Current mirror explanation

[Analog Kid]

I ran across this in my electronic document archive, part of an explanation of how current mirrors work:



OK, fine: I don't really understand how this (adding the two resistors) helps. They say

Due to the resistors, the function of base voltage to collector current becomes more predictable.

How, exactly, does it do that? What is the mechanism that accomplishes this? (Please, please, no Ebers-Moll! that stuff gives me a headache.)

Nothing said about the value of those resistors; I'm assuming it should be something low, something <100Ω.

[Benta]

It's simply because when making a current mirror, the V_BE of the two transistors should be equal and track each other over temperature, current etc. Only possible on a single substrate.
The emitter resistors equalize differences in V_BE. The resistor value is dependent on the current range in the mirrors, but a voltage drop of 50...100 mV across them is reasonable.

[Analog Kid]

The emitter resistors equalize differences in V_BE.

OK; how do they do that? Still mystified here.

[tggzzz]

I ran across this in my electronic document archive, part of an explanation of how current mirrors work:

(Attachment Link)

OK, fine: I don't really understand how this (adding the two resistors) helps. They say

Due to the resistors, the function of base voltage to collector current becomes more predictable.

How, exactly, does it do that? What is the mechanism that accomplishes this? (Please, please, no Ebers-Moll! that stuff gives me a headache.)

Nothing said about the value of those resistors; I'm assuming it should be something low, something <100Ω.

The mechanism is a variant of "emitter degeneration".

That is the simplest current mirror variant, and unsurprisingly it has non-ideal behaviour. Slightly more complex ones reduce the non-idealities.

To understand the non-ideal behaviour requires understanding the Ebers-Moll equations, or similar. No point in trying to sugar-coat a complex subject, and make it too simple. Understanding complex subjects requires a lot of hard work and effort; deal with it.

Key search term, other than "current mirror", is "translinear circuits".

[Analog Kid]

To understand the non-ideal behaviour requires understanding the Ebers-Moll equations, or similar.

You're missing the point. I'm not trying to understand the "non-ideal behavior" here; I'm trying to understand the basic mechanism by which those resistors (at least somewhat) equalize differences in V_BE.

[golden_labels]

Tggzzz did answer that in the first sentence: “the mechanism is a variant of ‘emitter degeneration’.”

The rest is a commentary giving you a wider perspective.

[langwadt]

To understand the non-ideal behaviour requires understanding the Ebers-Moll equations, or similar.

You're missing the point. I'm not trying to understand the "non-ideal behavior" here; I'm trying to understand the basic mechanism by which those resistors (at least somewhat) equalize differences in V_BE.

if one transistor has a slightly lower Vbe it'll draw abit more current, the extra current will cause a higher voltage drop over the emittor resistor, that extra voltage is subtracted from the voltage on the base

[TimFox]

The total voltage from each base to the negative rail includes a V_be and the voltage across the series resistor (R1 or R2) to the negative rail.
There will be a small variation from part to part on V_be, but the resistors can be matched to very tight tolerance.
If the voltage across the resistor is greater than the V_be, then it dominates the current mirroring and improves the match in the two emitter currents.
The emitter current of the right transistor will be determined by the voltage across its emitter resistor R2.
In the limit where the resistors are huge, where V(R1 or R2) >> V_be, then the current ratio becomes independent of the difference between the two V_bes.

[Kim Christensen]

FYI, matched pairs for current mirrors are pretty cheap to buy. BCM61B for a matched NPN pair in a single package or BCM62B for PNPs. Less than $0.50 at Digikey.

[Analog Kid]

There will be a small variation from part to part on V_be, but the resistors can be matched to very tight tolerance.

The source I quoted above goes on to say:

Now note that the two resistors need not be equal. If we consider the B-E voltage of the two transistors fixed,
then the ratio of output current to input current will be R1/R2. If R2 is half of R1, for example, then the mirror
will want to sink twice the current at OUT that you put in to IN. At high ratios you need to let the resistors drop
a bit more voltage to get reasonably linear operation since there will be more variation of the B-E drop between
the two resistors. It is a tradeoff between voltage lost in the resistors to accuracy, but this circuit will work well
enough for many purposes when built from discrete parts.

So I take that to mean that by intentionally not matching the Rs, you can build a proportional current mirror where the output (sink) is some multiple or fraction of the input current: is that correct?

[langwadt]

FYI, matched pairs for current mirrors are pretty cheap to buy. BCM61B for a matched NPN pair in a single package or BCM62B for PNPs. Less than $0.50 at Digikey.

I seem to remember someone xraying and thermal imagining some transistor pairs, and I think all of them where separate dies in a package, which might be good enough

[Analog Kid]

Heh; I've thought of hand-matching 2 Qs (breadboard a current mirror and measure how well it tracks), then epoxying the 2 Qs together to make a DIY "matched pair". ("With temperature compensation!")

Could sell them on eBay, maybe.

[David Hess]

The trick here has to do with the emitter resistance of the transistors, which determines the change in emitter current versus a change in base-emitter voltage:

r_e in ohms = 0.026 volts / I_E (emitter current in amps) or
r_e in ohms = 26 / I_E (emitter current in milliamps)

So 1 milliamp yields 26 ohms, 10 milliamps yields 2.6 ohms, etc.

Now the transistors have a mismatch in Vbe voltages, which when combined with the emitter resistances produces an error in the emitter currents, and thereby the collector currents.  Adding the external emitter resistors reduces this error because they add to the emitter resistances of the transistors so the difference in Vbe voltages has less of an effect.

Note that this also applies to noise.  The transistor operates as if a voltage noise source is in series with the base-emitter voltage, and the total emitter resistance converts this into a noise current, so adding the external emitter resistors also reduces the current noise.

[Analog Kid]

FYI, matched pairs for current mirrors are pretty cheap to buy. BCM61B for a matched NPN pair in a single package or BCM62B for PNPs. Less than $0.50 at Digikey.

Thanks. Unfortunately we only use through-hole parts here at AnalogKid Laboratories, GmbH.

[TimFox]

For non-precision applications, the left transistor Q1 is not actually necessary.  As mentioned above, the current in Q2 is determined by R2 and the voltage across it.
Q1 helps to reduce the temperature dependence of the mirror, since the two V_be voltages tend to track each other with temperature.
For precision applications, with non-huge resistors, IC designs often parallel n identical transistors at Q2 to get a higher mirrored current where I(Q2) = n x I(Q2), each component of Q2 having its own emitter resistor R2.
If the transistors are matched really well, since each transistor's emitter current is determined by its V_be, the mirror can omit the two resistors R1 and R2.
There is also a three-transistor current mirror circuit that improves performance, which you can find by searching the literature.

[David Hess]

Heh; I've thought of hand-matching 2 Qs (breadboard a current mirror and measure how well it tracks), then epoxying the 2 Qs together to make a DIY "matched pair". ("With temperature compensation!")

Could sell them on eBay, maybe.

It works fine, but I doubt you could sell them at any sort of profit.

I match the transistors by measuring their Vbe at a suitable test current.  It usually only takes a few minutes to get several matches better than 100 microvolts.

I seem to remember someone xraying and thermal imagining some transistor pairs, and I think all of them where separate dies in a package, which might be good enough

Separate dies in the same package works fine, and even has some advantages like less parasitic coupling between the transistors.  There are many circuits which work better with matched pairs rather than monolithic matched pairs because of parasitic elements in the later.

[golden_labels]

You may also approach this starting from an emitter follower configuration.

Ignoring small current to Q1’s base, all input current must go through Q2 and R1. This means voltage across R1 reflects input current. The base-emitter junction of Q2 offsets that by about +600 mV.

The right part, if seen as an emitter follower, outputs the same voltage, reduced by Q1’s base-emitter drop of -600 mV, on R2. If voltage on R2 is the same as on R1 and they’re the same value, by Ohm’s law the current on them must also be equal. And since that current on R2 is mostly drawn from Q2’s collector, Q2 pulls the same current as Q1 is fed.

This isn’t true, because components are mismatched. So they’re not going to be identical. However, an emitter follower configuration is highly resistant to variation in transitor’s amplification and base-emitter drop. Much, much more than the original configuration, which depends on both transistors having identical behavior. Since the emitter resistors can’t be too big, this is goign to be a very imperfect emitter follower, but in principle the benefits are still there.

[TimFox]

FYI, matched pairs for current mirrors are pretty cheap to buy. BCM61B for a matched NPN pair in a single package or BCM62B for PNPs. Less than $0.50 at Digikey.

Thanks. Unfortunately we only use through-hole parts here at AnalogKid Laboratories, GmbH.

In the current trend of discontinuing TH parts, it is hard to find dual transistors.
You might still find them on eBay, in TO-5 shaped TH packages.
Another interesting obsolescent part is the old RCA CA3046 transistor arrays and others in that series, which were originally in DIP.
Jameco is a reasonable source for obsolete semiconductors  https://www.jameco.com/z/CA3046N-Major-Brands-IC-CA3046N-LM3046N-DIP-14-General-Purpose-NPN-Transistor-Array_12079.html

[Analog Kid]

There is also a three-transistor current mirror circuit that improves performance, which you can find by searching the literature.

Yes, the Wilson mirror. Was just reading about it a minute ago in Horowitz & Hill.

[mawyatt]

There is also a three-transistor current mirror circuit that improves performance, which you can find by searching the literature.

Yes, the Wilson mirror. Was just reading about it a minute ago in Horowitz & Hill.

A good book on Current Sources & Voltage References is by Harrison.

https://books.google.com/books?id=03JmxpE39N4C&pg=PA87&lpg=PA87&dq=Wyatt+Current+Source&source=bl&ots=5zevJ8Ypzb&sig=ACfU3U3DvUbu9AotugZ_GWm2zi6zyEj6pA&hl=en&sa=X&ved=2ahUKEwicmoqB9Oj2AhWOSjABHRumDjYQ6AF6BAgYEAM#v=onepage&q=Wyatt%20Current%20Source&f=false

Orginial EDN Article.
https://www.edn.com/peaking-current-source-has-high-rejection/#google_vignette

See #2 mentioned below.
https://www.eevblog.com/forum/projects/some-free-bw-in-a-cascode-amp/msg3719158/#msg3719158

See #10 mentioned below.
https://www.eevblog.com/forum/chat/leds-as-a-low-noise-shunt-regulator/msg3899102/#msg3899102

Best,

[ejeffrey]

The emitter resistors equalize differences in V_BE.

OK; how do they do that? Still mystified here.

Consider a 10 mA current mirror.  Let's say Vbe of one transistor at equal current is 25 mV less than the other.  In the first circuit, the operating Vbe is constrained to be equal so that transistor will have 2.7 times the current flow or 27 mA. That's the ebers moll equation.

Now instead put a 10 ohm resistors on each emitter.  Now the emitters can float relative to each other, and the 25 mV difference in emitter current is now only 2.5 mA instead of 17 mA.

[ButchJames]

I don't really understand how this (adding the two resistors) helps

Some of the stability reasons are ...

*** The base-emitter junction needs to be biased correctly, and an emitter resistor Re makes that mathematically easier and safer for the base-emitter junction.
*** The resistors in the emitter are there - to prevent Thermal Runaway.
*** The resistors in the emitter leg of the circuit also generate effectively Current-Series (negative) feedback, and as mentioned - naturally helps prevent Thermal Runaway.

{Current-Series: A 'sample' of the output current generates an effective Voltage (in series, at the input) feedback}

[Analog Kid]

I don't really understand how this (adding the two resistors) helps

Some of the stability reasons are ...

*** The base-emitter junction needs to be biased correctly, and an emitter resistor Re makes that mathematically easier and safer for the base-emitter junction.

OK, fine: but how exactly? Without further explanation that just sounds like a lot of handwaving to me.

*** The resistors in the emitter are there - to prevent Thermal Runaway.

OK, kind of like you'd have in a push-pull emitter follower, right? Sometimes referred to as a "cushion".

*** The resistors in the emitter leg of the circuit also generate effectively Current-Series (negative) feedback, and as mentioned - naturally helps prevent Thermal Runaway.

{Current-Series: A 'sample' of the output current generates an effective Voltage (in series, at the input) feedback}

Again, this is insufficient explanation.

Thanks anyhow.

[TimFox]

Read reply 14 above.

[tggzzz]

I don't really understand how this (adding the two resistors) helps

Some of the stability reasons are ...

*** The base-emitter junction needs to be biased correctly, and an emitter resistor Re makes that mathematically easier and safer for the base-emitter junction.

OK, fine: but how exactly? Without further explanation that just sounds like a lot of handwaving to me.

*** The resistors in the emitter are there - to prevent Thermal Runaway.

OK, kind of like you'd have in a push-pull emitter follower, right? Sometimes referred to as a "cushion".

*** The resistors in the emitter leg of the circuit also generate effectively Current-Series (negative) feedback, and as mentioned - naturally helps prevent Thermal Runaway.

{Current-Series: A 'sample' of the output current generates an effective Voltage (in series, at the input) feedback}

Again, this is insufficient explanation.

Thanks anyhow.

I suggest you devote some time and effort to reading textbooks and literature. The authors of such material will have taken a lot of time and care to make complex and subtle points understandable in detail. Several previous posts have provided pointers.

If you think that someone making a quick post on a forum is going to spend their time poorly duplicating that, then you are deluding yourself.

Personally I will spend some of my life pointing someone in the right direction. I will not spend my life spoon-feeding someone with information readily available elsewhere. People that demand spoon-feeding are time-vampires, and I have little sympathy for them.