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I have the following circuit and I want to reduce the current consumption to the lowest possible. Basically, I have a series voltage regulator whose output is connected to a transistor to switch on the OpAmp using the micro-controller(uC).

enter image description here

Prior to connecting 11V (from emitter of Q1) to R3 , I get \$120\mu A\$; but once the 11V is given to R3, the current shoots up to \$1.2mA\$ even before triggering the base of Q2. Where does the current leak and How do I reduce the current?

I Found a similar question, in which they mentioned that the \$V_{BE}\$ of Q1 increases by \$V_{CE}\$ of Q2. Do I need to alter R2 for that?

Thanks for the help!

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    \$\begingroup\$ If you're after reducing power consumption, why use the antique LM358 series? \$\endgroup\$ – Marcus Müller Jan 6 '17 at 11:34
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    \$\begingroup\$ Your circuit doesn't make sense. D2 is a Zener 1.225 V Zener diode – meaning that the voltage between V+ and V- can never exceed 1.225V unless you drive that diode with a large voltage, in which case the diode will overheat and just fail. \$\endgroup\$ – Marcus Müller Jan 6 '17 at 11:36
  • \$\begingroup\$ Actually, I need to reduce the cost too. But even then, The OP-AMP will be ON only after it is triggered so I am interested in the Standby current (before triggering). D2 is a 9V Zener. I mis-typed the name. \$\endgroup\$ – Injitea Jan 6 '17 at 11:38
  • \$\begingroup\$ as said, your circuit doesn't make sense, D2 is definitely not doing anything useful here – unless it's not actually an AD1850 (1.225 precision voltage reference diode, see its datasheet). \$\endgroup\$ – Marcus Müller Jan 6 '17 at 11:39
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    \$\begingroup\$ Again, if you need to reduce the idle current of your Opamp, ditch the ancient LM358, and use an opamp with lower power consumption (probably FET-based) and an enable pin. \$\endgroup\$ – Marcus Müller Jan 6 '17 at 12:02
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So, I think I can conclude what's wrong with the circuit in an answer after having discussed it in the comments:

  • You say you need to minimize current consumption, yet you burn 13V (24 V - 11V) in your linear power supply. Don't do that. Use a switch mode power supply (step-down) to do e.g. 24 V -> 12 V (effectively halving the power consumption on the 24V rail) and do the rest (12 V -> 11V) using a LDO (low drop-out) regulator, if you need the exactness. You probably don't – opamps do have a supply voltage rejection that should eliminate the effect of supply voltage uncertainty on your signal.
  • D2 is an especially bad idea since it must lead to additional current being wasted, at least in "ON" state. Also, as Andy pointed out: 9V is even higher than what the LMX358 is specified for! So, this is even twice a wrong choice.
  • if you want to reduce current draw, use MOSFETs instead of bipolar junction transistors
  • A simple high-side MOSFET switch would do the job without you burning energy (and thus, 24V current) on the forward voltage of your 1970s BJTs as well as their base currents
  • LMX358 still has a relatively high quiescent current and no disable/enable pin. You could go to TI.com, and click through to "amplifiers -> Opamps -> Ultra low-power opamps", and pick one that directly works off the voltages you have, has low quiescent current, and a high supply voltage rejection (so you can directly connect it to your SMPS instead of the LDO). In the best case, it would also have an enable pin, so you can simply shut it down when you don't need it. In your case, the simplest solution would be to replace the LMX358 by an LP358, which isn't even more expensive, usually.
  • I highly doubt your requirements for low current if you have a linear supply, and some unnamed microcontroller that you use, as well as your 30 Ω R4. Also, 24V doesn't come from "anywhere". If this is from a lead-acid battery, or an array of alkaline batteries, or an array of NiCd/NiMH rechargables, compare your currents to the self-discharge currents. You'd be surprised. If this is coming from a grid (220V/110V...) adapter, then ignore the quiescent current, since the conversion efficiency for small loads will be terrible, anyway. If this comes from a solar cell, same applies.

Furthermore, since this is an analog signal processing circuit, there's a distinct lack of decoupling capacitors (and a distinct lack of consideration for the current these leak).

As usual, pick the right components for your job based on what they need to do. If you can restrict the bandwidth you need from the opamp to a couple kHz and the gain to a couple V/V, you can easily drop in a nA-quiescent-current opamp.

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  • \$\begingroup\$ Yup, some differences, but lots of similarities too. +1 \$\endgroup\$ – Olin Lathrop Jan 6 '17 at 12:26
  • \$\begingroup\$ Marcus, it's an LMX358 - it is typically not run on supplies higher than 7 volts. \$\endgroup\$ – Andy aka Jan 6 '17 at 12:29
  • \$\begingroup\$ @Andyaka hopefully, fixed it. \$\endgroup\$ – Marcus Müller Jan 6 '17 at 12:34
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    \$\begingroup\$ And don't forget the absurdity of running a 12 volt zener with a nominal 60 uA, That's only about 2-3 orders of magnitude too low. \$\endgroup\$ – WhatRoughBeast Jan 6 '17 at 12:49
  • \$\begingroup\$ Also R6 with 11V on one end and presumably a microcontroller IO (3.3V, 5V?) on the other will bias the micros input clamp diodes into conduction I could easily see a few mA going that way. Step back and look at the overall situation, but I would be investigating such things as CMOS opamps if the performance envelope suits the application, you might get away without switching it if you go there. \$\endgroup\$ – Dan Mills Jan 6 '17 at 14:42
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There are so many issue, it's hard to decide where to begin. To be clear, here is the circuit you propose:

Apparently you ultimately want to switch the low side power of a opamp, with the high side at about 9 V when on. Your circuit raises the high side of the power to about 12 V when off, but I'll assume that's just a accidental byproduct of the strange design and not a real requirement. The only available supply is +24 V.

Issues with your design are:

  1. You show the opamp as being a LMX358. As Andy pointed out, it has a maximum supply voltage of 8 V. This is apparently why you try to regulate down the 24 V. However, regulating it down to 9 V makes no sense when the opamp can't handle more than 8 V.

  2. The LM358 (not "X" in part number) can take up to 32 V supply absolute maximum, and 30 V is used for some of the operational specs. It can therefore certainly handle 24 V directly. There would be no need to create a separate 11 V or 9 V supply.

    However, the LM358 is a old opamp and draws up to 2 mA just for internal use. There are newer opamps that should be usable here.

    Ideally, use a opamp that can run from 24 V directly. That simplifies things considerably.

  3. If you're really worried about power consumption, then don't use a BJT low side switch since you have to keep feeding it base current. A N channel MOSFET requires no current to keep it on. The IRLML6344 is one possibility since it can be switched directly from a 3.3 V digital signal, and it can handle up to 30 V.

  4. Step back and explain what analog signals this opamp gets as input, and what it's supposed to produce from them. The best solution may be to step back a couple of levels and solve the problem with a altogether different topology.

  5. Power minimization needs to be consider with the whole circuit in mind. How low does the current really need to be? If this is battery operated, what is the self-discharge current of the battery? How much are other parts of the circuit drawing already?

    For example, it makes little sense worrying about a extra 500 µA when the battery self-discharge is 3 mA, or the microcontroller draws 10 mA. Again, you have to look at the whole circuit, and perhaps different topologies, not just different parts.

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  • \$\begingroup\$ ha, we basically wrote the same answer. Here, have an upvote! \$\endgroup\$ – Marcus Müller Jan 6 '17 at 12:18
  • \$\begingroup\$ Olin, it's an LMX358 - it is typically not run on supplies higher than 7 volts. \$\endgroup\$ – Andy aka Jan 6 '17 at 12:29
  • \$\begingroup\$ @Andy: Oops, I didn't catch that. I'll go update my answer now. \$\endgroup\$ – Olin Lathrop Jan 6 '17 at 14:16
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The LMX358 has an absolute maximum power rail specification of 8 volts so using a 9 volt zener diode is a poor choice. The circuit is fatally flawed (because the voltage applied will be greater than 8 volts) and this needs to be fixed first.

So, in the absense of any data about what the op-amp is intended to be used for there is no point trying to reduce power supply current. Fix the design by specifying what the design is intended to do, show the connections to the LMX358 and then we can proceed.

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