Hey everyone,
I’ve recently built myself a LM13700 based VCF supplied by a 9V battery and in the process started wondering about the pros and cons of different ways to create a bias voltage for this circuit.
I’ve followed this schematic („MS-420“ by CCTV) using a 3.3V zener diode to create a 3.3V bias voltage. In my experiments the actual voltage created by this circuit is about 2.5V:
Another very similar schematic floating the internet uses two led’s in series as some kind of voltage regulator which in my experiment create a bias voltage of around 3.4V (much closer to the voltage aimed for in the former schematic):
Furthermore in this video it is suggested, that a simple 100k/100k voltage divider can be used instead of any diodes, which of course would reliably result in 4.5V (or half of whatever voltage the battery actually provides): https://www.youtube.com/watch?v=AqhvmgtHi_8
Can someone enlighten me, why someone would choose to use diodes (LED or zener) instead of a voltage divider? Are there any benefits to this method? And why not use an actual voltage regulator (e.g. this one: LP 2950 ACZ3,3: Voltage regulator, +3.3 V, 0.1 A, TO-92 at reichelt elektronik) which also wouldn’t break the bank?
Don’t take my word for it, I’m not an expert with Zeners, but I think the 10k resistor is too large. The voltage drop across the resistor is 9 - 3.3 V = 5.7 V and with a 10k resistor the current is 0.57 mA. But with the BZX79-C3V3 the datasheet says I_Z = 5 mA. So I’d say you need more like a 1k resistor.
The possibilities you mention give different precision for the voltage. So which one to use depends on what kind of variations in the bias voltages you can tolerate.
The 3.3 V Zener, per the datasheet, should give 3.1 to 3.5 V.
A voltage divider would give different voltages depending on both the battery voltage and the output current. For instance with 5.1k on the top and 3k on the bottom, with a 9 V battery the output voltage is 3.33 V if there is no output current at all. But that sags to 3.31 V at 10 mA and 3.14 V at 100 mA, and if the battery drops to 8 V the zero current output drops to 2.96 V.
That voltage regulator gives 3.267 to 3.333 V. But a voltage regulator is designed to supply current at a given voltage to power a circuit. Here you just need to offset a voltage, not to power anything, and for that a voltage reference gives more accurate voltages (if it matters). For instance the LM4040. It comes in 2.048 V, 2.5 V, 3 V, 4.096 V, 5 V, 8.192 V, and 10 V versions. You use it just like a Zener diode though it’s actually an integrated circuit. Tolerances are 0.1%, 0.2%, 0.5%, or 1% depending on which version you want to pay for, with current up to 15 mA (vs. 1.4% and up to 100 mA for the regulator). The datasheet tells you how to calculate the resistor needed (p. 27).
There also are ICs designed specifically for this job, like the TLE2426, but that may be overkill.
Thanks for the answer! I replaced the 10k resistor with a 1k resistor and now measure arount 3.4V coming from the Zener circuit. So this seems to „fix“ it, although the circuit did work with a 2.5V bias voltage. As there seems to be quite some tolerance I might switch to using a simple voltage divider for future builds. I plan to build a few of these as gifts for friends and will soon share my very neat and compact stripboard layout for this
If you only need the reference voltage at one point in the circuit, than a simple resistive voltage divider will probably be sufficient. If you’re going to use it in several different places, it’s good practice to buffer it, usually with an op-amp voltage follower.
