I am building this VCO in an attempt to study it and I'm having trouble understanding the references section seen bellow.

enter image description here

I have used a KIA431 that I 'stole' from a broken switching power supply on the breadboard in place of the specified LM4041CZ-ADJ and it is giving me the +5V I was expecting. I am using the LM336Z-5.0 as a substitute in Multisim because I couldn't find a part model. Besides this, the rest of the circuit is quadruply checked to be as specified in the schematics and as seen in the Multisim snippet.

To make this as clear as possible, I need the following things: a) to know what this op amp/bjt configuration is meant to achieve and why a simple buffer (or amplifier with a gain of 2) wasn't used to acquire the +-10V reference b) the name of the bjt configurations c) why I am reading different voltages than what is specified (+-10V). I am reading +13.85V and -13.18V on the breadboard and +12V and -11.3V on the simulation

One thing that's also really bugging me to understand is that in both cases we see one diode drop difference between the positive and negative output voltages.

P.S. I am assuming that the only reason this op amp/BJT configuration is used is to feed the CA3280Es at the Tri to sine converter on page 6 of the linked PDF. I would appreciate a confirmation on my assumption (or the opposite) and a link to any resources that might help me get my head around this. (This is not part of the question in the title so it will not be considered in my answer acceptance.)


"I have used a KIA431 that I 'stole' from a broken switching power supply on the breadboard in place of the specified LM4041CZ-ADJ and it is giving me the +5V I was expecting. I am using the LM336Z-5.0 as a substitute in Multisim"

That's the root of your problem. I've never heard of the KIA431, but that LM336 is quite a different part to the LM4041. The latter allows you to use resistors to set the reference output voltage from 1.2V to 10V, while the LM336Z-5 generates a 5V reference that you can trim by only a few hundred millivolts either way.

With the pot in the midpoint the LM4041 will generate a reference of about 2.5V, which the opamp circuits will multiply by 4 to get +10 and -10. With the LM336Z-5 in its place, you'll get a 5V reference, which the opamps will try to multiply to +20 and -20 but fail because they can't get anywhere near it given the supply voltage and the circuit configuration.

Try using an LM336Z-2.5 instead if it's available in your simulation.

Oh, and the transistors provide current limiting on the output. The first is a current amplifier for the opamp output and it drives the load through the 10 Ohm resistor. If the current flowing through the resistor reaches about 600mA the second transistor will start to turn on, reducing Vbe of the first one and limiting the output.

  • \$\begingroup\$ Thank you. You have managed to read between the lines in my question and went straight for what the problem was. Your answer also explained to me in simple words the results i am getting - hitting the limits of the op amp. I was split between accepting your answer and HatimB's but the addition of the last paragraph made things really clear for me. \$\endgroup\$ – Schizomorph Mar 16 '17 at 15:48
  • \$\begingroup\$ The KIA431 is the KEC version of the TL431 adjustable shunt regulator. Going back to its datasheet to see if it can be set to 2.5V. \$\endgroup\$ – Schizomorph Mar 16 '17 at 15:57
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    \$\begingroup\$ Ah, yes, found the datasheet. You can use it with some minor tweaks.... \$\endgroup\$ – Finbarr Mar 16 '17 at 16:08
  • \$\begingroup\$ Confirmed. I took out R2 and swapped the trimmer for a 5k used as a variable resistor. I was able to set it to 3 decimal digits of accuracy quite easy. Thanks again. \$\endgroup\$ – Schizomorph Mar 16 '17 at 17:19

a) You'll have a higher output current capability with this configuration compared to the one without the op amp alone.

b) I don't know if this configuration has a specific name. I would call it "A BJT with current limitation"

c) Are you sure about R7/R5 and R9/R4 combinations? Those look like gains of 4. I'm guessing that the op amp can't output more (less) than a defined voltage (which is lesser than +15V or higher than -15V depending on which op amp you're looking at), not high/low enough to output the full +15V/-15V of your rail. That being said, you should make R7=R5 and 2xR4=R9 to get the targeted gain of 2.

Regarding the diode voltage drop you talked about, all I'm seeing on the simulation is an oscillating voltage (Vpp=11.3V and Vdc=-). Let us know if the problem persists after the gain correction to think about it.

  • \$\begingroup\$ Thanks HatimB. I had to accept Finbarr's answer instead of yours because he spotted the root of my problem. I appreciate your reply nevertheless. \$\endgroup\$ – Schizomorph Mar 16 '17 at 15:50
  • \$\begingroup\$ We're saying pretty much the same thing. I didn't argue with your choice of components because this schematic could work again just by adjusting your resistors. Anyway, I'm glad that these answers helped you to solve your problem ;) \$\endgroup\$ – HatimB Mar 16 '17 at 17:05

I agree with HatimB the opamp VDDs are a problem.

I also suggest you add a feedback cap, to change the phaseshifts of your opamp; a load resistor may also help, by reducing Rout of the emitter follower:


simulate this circuit – Schematic created using CircuitLab

Let's examine the world of stability. We have two delays in this circuit: the opamp, and the transistor. First we'll operate the transistor with 1uA current, because 1uA is a likely leakage current through an electrolytic on that emitter. The Rout (from the emitter) of a bipolar is 0.026V/Ie_amps, thus 0.026/1uA = 26,000 ohms. The delay of 10uF and 26,000 ohms is 0.26 seconds. This very slow behavior, being monitored by the OpAmp through R3, produces an oscillator, unless we provide a sneaky way to monitor the OpAmp's output, through C2.

What if you add an intentional current through that emitter, to greatly reduce the emitter's Rout? Make R4 be 10K ohms. Rout is 0.026v/0.001 = 26 ohms. We've sped up the delay by 1,000x to 0.26 milliSeconds. But the OpAmp is out at 1MHz, and acting very impatient with such long delays. Again, use C2; if R2 and R are each 20,000 Ohms, the Requivalent on (-) OpAmp pin is 20K||20K or 10Kohm. Strive to get the feedback response up to 1MHz, or 160nanoSeconds. Using $$Omega* Tau = 1$$, with $$Omega = 2*pi*frequency$$, and $$Tau = R*C$$, I know 10Kohm & 100pF -> 1uS, thus 10Kohm and 16pF -> 160nanosecond. Bigger caps are OK.

Here is BODE plot, along with OpAmp with NO Cfeedback and a large Cload. You can alter the params, and rerun yourself. enter image description here

  • \$\begingroup\$ This is used for a voltage reference, therefore DC and I can't understand what phase difference means in terms of DC. Is it correcting 'delay' in cases where the input voltage may vary (which it normally shouldn't)? \$\endgroup\$ – Schizomorph Mar 16 '17 at 16:03
  • \$\begingroup\$ The OpAmp responds out to 1MHz or faster. Delays --- that 10uF capacitor on the transistor's emitter --- become phaseshifts, and the world of BODE plots and feedback stability need attention. To provide that attention, add the C2 as I showed in the schematic. 1UF may be too big. \$\endgroup\$ – analogsystemsrf Mar 16 '17 at 16:46
  • \$\begingroup\$ Thanks. I think I don't have the theoretical base to understand this yet but I'll look into it and try it on the breadboard. \$\endgroup\$ – Schizomorph Mar 16 '17 at 17:11
  • \$\begingroup\$ You've now got a bit of theoretical basis. Go for it. \$\endgroup\$ – analogsystemsrf Mar 20 '17 at 9:35
  • \$\begingroup\$ That's great! Thanks for updating your answer. I think I got it (but I'll read it again tomorrow after some rest). Basically, it is about 'telling' the op amp early enough if there's any changes in voltage right? Can I ask you one last question? Where did you get the 1MHz figure? Is it the Unity-gain bandwidth? If so, it appears to be 2.6-3.2MHZ for the TL05xC. \$\endgroup\$ – Schizomorph Mar 20 '17 at 22:56

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