OPA860 ("perfect" transistor)

The "emitter" current is coming from +Vcc... the "collector" is a "replication" port.

Increasing the "base" voltage increases the "collector" current.

Where they foul up conventional thought is the "emitter" is really best thought of as a feedback node.

Yep ;-) ...Jim Thompson

Right, thanks. Say do you have a copy of AoE3? Figure 2.57 B the diamond transistor.

I would say the two current sources should not be on the same emitter follower. The upper one should fed the collector of the npn.

George H.

Not yet. I'm sure it's great! I'm just waiting for the reduced-errors printing ;-)

...Jim Thompson

No worries... I checked Win's errata page.. someone found the mistake.. but I think had the wrong solution.

George H.

It looks similar, but it's different. The circuit in Figure 61 of the OPA860 datasheet is a state-variable filter. We talk about them in AoE-III, see Figure 6.31, and discuss the attractive aspects of this configuration.

Jim's favorite circuit is a GIC type, or "generalized impedance converter", see Figure 6.14. This is a much more sophisticated filter, and has better performance than the state-variable type, as we show in the plots in Figure 6.15. It's easy to get a Q of over 1000.

Actually, the specific form Jim T has developed is a slight variant of the standard published versions, which he points out has even better performance at high frequencies, IIRC. I have six pages of hand notes and annotated drawings, from the lengthy s.e.d. discussions of these variants, and SPICE modeling I did at the time (July, 2002). Those were really fun times, with Sedra's graduate student, Peter Brackett, joining in, if I recall correctly.

In the notes I compared Jim's form to a few others.

Only a tiny bit of that work made it into AoE-III.

The first printing is sold out at most locations, and Amazon doesn't have many left. The 2nd printing with accumulated fixes is underway now, will probably take about a month.

No, it's correct on the errata page: Found by Anders Jellinggaard, AoE-III, page 100, Fig 2.57B. Our note says: "move lower current sink to emitter of input transistor"

Here's how to think about it, George. Figure 2.57.B

The diamond transistor starts with two complementary emitter followers, each biased with a current source. The two equal current sources dictate the Vbe voltages of the emitter followers, and this voltage is presented to the two output transistors and this dictates their current. These two currents are equal, except for any current going in or out of the "E" node. In the OPA860 the equal currents are mirrored and summed. This sum is zero, except for any E node current, which is thereby replicated at the output, as Jim said.

Unlike an ordinary transistor, the current is inverted.

Hi Win!

FYI, enclosing a URL in angle brackets stops it breaking for many readers:

Hey, what a great hint, thanks!

Thanks so much Win, I did predict that any error's I found would be mine, :^) (It does look fairly obvious in the light of day. My way would have input to the push-pull thing coming from the power rails)

You talk of the current sources biasing the followers, (setting the Vbe voltage) I tend to think of them as the load. (Active load in the common emitter amp) Maybe this is just two sides of the same coin... Or should I think of the current source differently when it's an emitter follower? (I've never done a transistor amp with a current source as load... which no doubt adds to my confusion. Hmm maybe some Friday afternoon.)

George H.