Current differential amplifier with different input impedances?

[Sioux12]

Hello!
Consider page 2 of the datasheet for current differential amplifier "LM2900/LM3900/LM3301 Quad Amplifiers", "Schematic and Connection Diagrams".
They are 4 identical differential amplifiers. The first stage gets two currents as its inputs: I_IN+ and I_IN-. Their difference (I_IN+) - (I_IN-) is the current input of the second stage.
As stated in the "General Description",

These amplifiers make use of a current mirror to achieve the non-inverting input function.

So, the "diode" connected to the +INPUT pin actually represents a diode-connected transistor, which is the master branch of a current mirror; the transistor of the slave branch is on his right hand side and it is explicitly shown.
Assuming that a DC operating point is defined, consider the small-signal equivalent circuit of the first stage. The small signal impedances from the two input pins, obtained considering just a small AC voltage and current variations around the DC operating point, look different to me.

The +INPUT pin faces two small-signal impedances in parallel:

• the one of a diode-connected transistor, which is very low (being the input of a current mirror);
• the one looking into the base of an ordinary BJT, which is higher.

The resulting overall small-signal impedance seen from the +INPUT pin will be similar to the first (smaller) one, so the +INPUT pin sees a low small-signal impedance.
The -INPUT pin faces something very different. The two parallel small-signal impedances seen by this pin are:

• the small-signal impedance seen from the collector of an ordinary BJT, which is very high;
• the small-signal impedance seen into the base of an ordinary BJT, which, as before, is high.

The resulting overall small-signal impedance seen from the +INPUT pin may be lower than the ones just listed, due to the parallel connection: anyway, it looks to me higher than the small-signal impedance seen from the -INPUT pin.

If that is true, the +INPUT and -INPUT pins are in a different condition, so the current entering into the +INPUT pin is not treated by the circuit the same way as the current entering the -INPUT pin.
Is this correct?
And, if yes, shouldn't a differential amplifier treat instead its two inputs the same way?

 

[D.A.(Tony)Stewart]

If that is true, the +INPUT and -INPUT pins are in a different condition, so the current entering into the +INPUT pin is not treated by the circuit the same way as the current entering the -INPUT pin.
Is this correct?
And, if yes, shouldn't a differential amplifier treat instead its two inputs the same way?

1. Yes, that is correct.

2. Well it it is not a differential voltage amplifier, rather it is a Norton amplifier so it has more rules to follow than a simple Op Amp.

I made a discrete SIM and you may see I have Acl=1000 voltage gain that you can control with Rfb and that reduces with Rload.
With the mouse you can inspect the V, I of every part and change any values. I chose hFE =200. Since the Vin- source current is low the stabilization time is long with the input Cap.

 

[danadakk]

6.8: Norton Amplifier

eng.libretexts.org

Ref material.

Also attached.

 

[FvM]

When LM3900 was launched in 1972, National Semiconductor also published a detailed application note https://www.ti.com/lit/an/snoa653/snoa653.pdf

It was the first OP like component in a quad package and quite popular for some years. It doesn't behave well in some of the advertized applications, e.g. turned out to be too noise for serious audio applications.

 

[Sioux12]

2. Well it it is not a differential voltage amplifier, rather it is a Norton amplifier so it has more rules to follow than a simple Op Amp.

What are the different rules for a Norton amplifier? Can I check them out?

I made a discrete SIM and you may see I have Acl=1000 voltage gain that you can control with Rfb and that reduces with Rload.
With the mouse you can inspect the V, I of every part and change any values. I chose hFE =200. Since the Vin- source current is low the stabilization time is long with the input Cap.

Thanks for the simulation, it gives a very useful overview of this circuit. I am at a basic level, so there are some details I can not understand:

• Did you choose a 1M resistor to limit the current and power dissipation from the +15 V DC source?
• Is there any specific reason why you chose to connect the +INPUT pin to a DC voltage and just the -INPUT pin to the actual signal?
• What is the role of the 10uF capacitor?

--- Updated Apr 2, 2024 ---

6.8: Norton Amplifier

eng.libretexts.org

Ref material.

Also attached.

Thank you!

• As regards the first source: why in Figures 6.8.2 and 6.8.4 also the in- pin is connected to a diode? What does this diode represent? It could represent the base-emitter junction of Q6, but then there is also the I_B current generator which probably also represents the same junction, so I'm confused.
  
• As regards the second attached source: are your referring to some specific sections/chapters? Because, even if there's a section dealing with Norton amplifiers starting at page 323, I did not spot the same circuit/device of this post.

 

[D.A.(Tony)Stewart]

What are the different rules for a Norton amplifier? Can I check them out?


Thanks for the simulation, it gives a very useful overview of this circuit. I am at a basic level, so there are some details I can not understand:

• Did you choose a 1M resistor to limit the current and power dissipation from the +15 V DC source?
• Is there any specific reason why you chose to connect the +INPUT pin to a DC voltage and just the -INPUT pin to the actual signal?
• What is the role of the 10uF capacitor?

DC bias from negative feedback for Vin+ becomes the collector current for the Vin+ bias current. The AC coupling is essential to not disturb this balance. If you observe the values , there is some current offset and this was to achieve a relative ok mid-scale DC output in order to maximize the output AC swing with a NFB R Ratio of 1000:1 for 60 dB gain. Unlike conventional Op Amps which bias input current with constant current sources somehow, The Norton OpAmp had a limited useful inexpensive range of applications for 6** transistors in a quad package. You can add or subtract to this simulation to try some other examples.

The diode shunt was simulated with a transistor with Vce=Vbe with an hFE of 10, would affect the bias as any hFE variations in the output quiescent voltage and these variations are reduced by the amount of negative feedback, compared to the reference Vcc = 15V.
There are other variations of output stages to buffer the output on this layout.

Added

I am just referring to my experience demonstrated by my example simulation. Other variables for output biasing were excluded. Any variations in hFE make this circuit sensitive to bias changes required.

The most common choices of values were based only on the datasheet and I adjusted values for good gain and DC operating point.

Sorry if this does not answer all your queries.

Regards
Tony

 

[Sioux12]

I am just referring to my experience demonstrated by my example simulation. [...]

The most common choices of values were based only on the datasheet and I adjusted values for good gain and DC operating point.

Sorry if this does not answer all your queries.

You provided a working example, I really appreciate it, it is valuable and helpful in any case. You used for example the current generators values from the application note, page 7 of the pdf.
Yes, I still have doubts about the issues in the previous post and some new ones, which I try to write down.
But feel absolutely free about this.

DC bias from negative feedback for Vin+ becomes the collector current for the Vin+ bias current. The AC coupling is essential to not disturb this balance.

I can't see how this feedback can reach the Vin+ input. The 560k feedback resistor is not connected to the Vin+ branch.

Unlike conventional Op Amps [...] The Norton OpAmp had a limited useful inexpensive range of applications for 6** transistors in a quad package.

I can't follow this point.

There are other variations of output stages to buffer the output on this layout.

The example output in the simulation is enough. I would like to focus on the two input pins, their small signal impedances and how to treat them.

 

[Audioguru]

The old LM3900 was a very unusual type of amplifier. Use a modern normal opamp instead.

 

[D.A.(Tony)Stewart]

If you analyze the Sim bias currents and signals at DC and AC you will understand more than what I can write. You can scope any node.

_560k biases Vin- not Vin+
_ I excluded the output stages for simplicity which is a self biased emitter follower
So I only followed the front end of Fig 6 and copied fig 12 with changes that lead to a very low Zin + 560 ohms. Fig 16 is better for a higher Zin-