I am trying to bias the lm13700 Input amplitude bias current pin with a current source rather than using a voltage source and current limiting resistor as shown in all the data sheet schematics, but cannot figure out how to connect a current source up to the pin. Seems like it should be simple, but I am getting confused with how to connect The IABC pin as the "load" of the current source.

Here is a forum post where Andy aka describes the op amp/NPN transistor combination voltage to current converter I am trying to use: forum post on voltage to current conversion

As a shortcut, here is an image of the current source circuit: Current source circuit

If I were to use this current source to drive LEDs, it seems straight forward: put all the LEDs in series between Vcc and the collector of the NPN transistor. Each of those components have a positive and negative side and together they all become the load. How would I use this current source though to connect up to the IABC pin though of the LM13700? Unlike an LED, the pin appears to have just one node and no obvious way to complete the "load".

Here is a link to the LM13700 data sheet: LM13700 data sheet

Thanks, Dave

______Additional information below on questions including visuals from LTSpice______

I didn’t realize that my initial circuit was actually a current sink rather than a current source. I took the circuit provided by Spehro and modeled it as best I could into LTSpice with the LM13700. It appears to be working (somewhat), but I discovered some more unexpected things that I’m hoping you or someone else may be able to shed some light on. Figure 1 is a picture of the entire LTSpice schematic to test the Voltage Controlled Current Source (VCCS). I’ve also labeled a few key nodes used to plot output signals.

Figure 1 - LTSpice Schematic

Ultimate Goal: I’d like to modulate the amplitude of the Carrier (V2) using the the LM13700. To do this, I’d like to establish a linear VCCS where the input at V1 (Modulator) goes between the limits of 0v and +5v, and the output of the VCCS, which for purpose of this post I have been measuring through Ix(U1:Ibias), would be between 0mA and 1mA respectively. I used the value 5k for R5 to make this relationship work. I got something wrong in here though because my voltage readings are unexpected in the LTSpice simulation. My application uses a dual rail supply.

Question 1: (Reference Figure 2 below) How come the following test voltages at V1 are yielding unexpected current outputs when measuring current at Ix(U1:Ibias)? I think the equation for this VCCS would be Ic(Q1)=V1/R5. I suspect that the PN junction in the Q1(2N3906) might be causing some of my issues, though I am not sure.

  • Test 1: V1 = 0V; I would expect the current leaving the VCCS to be 0mA, but I am measuring 1.8mA going into Ix(U1:Ibias).
  • Test 2: V1 = 5V; , I would expect the current leaving the VCCS to be 1mA, but I am measuring 800uA or .8mA going into Ix(U1:Ibias).
  • Test 3: V1 = 2.5V; I would expect the current leaving the VCCS to be .5mA going into Ix(U1:Ibias), but instead it is 1.3mA.

Figure 2 - Output Waveform

Question 2: (Reference Figure 2 above) How come the current at Ic(Q1) is the opposite polarity of Ie(Q1)?

Question 3: (Reference Figure 3 below) How come the current at Ic(Q1) is the opposite polarity of Ix(U1:Ibias)? I would think that they would be exactly the same as they are connected by a wire.

Figure 3 - Output Waveform

I think that’s everything. Many thanks for taking the time to look at this. If you’re into LTSpice, here is a link to the SPICE files required to view this simulation using LTSpice 4.23e. Just keep all files together in one folder when running the schematic. I used my LTSpice base folder with the executable.


1 Answer 1


The circuit you show is a current sink. For the LM13700 Iabc input you need a current source.

The above circuit just needs to be 'flipped'. Use a PNP transistor. The voltage source is relative to Vcc now, so you might use a shunt reference such as an LM4040 or TL431 if the power supply is deemed not accurate enough.

Be sure to respect the common mode voltage range of the op-amp and don't use too low a reference voltage.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thanks, Spehro for your answer. Sorry it's taken me a few days to respond. I wanted to do some additional research and provide a response with visual aids that was as clear as possible. I have edited my original post with the new information. I think I am within limits of the common mode voltage range of the LM741; the closest field I could find on the datasheet, "Input Voltage Range" seems to go from -12V to +12V. I am unsure of your comment about not using a VRef that is too low. Why is this? \$\endgroup\$ Commented Aug 20, 2015 at 19:57
  • \$\begingroup\$ The input voltage range goes from +/-12V when you have a +/-15 supply. The 741 won't work reliably from a +5 supply. You can't get closer than 3V to the positive supply and expect it to work reliably. You also can't get closer than 3V to the negative supply, so there is no input voltage that is acceptable with a 5V supply. If you have a higher voltage supply, the reference has to be minimum 3V, so the output compliance will be up to about 1.9V, which is probably acceptable for this application. A better (more modern) op-amp would be good. \$\endgroup\$ Commented Aug 20, 2015 at 20:04
  • \$\begingroup\$ I think I understand your comment now. If I were powering the op amp from a 5v supply, then in that example it wouldn't allow input voltages to be more than 3v because they get too close to the power rail and it isn't a rail to rail input op amp. In my particular circuit though, I am powering the 741 op amp from a dual power supply made from two 9v batteries. The LTSPICE schematic above shows this. So if the op amp is powered by a +-9v supply, 0-5v should be well within the input voltage range, right? -- thanks. \$\endgroup\$ Commented Aug 20, 2015 at 23:35
  • \$\begingroup\$ yup, that's right \$\endgroup\$ Commented Aug 20, 2015 at 23:44

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