I thought I understood transistors but now I'm confused again.

I have a basic opamp circuit to amplify the input of a electret microphone. The output of the opamp will vary within 1-4V. Now I want to connect this output to a load (laser) and this modulate the light intensity via the mic.

The driver of the laser is not a constant current one, and output does change if input changes, thus modulation is possible.

The thing is that I cannot connect the laser directly to the output of the opamp circuit because the current flow will be to high (+-300mA), so this means I need to use a transistor right?

So I thought I can just connect this output to the base pin of the transistor(2n2222), the load from +5V to collector, and emitter to ground. The base input then controls the current over the proportionality.

So basicly a input of 1-4V on base must control the collector current also from 1-4V but allowing a larger current to flow.

I read that the base-collector current must not be the same and this shuts down the transistor. I tried a basic circuit and obviously it didn't work.


  • \$\begingroup\$ It will be much easier to understand what you have done if you include a schematic. While editing the question, press ctrl-M to open the schematic editor. \$\endgroup\$ – Phil Frost Jun 6 '13 at 20:55
  • \$\begingroup\$ books.google.com/… \$\endgroup\$ – Phil Frost Jun 6 '13 at 21:09
  • \$\begingroup\$ I've always found dimming lasers to be somewhat difficult, until I started controlling them w/ PWM \$\endgroup\$ – Scott Seidman Jun 6 '13 at 21:14
  • \$\begingroup\$ @ScottSeidman I'm sending voice over the laser and trying to keep sender and receiver circuits as basic as possible. will look into PWM, thanks \$\endgroup\$ – Roelf Daling Jun 6 '13 at 21:43
  • \$\begingroup\$ @PhilFrost Sorry for not adding a circuit, just added one. Thanks for ctrl-M tip! \$\endgroup\$ – Roelf Daling Jun 6 '13 at 21:43

A couple of circuits that spring to mind. The first was actually for controlling the light output from a laser (amongst other things). Incidentally the laser was used to shine at a turbine blade spinning in a turbine and the optical cable had a return path for reflections back to a photodiode to monitor the timing of the blades: -

enter image description here

The crux of this circuit is the emitter resistor and the feedback control from the op-amp. If you have 1V on your +Vin input (5V before pot divider) then providing you have a collector load, the current will be 100mA. If you have 2V, the current through the laser will be 200mA. With 4V on the +Vin input there will be 400mA through the laser.

Make the emitter resistor bigger for lesser currents and of course dispence with the pot-divider on the +Vin input.

The 2nd circuit is this; it's a power output stage to an op-amp based arounf an NPN BJT transistor and feedback to the opamp to provide regulation: -

enter image description here

Quite simply the feedback on the op-amp tries to maintain the emitter at precisely the same voltage as Vin. This will give you the power for driving your laser.

Things to watch - the laser forward voltage needs to be accounted for as does the heat dissipation in the BJT and the emitter resistor in the 1st example.

  • \$\begingroup\$ I can't get this to work. Now I'm using this sparkfun.com/products/9964 as my signal. Its already amplified and outputs a bias of 2.5V. How do I connect this to transistor to give me the same voltage out but with a 350mA or more potential current flow? \$\endgroup\$ – Roelf Daling Jun 7 '13 at 0:56
  • \$\begingroup\$ @RoelfDaling which of the two circuits did you try to get working but couldn't? What op-amp device did you try? What power supply voltage did you use? \$\endgroup\$ – Andy aka Jun 7 '13 at 8:24
  • \$\begingroup\$ @RoelfDaling I know it's an old comment, but it might be helpful. I had the same problem. You need to see your opamp maximum current output. Usually it's few mA. Multiply that by beta of the transistor and if it's less than 350 then you need either higher beta (hfe in datasheets) transistor or darlington, or a mosfet at the output. Just keep in mind that darlington configuration will have bigger voltage drop - 1.2 to 1.5V. With a mosfet the gate charge (capacitance) could be a problem. You may need a series resistor on the gate. The more powerful the mosfet, the higher capacitance (usually). \$\endgroup\$ – NickSoft Dec 8 '16 at 8:00
  • \$\begingroup\$ @NickSoft if you are going to recommend a MOSFET and also recommend a gate resistor then you will need an integrating capacitor directly on the op-amp and an input resistor to the inverting input to prevent it turning into an oscillator (as happens quite often). Most op-amps get by without the gate resistor and don't need the other two components I mentioned. \$\endgroup\$ – Andy aka Dec 8 '16 at 8:35
  • \$\begingroup\$ @Andyaka You are right. I had a lot of problems with oscillation when using output MOSFET. Maybe the best solution is to have low power mosfet - 2N7002, driving the base of a BJT (Drain to Vcc, source to base). It could be PMOS with source to Vcc, drain to base, but NMOS are cheaper, better and more available. Still a cap from -IN to the optamp output + output resistor won't hurt. \$\endgroup\$ – NickSoft Dec 12 '16 at 9:11

If your goal is simply to increase the current sourcing capability of an op-amp, the canonical way to do it is:


simulate this circuit – Schematic created using CircuitLab

A transistor used in this configuration is known as an emitter follower. The configuration is also known as "common collector", since the collector is shared between the input and the output of the transistor.

Your feedback elements go in the dotted box according to what you want the op-amp to do, just as if you hadn't added the transistors.

By incorporating Q1 in the feedback loop, the op-amp's gain will make significantly reduce any non-linearities in the transistor. Further, if you already have an op-amp circuit that does what you want, except you need it to source more current, adding the transistor this way won't significantly alter the function of your circuit.

If you also need to sink current:


simulate this circuit

This configuration, where one transistor sources and the other sinks current, is known as push-pull.


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