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I am trying to figure out how to use a Zetex ZDS1009 IC. The chip contains 2 PNP and 2 NPN BJTs set up for current mirroring. I have learned about current mirrors, but I have not seen a 4 BJT mirror.

enter image description here

I have another chip that outputs a current, but it is weak and I would like to isolate it from the load. Any recommendations on how to hook up the ZDC1009 to take the current input and mirror it?

I have a 24VDC supply if needed.

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  • \$\begingroup\$ You don't need 4 BJT's to make a current mirror. The part you found seems to be meant for a specific current sensing application, as shown in the "typical application circuit" shown on page 1 of the datasheet. See my answer at electronics.stackexchange.com/questions/22849/… for a simple current mirror circuit. But without more detail about what you're trying to do I can't say if it would apply to your situation. \$\endgroup\$ – The Photon Apr 5 '12 at 21:12
  • \$\begingroup\$ Yes, please specify the problem more fully: your other chip that outputs a current -- what is its voltage compliance range? And what would you prefer for an output? \$\endgroup\$ – Jason S Apr 6 '12 at 14:51
  • \$\begingroup\$ And is it sourcing or sinking? \$\endgroup\$ – The Photon Apr 6 '12 at 16:40
  • \$\begingroup\$ My other chip outputs 4-20mA at about 10 volts, with worst case of 24 volts. It's load capability is 350 ohms. So I wanted to build a compact, preferably IC, current mirror so I could duplicate the chips 4-20mA twice. \$\endgroup\$ – Z2012 Apr 9 '12 at 17:34
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That's really a dual current mirror hooked up in a balancing circuit. The application circuit makes more sense than the four transistors by themselves:

enter image description here

What's going on here? Transistors 2 and 4 are really just diode-connected; transistors 1 and 2 form one current mirror, and 3 and 4 form another.

The causality chain (not that there really is one in electronics, it's more of a constraint system) is something like the following (Qn = transistor n, Icn = collector current in transistor n):

  • current through R2 lowers the voltage on Q2's base.
  • This pulls Q1's base down and sources current out Q1's collector so that Ic1 * R1 is approximately equal to I * R2. (in general transistors Q1 and Q2 will have a slight mismatch and they won't be exactly equal)
  • This flows through Q4 and into R4, producing an output voltage = I*R2/R1*R4, as explained.
  • It also pulls the base of Q3 up, so that current through Q3 is approximately the same as Ic4 * R4/R3 which approximately = Ic4 if you use the same resistor values for R3 and R4.
  • This current flows through Q2, so that Ic2 approximately = Ic1.
  • By making the currents through Q1 and Q2 approximately the same, this compensates for transistor mismatch, and the part is specified to produce a gain variation of less than 1% with an offset voltage of less than 4mV.

You don't need to use this circuit. There are several ways to make current mirrors, depending on the requirements.

The simplest way (which you've probably read from learning about current mirrors) is just with two NPN or PNP transistors hooked up in the same configuration as the upper half or lower half of this application circuit. For example, see below:

enter image description here

You source current into node X1 (Q4's collector/base), and the current is mirrored by Q3 into its collector current. The balancing resistors (e.g. R3,R4) are there to equalize the current; if they're equal to zero, then the currents form in a ratio of the transistor parameter Is (from the equation Vbe = kT/q * ln(Ic/Is)) and could be off by quite a bit. Adding the balancing resistors so that you see a drop in the 50-200mV range makes them a lot more equal. There are other configurations that use a 3rd transistor to remove errors in base current (finite beta) and output compliance.

Current mirrors are best used when you have a unidirectional current you want to flow somewhere, and you need a high-bandwidth current source characteristic, i.e. the output voltage can vary but you still want constant current. Transistors are great for that.

It might be the case, however (we'd have to know from your requirements) that what you really want is a transimpedance amplifier. In other words, you have a source of current that you want to convert to a voltage. If this is the case, use this circuit (from one of Bob Pease's columns):

This will work for bidirectional currents.

If you really do want a current output, and it's unidirectional, you can use a cascode circuit:

http://upload.wikimedia.org/wikipedia/en/2/27/MOSFET_Cascode.png

M2 and its voltage sources should be replaced by whatever your weak current source is. M1 is a current buffer (on the gate, use whatever voltage is required to turn the MOSFET on in normal cases, e.g. whatever Rdson is specified). Don't add Id; and attach the load to the node marked "Out".

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  • \$\begingroup\$ Yes I need a unidirectional current flow, and the voltage will vary but in need the current to be stable. Thank you for breaking that chip down for me. I was hoping for a current mirror chip but I might just get a chip that has uncomitted BJTs. \$\endgroup\$ – Z2012 Apr 10 '12 at 15:54
  • \$\begingroup\$ Nobody really sells matched BJTs anymore for an affordable price, even though they're a basic building block build into most bipolar opamps. It's sad. But if you have emitter resistors, the sensitivity on transistor matching is greatly reduced, at the cost of decreased dynamic range. \$\endgroup\$ – Jason S Apr 10 '12 at 16:02
  • \$\begingroup\$ @Jason : Didn't quite get your typical application circuit, could you please explain the Vsense calculation part. Anybody please. \$\endgroup\$ – Durgaprasad Mar 20 '13 at 18:46
  • \$\begingroup\$ Where can I find a more detail explanation about the typical APP circuit you have above, there are some questions I have to follow the full operation: 1) How is that the current in Q2's base lowers by the current in R2 2) Where can I find more information about this kind of circuit or the application note where you obtain the image above? \$\endgroup\$ – Mario Arenas Sep 18 '18 at 20:31

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