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I have a UM66 melody generator IC which requires a supply voltage in the range of 1.5-4.5 V according to this datasheet. I am using a 9 V battery to power the IC and as far as I have researched, people use a 3.3 V Zener diode along with a 1 kΩ resistor to supply 3.3 V to the IC, like the below diagram.

enter image description here

Instead of a Zener diode, can I use a resistor divider to power the circuit? For example, if I use a 2.2 kΩ resistor instead of 1 kΩ resistor in the above image and replace the zener diode with a 1 kΩ resistor, the voltage drop across the 1 kΩ resistor will be 2.8 V (for 9 V supply voltage) which is in the permissible supply voltage range and the series current will be 2.8 mA which is also greater than the operating current of the IC (60 uA). Can I use this configuration or should some other things need to be considered?

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  • \$\begingroup\$ if you don't have a zener just use 2 LEDs in series (so they light up). it will be close enough, \$\endgroup\$ Aug 19, 2023 at 3:56
  • \$\begingroup\$ 60uA? is there no more current than that flowing into Q1 base? \$\endgroup\$ Aug 19, 2023 at 4:03

2 Answers 2

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In general, it is bad praxis to supply a circuit by means of a voltage divider because the load's input resistance appears in parallel to the lower resistor of the voltage divider. The input resistance changes depending on the supplied circuit's state / current assumption, thus leading to a change in the circuit's supply voltage.

However, in your case the input resistance with a minimum value of \$R = \frac{2.8~\mathrm{V}}{60~\mathrm{uA}} = 46.6~\mathrm{k\Omega}\$ is very high compared to the resistances of your voltage divider. Thus, the minimum supply voltage would be \$V = 9~\mathrm{V} \cdot \frac{979~\Omega}{979~\Omega + 2.2~\mathrm{k\Omega}} = 2.77~\mathrm{V}\$.

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When a load (like your UM66) draws current from a resistor potential divider, it diverts current away from the divider, changing the currents in the resistors, which in turn changes the voltages across them. Consequently, a resistor potential divider cannot maintain a fixed voltage output under varying load currents, which will be the case for the UM66.

You can reduce the output voltage variation of the potential divider, under varying load conditions, by reducing the resistances in the divider, but you'll never eliminate that variation, and (worse) you'll have more current flowing, which wastes battery energy.

Here's a little demonstration of the problem:

schematic

simulate this circuit – Schematic created using CircuitLab

On the left (A) the divider produces an exact +4V output, as shown on voltmeter VM1, because there's no load current being diverted away from the resistors.

In the middle (B) I am pulling 1mA of current from the divider output, which causes a significant drop in output voltage, as shown on VM2.

In circuit C I mitigate the effect of load current by reducing the resistances by a factor of ten. This helps, VM3 shows a much smaller deviation of voltage from the desired +4V, but it's still less than +4V.

What we require is something that performs a similar function as the resistor potential divider, to obtain +4V (or whatever voltage you desire), but is able to automatically vary it's own effective resistance to compensate for changes in load current. That's the role of the zener diode that you mentioned having seen used in this application. Its resistance changes automatically to the right value in order to keep 3.3V across it regardless (well, almost regardless) of the amount of current taken away from the divider output.

I assume you are asking this question because you don't have a 3.3V zener diode. Fortunately there are other methods to achieve this. Since you have an NPN bipolar transistor, and I'll guess you have more than one, you can use them to make a device that behaves very like a zener diode. That's shown on the right here:

schematic

simulate this circuit

Instead of using voltmeters to show behaviour, I'll plot output potentials at OUT1, OUT2 and OUT3 as I sweep load current from 0 to 8mA. Below, the basic resistor divider output is blue, the zener diode circuit output is orange, and the transistor design is in tan:

enter image description here

As you can see, the resistor divider (blue) doesn't regulate at all. Its output falls linearly as load increases. The best behaviour is seen for the zener diode (orange). It's doing a great job of keeping the output fixed at 3.3V all the way up to a load of 6mA or so. Second best is the transistor design, which produces a respectable, but less "flat" output.

The best performance would be achieved using a linear regulator IC, such as the LM317L, the circuit for which would look like this:

schematic

simulate this circuit

This will produce almost exactly 3.3V for almost any load current, up to many hundreds of milliamps.

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