0
\$\begingroup\$

I need to "duplicate" a given current: it has to feed two IC's. The context is an OTA (operational transconductance amp) whose gain is controlled by a bias current. Concretely, I am considering the CA3280 (or the modern LM13700). Since I have at my disposal two amplifiers in the same IC, I would like to connect them in parallel in order to halve the noise.

The problem is how to send the same (given) bias current to both amps: if the bias is "incoming" I cannot use a current mirror, since it would behave as a "sink", thus producing a current coming out from the amps. I need a current source instead - kind of "inverting" the current mirror.

Hope the problem is understandable. Any idea?

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Matched PNP or PFET current mirror? \$\endgroup\$
    – John D
    Sep 21 '16 at 20:40
  • \$\begingroup\$ A PNP or PFET current mirror would suffer of the same problem - in order to have a current flowing out at the output, I need a current flowing out from the input as well. Is this correct? \$\endgroup\$
    – Enrico
    Sep 21 '16 at 20:51
  • 1
    \$\begingroup\$ Yes, but if you drive the PNP current mirror with an NPN current mirror you can have a current flowing in to the NPN mirror and a current flowing out of the PNP mirror. All this is best done on an integrated circuit, but if you could find matched transistors or use enough emitter degeneration resistance you could come up with something that works. \$\endgroup\$
    – John D
    Sep 21 '16 at 20:59
  • \$\begingroup\$ This is likely to be an answer to my problem, thanks. If you could add a reference for a schematic and post this comment as an answer I would accept it! \$\endgroup\$
    – Enrico
    Sep 21 '16 at 21:10
3
\$\begingroup\$

(Updated to show 2 output currents, since it seems like that might be what you need.)

Here's what I was thinking assuming my understanding of the problem is correct. (Specific components and values are just the editor defaults, you shouldn't necessarily use those depending on your other requirements.)

You have an NPN current mirror that takes a current sourced by something, and duplicates that current as a current source.

The emitter degeneration resistors help with transistors that aren't perfectly matched.

Of course if you can find matched pairs the circuit would work better. High beta is best to avoid base current errors, or you could explore FETs to eliminate base current errors.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
5
  • \$\begingroup\$ Probably this answers completely my question. I just want to request two more things: 1. may I use a PNP in common base configuration at the input in order to set the input voltage at about ground? 2. is it possible to simplify the circuit or do I really need all these transistors? \$\endgroup\$
    – Enrico
    Sep 22 '16 at 13:57
  • \$\begingroup\$ I don't think a common-base PNP buys you anything, the input is already only a diode drop above ground plus however much drop you want to allow across the emitter degeneration resistor. Not sure if there's a way to simplify it much more than it is, maybe others will have some suggestions. \$\endgroup\$
    – John D
    Sep 22 '16 at 16:07
  • \$\begingroup\$ Sorry, you're right - the input is already close to ground. I was asking to simplify it because in my application there's not so much room. Last question: do you think there exist any IC which can perform this task? \$\endgroup\$
    – Enrico
    Sep 22 '16 at 16:42
  • \$\begingroup\$ There are current mirror ICs available, for example the BCM61B and BCM62B (NPN and PNP). Not sure if you can find a 2 output current mirror, but there used to be transistor array ICs with 4-5 PNP or NPN transistors on the same die, maybe you can find something like that. \$\endgroup\$
    – John D
    Sep 22 '16 at 17:14
  • \$\begingroup\$ Uff... there's the CA3096, which is an array composed of 3 NPN and 2 PNP. But I haven't been able to locate an array with 2 NPN and 3 PNP. Any idea? \$\endgroup\$
    – Enrico
    Sep 22 '16 at 20:57
1
\$\begingroup\$

If you can make the incoming bias current to be 2x, then just feed that to a current mirror and let it split the current.

Following is just an example, all the variations to improve the current mirror are applicable.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • \$\begingroup\$ I think this would not work, since the collector of Q1 is forced to lie at a diode drop below the emitter. So the voltage in the left branch cannot swing. \$\endgroup\$
    – Enrico
    Sep 22 '16 at 13:56
  • \$\begingroup\$ Yes. You can put a resistor or a diode or a zener on the left side to give the right side some voltage to swing up. \$\endgroup\$
    – rioraxe
    Sep 22 '16 at 19:40
0
\$\begingroup\$

You may want a current sense amplifier. Here is one I used in a design: http://cds.linear.com/docs/en/datasheet/6102fe.pdf

You are interested in a unity current gain, i.e. G=Rsense/Rin (see page 7). I'm sure there are other approaches but this one works.

\$\endgroup\$
1
  • \$\begingroup\$ This is a way, but I wonder whether there is something simpler. \$\endgroup\$
    – Enrico
    Sep 21 '16 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.