the differential amplifier uses no constant current source:
followed by an "improved" version which has the constant current source at the bottom near the -15 volt supply rail.
I'm pretty new to electronics and I understand the reason to use a constant current supply in some cases (for example charging a capacitor in a linear fashion) but for the life of me I can't figure out the advantage of using it in this case, some input would be greatly appreciated.
-Oddball -
Constant current there improves linearity, as the input transistors do not have a changing bias current with input voltage, so the change in gain with current is considerably reduced, and thus distortion as well. As well the constant current increases output voltage, as it can still work when the one side is nearly cut off, and can pull the output almost to the negative rail ( to about 3 diode drops above negative, the resistor version will be saturated at about half supply rail ) if needed. It also reduces the effect of noise on the power rails, which otherwise would be added to the signal, improving PSRR always helps to keep noise down.
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This is a particularly important case, because if you take the differential output across two collector load resistors, it doesn't matter. But if you take a single output (either collector), you still have common mode gain.
Common mode being, if the two inputs move together in the same direction, does the output move?
If you use a current sink, the common mode gain goes way down, because the resistance of the sink is the divider in the gain equation.
Tim -
A diff. pair has the task to amplify the difference between the two input signals only. That means - if both inputs see the same voltage the output should not react. With other words - the common mode gain should be as small as possible.
Here are the formulas (Re: common emitter resistance; Rc: Collector resistance; g:transconductance)
* Diff. gain: Ad=(+/-) g*Rc/2
* Common mode gain Acm=-g*Rc/(1+g*2Re)
As you can see, for very large Re values (good current source: nearly infinite) the common mode gain approaches zero.
Example (for Acm=0): Equal diff. amplification for (Vin1=5.01V and Vin2=5V) and for (Vin1=0.01V and Vin2=0V).
Only the difference matters. -
Hello, I often see many differential amplifiers using constant current supplies on one side of their power supply rail. For example in the circuit below (not my design just grabbed it off a web page) the differential amplifier uses no constant current source
Lemme explain in words, as opposed to equations that clarify nothing...
A differential amplifier is ,in essence ,a balance. You apply a voltage differential ( one input sits at a different level than the other) to the inputs. That means, one of the two transistors will conduct more current than its twin.
In the first schematic : lets say the left transistors base sits at a higher level , in respect to ground, than the right one
That means the left one will pass more current. At the point where the three resistors meet the currents are added . The current coming from the left is added to the current from the right and that sum flows through the bottom resistor to the ground.
The problem is that that node is not 'hard' . If i send the left transistor more into conduction i get a larger current there , that causes the voltage across the 75 k resistor to go up. So the right transistor now has less voltage between its base and that 75k resistor.
The total amount of current going through the 75k resistor is not a constant in this case. So the real behavior of the balance is not fixed.
Think of it this way. You have a teeter totter . normally the legs extend a same distance from the pivot point. In the first circuit that center point shifts, depending on which leg is up above the other.
So , to fix that we enter the current source.
now that current is a constant. So the sum of the current coming from the left and the right is always equal to the current in the current source.
Let's say the current source is set for 1 milliamp. If the left transistor lets 0.8 milliamp flow out of its emitter then the right transistor can only let 0.2 milliamp flow. And vice versa.
There is no more impact from the voltage drop across the emitter resistors (the 1k resisotrs in the first schematic) . The centerpoint of the balance no longer shifts depending on which leg is up.
That centerpoint shift causes non linearity. The current source solves that.
Now, both the 75k resisotr and the current source have a voltage drop across them. We call that the common mode voltage.
If you apply input voltages that are lower than Vbe+Vcommon the balance no longer works as you cannot send the transistors in conduction.
With the first schematic that common mode point shifts a bit depending on input voltages. with a current source it doesnt shift anymore. if the common mode shifts , the amplification shifts .. For an amplifier that is not a good idea.
You will find opamps that have the current source sitting between the negative rail and inputs, and others have it sitting between positive rail and inputs.
We call these top driven or bottom driven . A top driven amplifier can work with signals that go all the way to its negative power supply , but stops working if the signals are at vcommon off the positive supply.
A bottom driven on works for signals from its positive supply to vcommon off the negative supply.
Rail-to-rail input amplifiers (dont confuse those with rail to rail output , that is something else, to many times people talk about rail to rail opamps but do not specify if they mean input,output or both ...)
So , rail to rail input means you can go from positive rail to negative rail (rail meaning power supply voltage) . Do these guys have no drop across the current source ? Yes they do, they just employ a trick using a few additional transistors so you do not see the effect of the cinternal common mode voltage. For an applied input range of v+ to v- , you actually, internally in the device stay within v+ to vcommon range. They simp,y scale you up , or down , depending on where the current source sits.
There are even opamps that can scale you well above their positive rail or negative rail. A simple matter of a voltage divide made from two resistors.
Such opamps are found in current sensing applications. They sense a small differential voltage but at a common mode well above , or below, their own power supply voltage.
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Constant current there improves linearity, as the input transistors do not have a changing bias current with input voltage, so the change in gain with current is considerably reduced, and thus distortion as well.
It is rather easy to demonstrate why the diff. amplifier provides much less distortions if compared with a single BJT stage (class-A operation).
The input-output transfer characteristic Ic=f(Vd) can be calculated to be (Vd: Diff. input voltage):
Ic1=(1/2)Ie*[1+tanh(Vd/2Vt)]
with Ie: DC current in the common emitter path,
and Vt=temperature voltage (app. 26mV).
(Ic1 belongs to the left transistor; for the most right transistor with Ic2 the "+" sign changes to "-").
It can be shown that the tanh function is much more linear than the exponential function Ic=f(Vbe) which applies for a single transistor stage only (Shockleys equation).
(Remark: As you - and free-electron (his first sentence in reply#4) - can see: formulas can clarify a lot).
However, it is to be noted that the given relation Ic=f(Vd) was calculated based on Shockleys well-known equation for the pn junction of one transistor
Ic=Ic,o*[exp(Vbe/Vt)-1)]
(Remark: As one can see, both formulas demonstrate the voltage-control mechanism for the BJT). -
Try building both circuits.
Use two 10k potentiometers for the input voltages, adjust them and measure the output. You may find the results easier to understand by referencing the output to the positive rail (put the meter's + connection on +V and - on the output). -
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the exponential function Ic=f(Vbe) which applies for a single transistor stage only (Shockleys equation).
This has been pointed out for you countless times and I don't think you have any firmer grasp of it so I will say it again, likely in vain: in the said circuit, there is significant (current) negative feedback provided by the Re resistors. As a result, the diff and non-diff version will perform comparably (or identically if identical transistors were to be used).
The diff amplifier can be considered, in the case of two identical transistors, two SE stages in parallel). -
Another thing you can do to make it easier to understand, is make the output differential by adding a collector resistor to the left hand transistor.
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So I understand in effect that the two transistors in which the inputs connect to (lets call them Q1 for the left and Q2 for the right) are acting like current sources set by the resistors at their emitter. And that when both inputs are tied to ground the transistors stay in the forward active region because of the biasing on their emitters from the -15 volt supply. So if Q1 and Q2 are in effect current sources where their current changes with voltage applied at the base terminal.
Then is the reason for the current sink is to prevent changes in Q1 and Q2 beta from outside forces (such as heat) changing the delicate current balance between them? Secondly could one thermal couple these transistors together and get the same benefit of the current sink while using one less transistor? Or am I skipping over something? -
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I skipping over something?
What you said has nothing to do with what everyone has told you here.
For now, forget about Vbe and all that, just assume that Vbe is a constant. The emitter and collector current with depend on the input voltage. As such, when both input voltage change in tandem (=common mode input), the collector current with change, and the output voltage will change -> that's a bad thing.
If the resistor is replaced with an ideal current source, that's not going to happen.
It also provides a bias for the input pair that's independent of the rail voltage -> high psrr.
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Why would I use a differential input pair as the first stage of a audio power amplifier? Asking the question differently, how does taking an input signal, say the left channel input from an audio preamp, then "splitting" and amplifying the differences only, improve things?
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Why would I use a differential input pair as the first stage of a audio power amplifier? Asking the question differently, how does taking an input signal, say the left channel input from an audio preamp, then "splitting" and amplifying the differences only, improve things?
I don't know exactly what you're asking.
A differential input can be used to create a balanced input which rejects common mode signals. Take a long twisted pair with a microphone connected to the end. The voltage of both conductors with respect to earth, may have a few volts of mains ripple on them, but the ripple voltage on each conductor is the same and in phase. A differential amplifier takes the voltage difference between each conductor, thus removing the mains hum. -
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the exponential function Ic=f(Vbe) which applies for a single transistor stage only (Shockleys equation).
This has been pointed out for you countless times and I don't think you have any firmer grasp of it so I will say it again, likely in vain: in the said circuit, there is significant (current) negative feedback provided by the Re resistors. As a result, the diff and non-diff version will perform comparably (or identically if identical transistors were to be used).
The diff amplifier can be considered, in the case of two identical transistors, two SE stages in parallel).
I am sorry - but both statements are simply false:
1.) "the diff and non-diff version will perform comparably (or identically if identical transistors were to be used)."
No - a single transistor follows Shockley`s exponential equation (is there anybody who will deny this fact)? And the differential pair undoubtly follows a tanh-function (can be found in each relevant book).
2.) "The diff amplifier can be considered, in the case of two identical transistors, two SE stages in parallel)"
The diff pair - as seen from each of the both input nodes - can be considered as a series connection of a common-collector and a common-base stage. This can be verified very easily: A positive input signal at the left transistor T1 (base node for T2 grounded) causes a positive current change in T1 and a corresponding positive voltage change at the emitter (due to the common emitter resistance). This positive voltage change acts as an input signal for T2 at the emitter node (common-base because the base is grounded) and causes a negative current change in T2. Hence, we have to symmetrical current variations - but with opposite directions.
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Conclusion: This series combination of a common-collector and a common-base stage leads to the mentioned transfer characteristic following a tanh function.
If desired, I can show the derivation of the function.
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Why would I use a differential input pair as the first stage of a audio power amplifier?
Here is one reason (I don`t know if there are others):
As mentioned already in reply#1 - a diff. pair has a much better linearity if compared with a single transistor input stage.
With other words, you can allow a larger amplitude swing at the input terminal for the same distortion level (THD).
This is due to the transfer curve that follows a tanh-function (as I have explained earlier). -
Why would I use a differential input pair as the first stage of a audio power amplifier?
Here is one reason (I don`t know if there are others):
As mentioned already in reply#1 - a diff. pair has a much better linearity if compared with a single transistor input stage.
With other words, you can allow a larger amplitude swing at the input terminal for the same distortion level (THD).
This is due to the transfer curve that follows a tanh-function (as I have explained earlier).
I understand the linearity part.
How does the differential amplifier work on the schematic attached?
No need for equations, just a high level explanation.
Please note that there's feedback from the output back to the base of Q104.
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Let's not turn this into yet another argument about how transistors work...
I agree - but it is impossible to explain one of the advantages of the diff. pair (tanh tranfer characteristics) without mentioning the reason and the background of this property. External properties of the circuit are determined by the active parts and their parameters. -
An increase on Q102's base will open it up more, causing Q102's collector / Q106's base to drop. This closes Q106, causing its collector to rise. This causes output device's base to rise. That, through the feedback network, causes Q104's base, thus Q102's base too, to rise. This increase counters the increase in Q102's base -> ie. it is a negative feedback.
It is a pnp differential input stage, with Q110 sets the bias current. You can work out the bias points fairly easily.
Nothing unusual about the design, with the exception that the VAS runs very hot (@20ma or so). -
Checkout the site linked below which gives a good introduction to differential pairs.
http://www.4qdtec.com/opamp.html
The obscure complementary differential pair in that amplifier is also shown and a brief description given. Although at first glance it looks completely different. It works on the same principle.
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Although it really doesn't, because a long-tailed pair (I intentionally avoid the use of "differential pair" here) must divert a constant current, whereas the complementary case is only "more or less on", no better than a conventional common-emitter circuit. Which is to say, it has very different response when you include operation out to cutoff and saturation -- specifically, there is no limit to saturation, whereas the LTP makes, well -- that smooth TANH curve it's so good at.
The last time I used that motif, I inserted it as an extension of a common-emitter (with offset) amplifier:
http://seventransistorlabs.com/Images/LED_Light2.png
The problem is, the control circuitry requires low bias current, to be able to start up (see the 100k 1/2W from +160V). Ordinarily, I would simply drive the emitter from a (lower ohm) voltage divider. This has a comparator effect (think of it as a literally cut-in-half diff pair), while limiting the current (the "tail" is still relatively "long", thus limiting collector current in a reasonable way for most base voltages). The emitter follower, with base voltage divider, sets this common-emitter voltage more soundly (without great expense to supply bias current), but does make (nearly) unlimited current available; the "tail" resistor is still required, and here ends up in the collector. Which is fine, because the "follower" transistor can saturate hard (which means it will turn off slower), and I don't mind (if anything, that improves the snap action of this circuit).
I could use a diff pair just as well, but, it would require bias current all the time, which stinks (this only requires it during the "on" pulse). I don't require the low offset: only a very crude comparison is required here. The timing loop (which it controls) is within an outer loop, so it is self-adjusting.
So, it's another neat circuit fragment to have on hand, not nearly as useful as the primary example, but it does show up from time to time.
Another place you'll see a motif like that is in ICs with input bias compensation (eventually, one of the emitters is a current mirror to nearly cancel base bias current), or "over the top" input capacity (the input stage becomes powered by the input voltage itself, so input bias jumps up a bit, but it's not usually much).
Tim -
An increase on Q102's base will open it up more, causing Q102's collector / Q106's base to drop. This closes Q106, causing its collector to rise. This causes output device's base to rise. That, through the feedback network, causes Q104's base, thus Q102's base too, to rise. This increase counters the increase in Q102's base -> ie. it is a negative feedback.
It is a pnp differential input stage, with Q110 sets the bias current. You can work out the bias points fairly easily.
Nothing unusual about the design, with the exception that the VAS runs very hot (@20ma or so).
Ok. If a sinusoid signal is inputted, the amplitude of the differential signal (Vd) is held "fixed" for a given input at pin #1 and, under ideal conditions, is a function of the magnitude of the input signal at the pin #1 only.
So the differential input stage operates to maintain a voltage at the output section of the amplifier that is determined by voltage variations at the input alone(?).
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So the differential input stage operates to maintain a voltage at the output section of the amplifier that is determined by voltage variations at the input alone(?).
I am not sure exactly in what context you are asking the question.
For the differential amp that the OP was asking about, it operates in open loop and the output signal, when a resistor (=less than ideal current source) is used, the output signal is a function of both the differential signal as well as the common mode signal -> thus its downside we have been all talking about.
For the amplifier I was talking about that utilizes a less than ideal differential input stage, it operates in a closed loop. As such, you do not have a differential signal. As a matter of fact, any input signal will be amplified and recreated exactly on the other input pin of the differential stage -> that's just how any opamp works.
So in this case, the differential input stage sees nothing but common mode signal and its output is entirely driven by the common mode signal, because of the amplifier's high gain and the feedback loop. -
a long-tailed pair (I intentionally avoid the use of "differential pair" here) must divert a constant current
Who says?.
BTW, without a resistor for emitter degeneration (or two emitter resistors into a current source) the amplification is of course way too high to be used without feedback to the inverting input ... which makes it generally useless for audiophoolery, might as well use an opamp at that point. They tend to want feedback obscured or absent. -
a long-tailed pair (I intentionally avoid the use of "differential pair" here) must divert a constant current
Who says?.QuoteThe classic differential pair amplifier is formed from at least two identical transistors, configured with the emitters for BJT transistors or the sources for FETs connected together. A long-tailed pair (LTP), or emitter coupled (source coupled) pair, is a pair of transistors where the shared emitter or source node is supplied from a more or less constant current source/sink [Ed: emphasis]
QuoteBTW, without a resistor for emitter degeneration (or two emitter resistors into a current source) the amplification is of course way too high to be used without feedback to the inverting input ... which makes it generally useless for audiophoolery, might as well use an opamp at that point. They tend to want feedback obscured or absent.
For audio applications, in and of itself, this is true -- the gain is high, and not very well controlled (it has a TANH shaped transfer function). Within a loop, however, the high gain is beneficial for reducing total distortion, so emitter resistors are often omitted for such applications (such as the traditional diffamp - volt amp stage - emitter follower scheme of an audio power amplifier). And, obviously, for purposes where high gain is desirable, like comparators (which operate with no loop, or even with positive feedback).
It's noteworthy that emitter degeneration must be avoided in applications where the TANH behavior is desirable: mixers and analog multipliers.
Tim