Cute amplifier - bootstrapped

Yeah, because that's what SPICE is measuring. You then look at the AC reference to see the denominator which, for a voltage source, usually defaults to 1V. That gives you the common V/V gain ratio.

SPICE trying to make sense of a voltage/current ratio with who knows what set as the reference value. 1A?

It's a TIA... Trans-Impedance Amplifier

Yes. I could have normalized the plot to Ohms. ...Jim Thompson

Yes. He's determined to call this a TIA, so he applied a 1 amp AC current source directly to the base. Using linear transistor models, he got about 500 kilovolts or so at the output. The proper engineering unit for transimpedance is ohms, not volts, not V/V.

The transimpedance is about 220K * 1290 / 470 ~~ 600K. And it's doesn't greatly change if you bootstrap R2 or not. So he renamed the circuit and changed the rules to prove that bootstrapping doesn't increase the gain, defending his initial misunderstanding.

But of course when it's used as a voltage amplifier, bootstrapping does increase the gain.

That's not true. A high pass filter is not the same as a differentiator. Even I know that. They may be approximately the same in a limit condition, but they won't behave the same in a real circuit.

Rick

bandpass

for

Why? Because current flows into the base of a bipolar transistor? That's not 'news', Jim, and I'm not convinced that a current source is a proper simulation of a capacitive transducer. The ones I've seen, at least 'simple' models, are a voltage source driving the backside of a fix C, just like your first simulation with 500V/V.

However, that's moot to what the OP found 'cute', which was the bootstrap increasing the effective resistance of the collector load; and that's true regardless of how you imagine the 'input' works.

current

bandpass

for

Nope. The signal current DOES NOT flow into the base of the transistor... it flows thru the 220K.

Sheeeesh!

The "news" is how poorly your analysis skills are.

It would be properly modeled as a time-varying capacitor, so that works out being a current when the voltage across the capacitor is constant.

Try thinking instead of guessing and refer to...

Subject: Poor Man's Gyrator Date: Wed, 23 Jan 2013 11:19:41 -0700 Message-ID:

...Jim Thompson

JT is doing what a lot of people do, so we may as well give it a name: Babbling in Spice.

Don't know about him, but there is definitely a name for the s*it you pull.

Babbling

Fetish again.

--

John Larkin Highland Technology Inc

Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators

In real life, there is no such thing as a true differentiator. The circuit that James posted is AC coupled, and JT (and not I) called it a differentiator... just above.

My point was that all AC coupled amps behave sort of like differentiators. James' circuit is no different.

Had a sudden flash of memory of an old buddy at PSpice Support...

He made a typo on equation (2a). It should be:

dQ = C(V) dV

Then simply dividing both sides by dt provides equation (4) trivially, as shown.

Jon

So you agree that shunting current to ground through C1 isn't a good model.

No. After I sent the link to Brian's stuff, I then read it. It's really not about a transducer, it's about a capacitor than varies with the voltage across it.

I see it as I = d/dt (C*V) = C*dV/dt + V*dC/dt

Both of these terms ---------^^^^^^^---^^^^^^^ are current. The second term is the "modulation". ...Jim Thompson

Just found this on page 85 of "Small Signal Audio Design",

V+ | | | \ / R4 \ 47k / | C2 | || 47u +---||---, | || | | | V+ \ | | / R2 | | \ 47k | | / | | | | | | | |/c Q1 +------------| | | |>e | | | | | | C1 | | | || | | +----||----+--OUT | '-----+ || | C4 | | | || R6 |/c Q2 | | IN---||----/\/\--+------| | | || 68k | |>e | \ | | | / RLoad | | | \ "" | | R5 | / +----------------/\/\---+ | | | 220k | | | | | | | +------, | | | | | | gnd \ | | | / R7 \ | \ \ 82k / R3 --- C3 / R1 / \ 33k --- 47u \ 2.7k | / | / | | | | | | | | | | | | gnd gnd gnd gnd

Described as a 2-transistor shunt feedback stage with bootstrapping added to the first stage to improve linearity.

Jon

Just because it's by Douglas Self (*) doesn't mean he understood it... it "bootstraps, but the effective load is inductive. I've shown that both mathematically and by simulation.

Note the other important added ingredient... R6.

What does that make the gain?

Well? Duh! Try R5/R6 >:-}

All that s*it to get a gain of 3.235 ?:-(

(*) Just another audiophool that probably prefers toooobs.

I do hope everyone will simulate (and calculate by hand, if you're capable), and compare the various things that have been tossed out here. ...Jim Thompson

Too bad. Keep working on it and you'll eventually figure out that shorting the signal to ground is just plain obviously 'wrong'.

Current is a term in every component but that doesn't make them all 'current sources'.

With no 'input' the cap is stable at Vbias, say 600mV or so. Now disconnect the cap from the base. Now move the plates further apart some distance. What happens? The voltage increases (and C decreases).

Shazzam!

Now, keeping the plates static at the new distance reconnect the cap. What happens? Current flows but it is anything but 'constant'. As current flows the voltage drops and so does current.

Now, instead of moving the plates change the voltage on 'the other side' of the cap. The rest of the circuit has no idea, nor does it care, whether you 'move the plates' to change voltage or if it is caused by 'the other side' changing in voltage. The end result is essentially the same.

As I said before, the 'simple' simulation for a capacitive transducer is a voltage source on the 'input' side of the cap.

Nonsense.

You love writing equations but if you don't understand them then it's just obfuscation and, in a bootstrap, the C is large so that it is, for all practical purposes, an AC short at the frequencies of interest.

If you had bothered to read the description you might have gleaned that the 'purpose' was to "improve linearity" and it does that by increasing gain so there's lots of negative feedback.

That's a bold statement coming from someone who still thinks shorting the input signal to ground is a good idea.

flipper, Can you provide a citation for the development of that form of modeling a capacitor microphone or similar transducer.

V*dC/dt fits your descriptive of moving plates. In the shunt fed-back case at hand, V is essentially constant.

I'll have to play the math and see if I can rationalize a voltage source in series with a cap. ...Jim Thompson