How is this a current mirror?

Question

This article is talking about floating an IC's GND to use it at a common mode voltage higher than it is rated for. It then says it level shifts the IC's current output to ground by using the Q1-Q2 current mirror.

I thought current mirrors BJT bases needed to be connected together and the reference side collector and base should be connected together also. What am I missing?

That article is strikingly similar to [Maxim app note 3331](https://pdfserv.maximintegrated.com/en/an/AN3331.pdf). Don't see a current mirror in either, the app note at least not claiming Q1 is involved. — greybeard, Jun 11 '23 at 22:11
Huh. It looks like it's almost the exact same text in the app note. Except the app note mentions "the current mirror consisting of Q2 and R2." The article I linked to said "the Q1-Q2 current mirror." Looks like neither make sense. Thanks for the sanity check. — TotesMyGoats, Jun 11 '23 at 22:24
@TotesMyGoats \$Q_1\$ just creates a floating ground that floats about 24 V below the top rail voltage and sinks the required IC current straight down to the actual ground level. Ignore \$Q_2\$ when looking at that fact. Now that this is cleared up, \$Q_2\$ is just a cascoded BJT with its base nailed to the floating ground. IC output is a current source. So \$Q_2\$ just passes that current through to the grounded resistor to generate a ground-referenced voltage output. "Current mirror" here is probably just the author's way of talking about \$Q_2\$'s cascode behavior. Loose talk. Happens. — periblepsis, Jun 11 '23 at 22:57
A more accurate term could be "current buffer" instead of current mirror. — jpa, Jun 12 '23 at 09:36

score 8 · Accepted Answer · answered Jun 12 '23 at 03:45

Q1 is an emitter follower, with its base set to 24V lower than the high voltage supply. Its emitter is therefore at 23V or so, and is providing the GND supply for the IC. The same could be achieved with just the zener diode, but that would require a lot of zener current, and would not regulate very well against varying IC supply current. Employing a transistor in this way keeps zener current much smaller and more consistent, and improves regulation of the IC's power supply:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The role of the other transistor, Q2, is to act as a surrogate current source, to prevent the voltage at the IC's own current source output from falling too far in potential. That voltage could typically be only a volt or so above physical ground, and the potential difference between the IC's own ground and real ground could be huge, like 77V with a 100V supply. That's enough to cause damage to the IC:

schematic

^{simulate this circuit}

Q2 develops most of that potential difference across its own C-E, while still permitting current from the IC's output to pass through R2. Q2 is configured with common base, where the base is held at a very high potential, and the emitter settles slightly above that:

schematic

^{simulate this circuit}

With Q2's emitter (the IC output) so close in potential to the IC's own idea of ground, the actual potential at the IC's output is no longer a threat.

Naturally, some emitter current will be sunk via the base, which introduces a small current difference (error) at the collector, but the error will only be about \$\frac{1}{\beta}\$. Obviously, higher \$\beta\$ would be better in this respect.

Q2 essentially operates as a so-called current conveyor. (Specifically, a CCII type current conveyor.) — Jonathan S., Jun 12 '23 at 19:01

Kuba hasn't forgotten Monica · Answer 2 · 2023-06-11T22:52:47.687

3

How is this (Q1-Q2)[ed.] a current mirror?

It's not a current mirror.

Q1 is a series voltage regulator for the negative supply voltage supplied to the GND terminal.

Q2 is a cascoding element that decouples the highly negative voltage at the load R_2 from the voltage at the OUT pin. The principle of its operation is that the emitter and collector currents are approximately the same.

Q2 could be replaced with a suitable high-voltage small-signal mosfet. That would zero-out the DC current error, but would increase the AC errors due to gate-source capacitance.

Whether the DC error due to Q2 is significant is application-dependent of course.

A MAX4172 is overkill for this application, though. TL071, TL081, or any other op-amp with common mode including V+ will work great for this, and cost many times less. TL071,081 have offset trim so can be nicely zeroed in this application.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

I2 is only needed to model the op-amp VSS current and is not part of the physical circuit.

edited Jun 11 '23 at 22:52

answered Jun 11 '23 at 22:36

Kuba hasn't forgotten Monica

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Can you help me understand what R4 and R5 are for in your circuit? – TotesMyGoats Jun 12 '23 at 00:30
1

although they are not the traditional current mirror circuit, common-base amplifiers do mirror current, so that's probably what they're getting at. – user253751 Jun 12 '23 at 08:09
@TotesMyGoats The ratio of R4 and R5 sets the gain of the amplifier. – Kuba hasn't forgotten Monica Jun 12 '23 at 12:20
@user253751 It's a matter of perspective. Current mirror has a certain meaning, it's not just a general term for "anything that maintains similar current on two terminals". For example, it'd be mighty confusing if someone called a current conveyor a "current source", even though technically it's not a wrong thing to say. But then you can call voltage sources current sources, since they also can source current. So it's best to call things what they are: a current mirror is a very particular kind of a circuit, with the sum of two mirrored currents flowing out of it via a common set of terminals. – Kuba hasn't forgotten Monica Jun 12 '23 at 15:24
as always, the meaning of words depends on the intention of the person who said them. – user253751 Jun 12 '23 at 15:35

score 2 · Answer 3 · answered Jun 11 '23 at 22:00

2

You are missing what MAX4172 is designed to do.
Q1 is sort of supplying power to it,
Q2 acts as a (voltage) level shifter with \$I_C \approx I_E\$.

answered Jun 11 '23 at 22:00

greybeard

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I think the purpose of Q1 is to prevent the majority of MAX4172's current from going through R1 and changing the Zener diode's voltage. And I can see how Q2 would be doing the same thing....but is this a current mirror like the article (that I've now fixed the link to) says? – TotesMyGoats Jun 11 '23 at 22:11
@TotesMyGoats That's correct, although of course R1 could be lower in value - it'd dissipate more. As it stands, Q1 dissipates more instead, but it only dissipates as much as needed, whereas setting a conservative R1 value would require some safety margin and thus higher dissipation on R1. – Kuba hasn't forgotten Monica Jun 12 '23 at 15:21

How is this a current mirror?

3 Answers3