please help me decide if this will work:

Primary objective: Amplify 0~3v input signal to 0~12V output signal.

enter image description here

I understand that this whole setup is a bit unusual, since I can simply connect the +input of opamp to 12VDC, and with my resistors' values, 3v input could be amplified to 12V (I'm 80% sure). But nevertheless I want to know if this odd scheme will work? I think I'll have to be very careful with my OpAmp output, which dictates the 3904?

Edit: My most sincere apologies there SHOULD BE a pullup resister between the collector and the 12V, for some reason this slipped through my double check, incredible, I'm new to eagle and the copy paste in it isn't that intuitive.

The resistor has a value of 2K.

Edit #2:

Sorry for the delay but a few things to clarify, apologize again if I overlooked your question, there are too many and I'm kind of really busy:

  1. 200mA is supposed to be the current drawing in. This whole setup is kind of to sink current from lower-tier components that will be connected to this interface, so 12v/2K = 6ma isn't what I'm talking about.

  2. The resistor should be 90k/30k, as some of yous suggested, my mistake.

  3. Oscillation: I believe your concerns are quite legit but please be advised that the input will be mostly "static" voltage signals, meaning it rarely changes more frequently than 5 times per second, so I'm skeptical that it will be that huge of a problem, I could very well be very, very wrong.

    • Matt
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    \$\begingroup\$ It is not clear what you are doing. Do you have a +12 volt power supply? \$\endgroup\$ Commented Apr 12, 2017 at 2:57
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    \$\begingroup\$ It won't do anything at all useful. The output will always be +12 unless the power supply loses the fight destroying Q4. \$\endgroup\$ Commented Apr 12, 2017 at 3:10
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    \$\begingroup\$ Just use a 12V op amp non-inverting amplifier with resistor ratio 3:1 to get a gain of +4. There are many problems with your general approach, not the least of which will be compensation to prevent oscillation, since you are intending to add gain. \$\endgroup\$ Commented Apr 12, 2017 at 3:43
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    \$\begingroup\$ It is important to write better specs before presenting a design. What load, current and BW? This transistor pull down only sinks current and needs a pullup R and Vin+- reversed, but will be less stable (due to added GBW phase shift) than opamp which can drive 600 Ohms \$\endgroup\$ Commented Apr 12, 2017 at 3:45
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    \$\begingroup\$ @TonyStewart.EEsince'75 assuming the transitor has a resistor to +12V it inverts the output of the op-amp so over-all the feedback is negative, and but for C2 the circuit should be well-behaved.. \$\endgroup\$ Commented Apr 12, 2017 at 4:12

3 Answers 3


No, your circuit will not work. The main problem is that 12V is connected directly to Vout. And Q4 is going to short 12V to ground (which will destroy Q4).

Look at this schematic. Just for inspiration.


simulate this circuit – Schematic created using CircuitLab

I changed value of R2 from 120kΩ to 90kΩ. It is wrong in question too. If OP wants to amplify from 0-3V to 0-12V than there should be 1:3 ratio for R1:R2

This circuit makes Vout = 4* Vin

If you increase Vin, output of U3 will go lower, which will make mosfet M1 more open (Vgs is more negative = Rds is lower) and thus Vout increases. It will increase until voltage at non-inverting input of U3 is equal to inverting input (which is Vin).

If you increase Vin, output of U3 will go higher, which will make mosfet M1 less open (Vgs is less negative = Rds increases) and thus Vout decreases. It will decrease until voltage at non-inverting input of U3 is equal to inverting input (which is Vin).

  • \$\begingroup\$ Unfortunately, OP stated in comments that most of his proposed circuit is a holdover from a previous revision of a project, and he's supposed to reuse it with as little modification as possible...which rules out this option, or my original suggestion to simply use a non-inverting topology to start with. \$\endgroup\$ Commented Apr 12, 2017 at 7:19
  • \$\begingroup\$ @RobhercKV5ROB I do not think it is possible to reuse only those components and make working circuit that comply requirements at the same time. \$\endgroup\$ Commented Apr 12, 2017 at 7:27
  • \$\begingroup\$ You switched the inputs of the op-amp.. \$\endgroup\$
    – m.Alin
    Commented Apr 12, 2017 at 7:28
  • \$\begingroup\$ @m.Alin no, if you look closely, he jst drew his opAmp in the other orientation...the Vis is still connected to U3's inverting input, and feedback is to the non-inverting. \$\endgroup\$ Commented Apr 12, 2017 at 7:31
  • \$\begingroup\$ @m.Alin I also changed NPN BJT for P-channel mosfet. I think I wired opamp correctly. \$\endgroup\$ Commented Apr 12, 2017 at 7:31

Your gain is 1+ since you have inverted the feedback the Vin- is now the positive input so Av=1+R2/R1 assuming you use something 1k Pullup and you REALLY NEED the transistor to pull down a charged cap with high current and small pullup current.

This a poor inverting negative peak detector with storage cap and missing pullup resistor with a gain of +5?

Where are the rest of your design specs?


It's a good start, but I see two flaws.

1: your output terminal is connected directly to 12VDC, you probably forgot to draw a resistor between the 12VDC and the transistor. given your choice of 2K for the base resistor, 1K seems about right in this place.

2: C2 between output and ground seems likely to cause trouble (amplifier oscillation). you probably don't want it. or perhaps you want it between 12VDC and ground.

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    \$\begingroup\$ It doesn't look pretty at all. Output is connected directly to 12V and transistor Q4 is going to short 12V to ground. \$\endgroup\$ Commented Apr 12, 2017 at 4:48
  • \$\begingroup\$ did you read past the first sentence? I'm assuming that missing resistor is a draughting error, because without it the circuit makes no sense at all. \$\endgroup\$ Commented Apr 12, 2017 at 4:51
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    \$\begingroup\$ So, your first sentence was just ironic? \$\endgroup\$ Commented Apr 12, 2017 at 4:54
  • \$\begingroup\$ The guy's either a beginner or he's "having a bad day" and but for the resistor separating the collector and output from 12V and C2, it looks good to me. he got the resistive divider right, and connected it correctly to the non-inverting, input. even included power supply by-passing capacitors. He's clearly made an effort, and some praise is due. \$\endgroup\$ Commented Apr 12, 2017 at 5:04
  • \$\begingroup\$ If somebody is beginner, he needs honest feedback. Yes, he made pretty good effort. But the result schematic is not good, this is the honest information. Your suggestion to add resistor between 12V and collector doesn't seem good to me, because you don't know what amperage is required and you will not meet the requirements for 0-12V output. \$\endgroup\$ Commented Apr 12, 2017 at 5:25

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