# Specific current mirror explanation-source

I am trying to understand the following current mirror, though I haven't found any book that describes it. Is there any source in which I can see how it works and its pros and cons? The drawing is from https://patentimages.storage.googleapis.com/93/7e/df/722ee75b28e661/US4550284.pdf in which they do some sort explanation but I think it need more of description,

Thank you,

## 2 Answers

Here is how I understand this current mirror:

The upper two transistors are naturally designed to operate in the saturation region since the upper left transistor is diode connected (the upper right is not diode connected, but assuming there's a high enough voltage on its drain terminal, it will always be in saturation due to sharing a gate connection with the upper left transistor).

But the lower two transistors must be operating in triode. This occurs because the gates of the lower two transistors are pulled up to the same voltage on the gates of the upper two transistors (unlike in a normal cascode configuration). Thus, the overdrive voltage ($$\V_{GS} - V_{th}\$$) of the lower two transistors is higher than the upper two. Now note that all 4 transistors are running the same drain current, and they are all ideally matched. A higher overdrive voltage on the lower transistors will make them, should they operate in saturation, want to pass more current than the upper transistors. That is clearly impossible, so they must be operating in saturation.

Since the bottom two transistors are operating in saturation they look like the upper transistors, like the $$\R_{DSon}\$$ of the transistors. This has a similar effect as the degeneration resistor in a common-source amplifier, and increases the effective output resistance of the current mirror. Now because the $$\R_{DSon}\$$ of a MOSFET is usually in the $$\m\Omega\$$ region, this degeneration does virtually nothing.

A proper cascode circuit uses a similar configuration to achieve a similar effect, except the upper right transistor would see roughly the $$\r_o\$$ of the lower transistor instead of the $$\R_{DSon}\$$ -- a much greater quantity (many orders of magnitude).

I ran a couple SPICE simulations in LTSPICE to demonstrate the effect.

Here's the schematic. On the left is the standard current mirror circuit. The middle is the "cascode" described in the patent. The right is the textbook cascode circuit. The reference current is set to 1mA, and the mirror is sinking that 1mA into a 5k load.

Now here's the results:

The blue line is the ordinary current mirror, the green line in the patent current mirror, and the red line is the cascode current mirror. The current accuracy of the patent current mirror is significantly worse compared compared to the classic cascode mirror at this output voltage. In this case, the patent current mirror is even worse than the classic current mirror, but this may not be true with all transistors and at all output voltages. A cascode mirror is the clear winner here.

The poor performance of the patent mirror is, I believe, due to the voltage at the drain of the upper left transistor being different to the top right. Thus the bottom left transistor will see a different drain voltage than the bottom right transistor. This is seen in the simulation. This is, of course, the sort of behavior a cascode mirror is designed to rectify.

Lastly to demonstrate that the lower transistors of the patent cascode are in triode, here's the voltages from drain to source:

So their $$\V_{DS}\$$ is on the order of millivolts -- what you'd expect for a triode MOSFET. $$\V_{DS}\$$ is much smaller than $$\V_{GS} - V_{TH}\$$, so they have to be in triode.

• Nice answer. So would yo I say that the parent is just a poor design alternative to cascode current mirror? – Leoman12 Apr 25 '20 at 16:29
• That is exactly what I'd say! I can't see any particular use for the patent cascode mirror since it performs worse than an ordinary mirror with slightly less headroom. In general, I wouldn't use patents as a reliable source of information since it's not in the interests of the authors to teach engineers, but rather to simply protect their intellectual property. – LetterSized Apr 25 '20 at 16:33
• Everybody give an applause to the real cascode current mirror. That’s the real OG. – Leoman12 Apr 25 '20 at 16:38
• Surprisingly often, patents are not necessarily good ideas, but an adequate idea that avoids somebody else's patent. – Neil_UK Apr 25 '20 at 16:44
• How can you say this: The current accuracy of the patent current mirror is clearly awful compared to even a standard current mirror when you are plotting current against time and not looking into the stability of the current vs temperature? You are not using the Z/L ratios in the OP's question either because all your transistors are fixed at device = BSP89 - how can this be any form of comparison? – Andy aka Apr 25 '20 at 17:43

That circuit looks very close to a current mirror called the Cascode MOS current mirror from a book called Microelectronic Circuits 6th edition by Sedra/Smith (page 538). I haven’t studied this circuit enough to give you an explanation but I’ll provide you with the picture of the page. I’ll let the author be the information giver.

Note: Source of information

Microelectronic Circuits 6th edition by Sedra/Smith (page 538)

I give full credit authors Sedra/Smith. Providing information for academic reasons.

• It isn't the same - you have drawn a Wilson current mirror - see my comment under the OP's question. I will say that I haven't seen the one in the question either. – Andy aka Apr 25 '20 at 15:05
• Your answer consists almost entirely of copies of a copyrighted textbook. I think this goes beyond "fair use". – Elliot Alderson Apr 25 '20 at 15:07
• In that document they mention cascode mos current mirror. Though it really looks like some alternate version. – Leoman12 Apr 25 '20 at 15:08
• @andy aka, should I remove it then? Not trying to steal his work just provide references material. – Leoman12 Apr 25 '20 at 15:10
• It's up to you. I think there's a chance you have misinterpreted the question. I'm not downvoting because you are showing a genuine reason to help and that is good. However, somebody has!! – Andy aka Apr 25 '20 at 15:13