Hello,
I'm trying to find transconductance of an input differential pair because in Douglas Self's book, he starts calculations for VAS capacitor, knowing the transconductance (gm):
High Freq Gain = gm / (2*pi*f*Cvas)
When I simulate a simple pair in SPICE, there isn't problem, as you can see in attached files. If I simulate an active load pair, in the same configuration, the plot is wrong (see last picture)
The question is:
How can I find transconductance of a differential input transistor pair with active load in SPICE?
Thanks in advance.
I'm trying to find transconductance of an input differential pair because in Douglas Self's book, he starts calculations for VAS capacitor, knowing the transconductance (gm):
High Freq Gain = gm / (2*pi*f*Cvas)
When I simulate a simple pair in SPICE, there isn't problem, as you can see in attached files. If I simulate an active load pair, in the same configuration, the plot is wrong (see last picture)
The question is:
How can I find transconductance of a differential input transistor pair with active load in SPICE?
Thanks in advance.
Attachments
Last edited:
You have jumped through many stages in two simulations.
What about looking at one change at a time?
What about looking at one change at a time?
The GM for differential pair with a current mirror load is equal to
gm = 1/re = Ic/26mV = 1mA/26mA = 38.46mS - without emitter degeneration resistors.
If you add emitter degeneration resistors gm is now equal to
gm = 1/(re + RE) = 1/(26 + 22) = 1/(48) = 20.84mS.
gm = 1/re = Ic/26mV = 1mA/26mA = 38.46mS - without emitter degeneration resistors.
If you add emitter degeneration resistors gm is now equal to
gm = 1/(re + RE) = 1/(26 + 22) = 1/(48) = 20.84mS.
Oops, you have plotted the collector current of Q10, and the derivative of Ic(Q10) with respect to Vin. But that's not quite right: what you really want is to look at the output current of this four transistor stage. Output current is difference between Ic(Q10) and Ic(Q13). Transconductance is the derivative of the output current with respect to Vin.
One way to plot these with SPICE is to use a 0V voltage source as a current meter ("Vmeas" in the schematic below). Connect it to the correct bias voltage -- node ww is biased at the same voltage as node yy -- and Bob's your uncle.
Now plot the output current {namely I(Vmeas) } and the derivative of the output current {namely d(I(Vmeas))/d(Vin) }. Presto, you get the expected results, matching jony's hand calculation within the error tolerance of his (kT/q) number. When launching your SPICE simulation, don't forget to use a very small voltage stepsize (100 microvolts in my example) to ensure a smooth output curve.
One way to plot these with SPICE is to use a 0V voltage source as a current meter ("Vmeas" in the schematic below). Connect it to the correct bias voltage -- node ww is biased at the same voltage as node yy -- and Bob's your uncle.
Now plot the output current {namely I(Vmeas) } and the derivative of the output current {namely d(I(Vmeas))/d(Vin) }. Presto, you get the expected results, matching jony's hand calculation within the error tolerance of his (kT/q) number. When launching your SPICE simulation, don't forget to use a very small voltage stepsize (100 microvolts in my example) to ensure a smooth output curve.
Attachments
I’m very grateful to you for your answers.
AndrewT,
You’re absolutely right. One of my problems was I had not set the required sampling precision necessary for the derivative.
Jony,
One of the reasons why I simulated the circuit in a computer, is because I haven´t found information for hand calculation. I looked up in Douglas Self, Randy Slone, Hambley’s book… but I didn´t find anything. Could you tell me where those equations are?
Transistormarkj,
Your post is impressive. You have a total knowledge of the issue; I understand the reason why you are in Silicon Valley. Thanks a lot for your time spent in the post.
Thanks to you, I have got the right figure as you can see in the attachment. (The schematic is slightly different because I’m using OrCAD 16.2 software)
AndrewT,
You’re absolutely right. One of my problems was I had not set the required sampling precision necessary for the derivative.
Jony,
One of the reasons why I simulated the circuit in a computer, is because I haven´t found information for hand calculation. I looked up in Douglas Self, Randy Slone, Hambley’s book… but I didn´t find anything. Could you tell me where those equations are?
Transistormarkj,
Your post is impressive. You have a total knowledge of the issue; I understand the reason why you are in Silicon Valley. Thanks a lot for your time spent in the post.
Thanks to you, I have got the right figure as you can see in the attachment. (The schematic is slightly different because I’m using OrCAD 16.2 software)
Attachments
You shouldn't start to read these type of a books if you don't have anybasic knowledge about basic BJT amplifier (CE , CC and CB amplifers).
Because everyone knows that the BJT transconductance is equal to
gm = 1/re or if you use emitter degeneration resistors
gm = 1/(re + RE)
Where
re = 26mV/Ic
Re - emitter resistance.
So for the simplest differential pair gain is equal to
Av = Rc/2re = Rc*gm
gm = 1/(2re)
But you use current mirror as a load for a differential pair. So the gain is two times larger because the transconductance is two times larger.
gm = 1/re
Try read Bob's book
Designing Audio Power Amplifiers: Bob Cordell: 9780071640244: Amazon.com: Books
Or for example this one
Fundamentals of Microelectronics: Behzad Razavi: 9781118156322: Amazon.com: Books
Use DC operation point analysis.
I suggest it's better to memorize two facts rather than one
There's a wonderful lecture demonstration of this, which I've had the great pleasure of attending. The circuits professor brings out a portable radio, two loudspeakers, and two amplifier circuit boards. One amplifier is made of vacuum tubes and the other amplifier is made of silicon transistors. Both play music quite nicely out of their loudspeakers. Then the professor lights a blowtorch and blasts flame on the transistor amplifier; enough to ruin the musical sound but not enough to melt the components. Finally he shifts the blowtorch to the vacuum tube amplifier. The musical sound remains just as good, even when blasting flam directly on the tubes. Conclusion? Transistors aren't always better than tubes for all applications in all environments. A memorable demo.
- re = ((kT/q) / Ic)
- (kT/q) = 25.8 millivolts at room temperature (T=300K)
There's a wonderful lecture demonstration of this, which I've had the great pleasure of attending. The circuits professor brings out a portable radio, two loudspeakers, and two amplifier circuit boards. One amplifier is made of vacuum tubes and the other amplifier is made of silicon transistors. Both play music quite nicely out of their loudspeakers. Then the professor lights a blowtorch and blasts flame on the transistor amplifier; enough to ruin the musical sound but not enough to melt the components. Finally he shifts the blowtorch to the vacuum tube amplifier. The musical sound remains just as good, even when blasting flam directly on the tubes. Conclusion? Transistors aren't always better than tubes for all applications in all environments. A memorable demo.
A transconductance is a voltage controlled current source. Since a current source, by definition, outputs a current independent of its terminal voltages, transconductance should be constant and independent of bias voltage, to the first order.
Commanding LTSPICE to vary the bias voltage via the "STEP PARAM" statement, we see this is indeed so (simulation results attached): Transconductance is constant and independent of bias voltage, to the first order. Second order effects, such as finite output conductance ("the Early effect") and finite Beta, cause transconductance to vary slightly with bias point. Slightly.
Commanding LTSPICE to vary the bias voltage via the "STEP PARAM" statement, we see this is indeed so (simulation results attached): Transconductance is constant and independent of bias voltage, to the first order. Second order effects, such as finite output conductance ("the Early effect") and finite Beta, cause transconductance to vary slightly with bias point. Slightly.
Attachments
transistormarkj,
I have solved all my main doubts concerning the differential pair and I have learned many things.
Thank you for sharing, in such enthusiastic way, your valuable point of view and experience.
I have solved all my main doubts concerning the differential pair and I have learned many things.
Thank you for sharing, in such enthusiastic way, your valuable point of view and experience.
jony,
I am studying electronics and one of my student's book was "Electronics" by Hambley. I have readed the issue again and I have not found how to find the transconductance of a differential pair. It's obvious that we haven't covered the issue in such deep way as you have covered it, probably, instead of other topics.
I do questions when I don’t understand anything, because in my opinion, it's the best way to learn although it can be a simple question.
I´m going to buy Bob Cordell´s book, and I have readed many reviews about Razavi´s book and it looks like a very interesting book. I think it's just the book to go deeply into electronics I need.
Thanks a lot for your recommendations.
If you pay some attention, you'll see in the figure that I have already run a DC analysis, and the problem was, it gave two different Vbias voltages depending on the leg of the pair.
In any case, I am very grateful to you for your help.
And some books use r_pi - small-signal input resistance of between the base and the emitter as seen looking into the base terminal.
r_pi = Beta * re
But most engineers use this
re = 26mV/Ic - small-signal intrinsic emitter resistance, acts as if it is in series with the emitter in all transistor circuits
gm = 1/re
Little re images
r_pi = Beta * re
But most engineers use this
re = 26mV/Ic - small-signal intrinsic emitter resistance, acts as if it is in series with the emitter in all transistor circuits
gm = 1/re
Little re images
Attachments
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Hi sameerdhiman,
You can see the performance as regards frecuency in this way:
You draw the schematic with “Vac Source” (not “Vsin”).
Run “Ac Sweep/noise” (remember to set RELTOL = 1E-10 in simulation options)
And finally, you can see gain and phase, if you plot this curves:
Gain -> db((V(Us1)-V(Us2))/ V(Ub1))
where Us1 and Us2 are voltage across collector of differential transistors and Ub1 is voltage across base of left leg transistor as you can see in schematics
Phase -> p((V(Us1)-V(Us2))/ V(Ub1))
You have to obtain curves in a similar way they are in the attachment
As far as I know, the important thing is to know the transconductance, because a differential pair is a transconductance amplifier, not a voltage amplifier and the goal is to obtain a fine and linear current to drive VAS stage. VAS stage is a transresistance amplifier, it absorbs current and gives voltage (The opposite of transconductance). I suppose this analysis is better for knowing frequency performance.
I’m sure other people (transistormarkj, jony…) in this forum can give you a better info.
You can see the performance as regards frecuency in this way:
You draw the schematic with “Vac Source” (not “Vsin”).
Run “Ac Sweep/noise” (remember to set RELTOL = 1E-10 in simulation options)
And finally, you can see gain and phase, if you plot this curves:
Gain -> db((V(Us1)-V(Us2))/ V(Ub1))
where Us1 and Us2 are voltage across collector of differential transistors and Ub1 is voltage across base of left leg transistor as you can see in schematics
Phase -> p((V(Us1)-V(Us2))/ V(Ub1))
You have to obtain curves in a similar way they are in the attachment
As far as I know, the important thing is to know the transconductance, because a differential pair is a transconductance amplifier, not a voltage amplifier and the goal is to obtain a fine and linear current to drive VAS stage. VAS stage is a transresistance amplifier, it absorbs current and gives voltage (The opposite of transconductance). I suppose this analysis is better for knowing frequency performance.
I’m sure other people (transistormarkj, jony…) in this forum can give you a better info.
Attachments
Yesterday, I was doing some comparison on Resistor vs Current Mirror load configurations. I was able to plot Trans-conductance (Gm) and Gain Av (RC * Gm) in Resistor load (many thanks to this thread contributors for showing a nice trick to plot Gm) but I got stuck with Current Mirror configuration where Gm got plotted but I did not find the way to plot Gain.
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