2
\$\begingroup\$

More specifically, my question is why does the natural frequency of the canonical (basic) symmetric transistor astable multivibrator (STAM) vary so strongly on the power supply voltage? As a follow-up question, what can be done to reduce that dependence?

Background:

I have constructed and measured the frequency (F) of symmetric transistor STAM circuits using a pair of 2N3904 NPN transistors for the oscillator, pairs of red LEDs (VF = 2.0 V) driven through respective pairs of resistors for the load, a common (single) power supply for the oscillator and load (Ref. 1). Values of the resistors and capacitors for the oscillator were selected to produce relatively low repetition rates in the nominal range of 30 to 120 cycles per minute (0.5 to 2 Hz).

Circuit Layout and Measurements:

The STAM circuits illustrated in Figs. 1 and 2 were constructed using consumer-grade components. The circuit in Fig.1 is the canonical STAM with the loads connected directly to the oscillator's transistors, and the circuit in Fig.2 is a modified version that is less sensitive to the load current, but still relatively strongly-dependent upon the source voltage.

Figure 1

Measurements of the frequency of oscillation were made for supply voltages, VS, in the range of 3-10 V with an accuracy of approximately 0.1 V, symmetric oscillator capacitance values, C1 = C2= CO, in the range of 10 uF – 100 uF, symmetric oscillator resistance values, R1 = R2 =RO, in the range of 75 kOhm to 300 kOhm, and for symmetric load resistances, RL1 = RL2 = RL, of nominally 150 to 600 Ohm (individual LED load current of 2-10 mA). Combinations of capacitance and resistance values were chosen to produce repetition rates in the range of 30 to 120 flashes per lamp per minute (fpm), which can easily be counted by eye using a stopwatch with an accuracy of +- 1 flash/lamp/minute. (Note: the numerical values of the repetition rates are reported here in the measured units of fpm.)

The capacitors are electrolytic with a voltage rating of 25 volts. The resistors are metal film with quoted tolerances of 1% and power rating of 1/4W. The average total current, I, flowing through the voltage source was measured using a Simpson 0-25 mA analog ammeter. A Lambda LQ-520 was used for the voltage source and the voltage was measured using a digital multimeter after the ammeter and between the rails of the STAM circuit. The source voltage was checked and adjusted, if necessary, between measurements.

Observations:

  1. My first observation was that the conventional (i.e., often quoted) formula for the frequency, F, of oscillation of the canonical STAM, F = 1/{2 ln(2) RC} with R = RO and C = CO, does not accurately predict the observed flash rate, with the model calculation being in error by more than 50% to as much as a factor of three, or more, depending upon the load current or the power supply voltage (Ref. 1). For example, for VS = 3.0 V and a low value of the total load current, ID, of about 1.8-2 mA, the frequency was measured for several values of the RC product. A sample of the results are:
CO (uF) RO (kOhm) F-meas (fpm) F-calc (fpm)
10 100 69 43
10 150 51 29
10 220 32 20

Table 1

The discrepancy between the model and empirical values for the frequencies was typically significantly larger than the 20% tolerance expected for the electrolytic capacitors.

  1. My second observation - again using the circuit of Fig. 1 - was that the frequency also varied strongly on the value of the load current. For these measurements the total load was varied by adding additional pairs of LEDs and a corresponding load resistor for each in order to maintain a nearly identical current through the individual LEDs. For example, again for VS = 3.0 V and with C = 10 uF, R = 150 kOhm fixed:
ID (mA) F-meas (fpm) F-calc (fpm)
2.1 51 29
4.2 65 29
8.4 100 29

Table 2

  1. My third set of observations were an investigation to reduce the dependence of the flash rate on the load current. I modified the canonical circuit by introducing a separate pair of buffer transistors, RB1 = RB2 = RB (also 2N3904), operated in open collector mode between the oscillator transistors and the loads. This circuit is illustrated in Fig. 2.

Figure 2

The respective collector outputs of the oscillator transistors were connected directly to the base inputs of the buffer, or (load) drive, transistors and with pull-up resistors, RP1 = RP2 = RP, of 5.1 kOhm to VS. This worked very well to decouple the oscillation frequency from the load current, as was anticipated. For example, the frequency was measured as a function of the total load current for CO = 22 uF and RO = 110 kOhm and again VS = 3.0 V:

ID (mA) F-meas (fpm) F-calc (fpm)
2.1 54 18
4.2 52 18
8.4 49 18

Table 3

As an aside, note that the discrepancy between the model calculation and empirical data has increased.

  1. Remaining is the question of the dependence of the oscillator frequency on the supply voltage, which is the main subject of this post. For these observations I used the same modified (with buffer output transistor) circuit as was used for the third set of observations (Fig. 2), above. Again using CO = 22 uF and RO = 110 kOhm and using two pairs of LEDs having an effective load resistance of 280 Ohm, the frequency was measured as a function of the power supply voltage. The measurements were:
VS (V) ID (mA) F-meas (fpm) : F-calc (fpm)
3.0 4.2 52
3.5 5.7 59
4.0 7.4 67
4.5 9.2 73

Table 4

In comparison to the third set, where increasing the load current by four-fold resulted in only a 10% reduction in the frequency, in this fourth set of measurements, changing the source voltage by 50% and the corresponding load current by about a factor of two increased the frequency by 40%, or nearly 4 times as much.

Next, I tested the circuit by adding resistors, RB1 and RB2, with RB1 = RB2 = 5.1 kOhm between each of the collector terminals of the oscillator transistors, TO and TB, and the drive transistors, TB1 and TB2, as shown in Fig. 3. (Note: No other additional components illustrated in Fig. 3, such as signal or Zener diodes, were added for these measurements.) Initially these measurements appeared to show a large change in oscillation frequency with drive voltage, VS. However, two respondents to my question pointed out that in the absence of the base resistors the output drive transistors, TB1 and TB2, have the effect of clamping the oscillator transistor’s (TO1, TO2) collector voltages to 0.7 V. Consequently, I reconstructed the circuit and repeated those measurements. Evidently, something was amiss with the addition of the base resistors in my earlier breadboard, as this time I observed very little variation of frequency with the drive voltage. These newer measurements are reported in Table 5.

RO (k Ohm) CO (uF) RB (kOhm ) VS (V) ID (mA) F-meas (fpm) F-calc (fpm)
110 22 0 3.0 4.5 53 18
110 22 0 4.5 10.5 79 18
110 22 5.1 3.0 4.5 23 18
110 22 5.1 4.5 10.0 24 18
110 22 10 3.0 4.5 20 18
110 22 10 3.0 6.7 20 18
110 22 10 4.5 9.8 20 18
110 22 10 4.5 15.5 20 18
100 10 10 3.0 4.2 48 43
100 10 10 3.0 6.5 48 43
100 10 10 4.5 9.1 50 43
100 10 10 4.5 15.5 50 43

Table 5

As the respondents suggested, the base resistors, RB1 and RB2, are expected to have a significant effect on the oscillation frequency and its dependence upon drive voltage, as those resistors affect the voltage at which the STAM changes state. As can be seen in the Table 5, inclusion of these base resistors significantly improves the stability of the oscillation frequency. In the measurements with RB1 = RB2 = 10 kOhm, there was only a 4% increase in oscillation frequency when the voltage was changed from 3.0 V to 4.5 V. As well, the simple model is now in close agreement with the empirical observations – differing now by roughly only -14%. We also observe that within the accuracy of the frequency measurements, changing the load current did not change the oscillation frequency.

The various sets of measurements suggest the astable multivibrator is relatively non-linear, which is perhaps not unexpected considering the nature of transistors. Comments to this post have described and illustrated how changes in the source voltage can alter the shape of the rise and fall of the voltage to the transistors and thereby alter when the change of state is triggered. Some of the comments have also addressed the question of whether the circuit can be modified by the addition of either resistors, capacitors, diodes or transistors (e.g. PNP) to reduce the relatively strong dependence of the frequency on the source voltage.

Since asking the question, "how might the circuits be modified to reduce the sensitivity to the source voltage," I have read and it has been suggested in comments to this post and described in another post on this forum that selectively inserting diodes and/or Zener diodes into the circuits, to control currents to, and potential reverse biasing of, the oscillator transistors, as illustrated in Fig. 3, can further stabilize the oscillator and extend the reliable operational voltage range (Ref. 2, Ref. 3, Ref. 4, Ref. 5). These recommendations are collectively illustrated in Fig. 3. I plan to explore these possibilities empirically.

Figure 3

1 Wikipedia, Multivibrator, https://en.wikipedia.org/wiki/Multivibrator, Accessed December 7, 2023.

2 R. Marston, Bipolar Transistor Cookbook, https://www.nutsvolts.com/magazine/article/bipolar_transistor_cookbook_part_6, 2021, Accessed December 7, 2023.

3 Bastien, BJT astable multivibrator frequency increasing with voltage, Electronic Stack Exchange, December 5, 2021, and questions and comments therein, BJT astable multivibrator frequency increasing with voltage, Accessed December 7, 2023.

4 B.R. Wood, Adding diodes to transistor astable multivibrator, …, Electronic Stack Exchange, February 22, 2019, and questions and comments therein, Adding Diodes to Transistor Astable Multivibrator; 1N5818 Schottky diodes work but 1N4148 diodes don't; can't figure out why, Accessed December 8, 2023.

5 S. Rotos, Astable multivibrator diode improvement, Electronic Stack Exchange, July 6, 2018, and questions and comments therein, Astable multivibrator diode improvement, Accessed December 8, 2023.

\$\endgroup\$
9
  • 3
    \$\begingroup\$ That's a lot of empirical data! But have you tried doing a time-domain simulation of any of those configurations to understand how the capacitors actually charge and discharge? You can draw each of your circuits using the tool build right into this website, and then run simulations while varying some of the parameters. I think that would provide a lot of insight. The math really isn't all that difficult. \$\endgroup\$
    – Dave Tweed
    Dec 6, 2023 at 4:35
  • 1
    \$\begingroup\$ One general idea with transistors in stable circuits is to have a constant amplitude of collector current and a constant collector current bias. Replace collector resistors with current sources and see what happens :) In any case, the circuit is not at all designed to be stable with variable supply. It needs a regulated supply. For unregulated supplies you need a different circuit. \$\endgroup\$ Dec 6, 2023 at 12:22
  • \$\begingroup\$ Please show the circuit diagram and a picture of the physical circuit you made measurements on. It helps in answering questions. Questions about a circuit without a circuit anywhere in sight are let’s say an interesting idea. And there’s nothing too “conventional” - the world is a big place and your textbook that mentions a “conventional” formula may well not be my textbook. Don’t refer to pseudo-convention; be explicit: show the circuits and formulas. This site has a nice circuit editor and formula formatter (mathjax) built in. \$\endgroup\$ Dec 6, 2023 at 12:24
  • 1
    \$\begingroup\$ Note that in fig 2, the two TB transistors' Vbe clamp the two TO transistors' collector voltages to 0.6 V, while in fig 1 the collector voltages could be almost equal to VS. This will have a large impact on the operating frequency. Also, revisit the concept of unique reference designators (R1, R2, R3, etc.). \$\endgroup\$
    – AnalogKid
    Dec 7, 2023 at 13:53
  • 1
    \$\begingroup\$ Regarding the “NPN buffer” for driving Led -> the BE of NPN clamps the collector voltage. You can put a base resistor to “buffer” transistor but it still decreases the multivibrator collector voltage. Best is use PNP with base resistor for buffer. With PNP as buffer it decreases nothing. \$\endgroup\$ Dec 9, 2023 at 0:30

5 Answers 5

2
\$\begingroup\$

The point to hold the freq. stable at wide Vcc range is to use capacitors voltage symmetry.

The same voltage has to be charged, same discharged and flip exactly when cap has 0V.

In schematic below this condition is met:

enter image description here

enter image description here

The caps are charged to Vcc-0.7v , discharged by Vcc-0.7v and because of diodes the transistors flip at Vcap=0V. Another words a perfect symmetry is fulfiled.

As others mention don’t use higher Vcc then 6V because the BE breakdown can occur. The circuit can be modified for high voltage but more parts will be needed.

\$\endgroup\$
6
\$\begingroup\$

What does "fpm" stand for?

You make statements about the "conventional formula" for an astable circuit without stating the formula that you are using. Lacking that information, my guess is that you are misapplying it to the circuit.

You also did not post your schematic. Lacking that information, there are many possible reasons you circuit might not be conforming to your expectations, including base-emitter junctions going into reverse breakdown.

Please update your question with your schematic. Also, make sure each component has a unique reference designator, since this discussion will center around the behavior of specific parts.

To your last question - no. The idea of NPN and PNP transistors somehow offsetting and cancelling out some parameter of each other is not a bad one, but not correct for this circuit.

In very round numbers, the reason the invisible circuit's frequency changes with applied voltage is that the larger the operating voltage, the larger the voltage swing across the timing capacitors - while the base-emitter voltage of the transistors (and LEDs, depending on where they are in the circuit) is basically constant.

\$\endgroup\$
3
\$\begingroup\$

Here's an example of the kind of simulation you can run. Note that I've made R3 and R4 very slightly asymmetrical, and also set the "Skip initial:" box to "Yes" in the simulator. Otherwise, the simulator thinks the circuit is perfectly symmetrical and it ends up in a quasi-stable state without oscillating.

schematic

simulate this circuit – Schematic created using CircuitLab

Here are the voltage waveforms for the two pins of C1:

voltages on C1 terminals

... and here is the current through C1:

current through C1

Think about why the voltages are what they are at the beginning and end of each half-cycle. How does the charge change on C1 when Q1 is on and Q2 is off? Which circuit elements are involved? How about when Q1 is off and Q2 is on? What controls the time constant of the charging curve in each case?

\$\endgroup\$
3
\$\begingroup\$

Things like LEDs and base-emitter junctions have roughly fixed potential differences across them. These devices all "offset" potentials by fixed amounts, meaning that when you double the supply voltage, you aren't doubling the potentials everywhere.

For direct proportionality between supply voltage and switching frequency, you would require that collector loads be perfectly resistive, purely ohmic. If you are using LEDs there, this isn't the case. They would introduce the same 2V offset to collector potential, whether your supply is 5V or 10V.

You would also require that the transistors' switching thresholds (base potentials) also change in proportion with the supply. They don't; \$V_{BE}=0.7V\$ regardless of supply voltage.

Those base-emitter junctions are zener diodes. When reverse biased to about 5V (could be more, but 5V is typical) they begin to conduct heavily. I haven't studied the STAM in any great depth, but I suspect that base potential becomes significantly negative for short periods, and this means that the base-emitter junctions will become conductive for certain parts of a cycle, if the supply voltage is over 7V or so. That's going to introduce a kink in the graph of frequency vs. supply voltage.

There are too many non-linear elements in a STAM for it to be reasonable to expect a linear relationship between frequency and supply voltage, much less a proportional one.

Just to illustrate, watch what happens when we double the supply voltage of a simple RC delay system, employing a single transistor:

schematic

simulate this circuit – Schematic created using CircuitLab

Both circuits have the same timing elements, 1kΩ and 1μF. You might argue that since the power supply on the right is twice that on the left, C2 should charge twice as fast as C1. This would be the case, except for two things:

  1. There's an LED in that charging path, and therefore the voltage across the RC pairs in each each are not different by a factor of two. This is shown on voltmeters VM1 and VM2.

  2. Base current in each case will be different. There's a path into the transistor bases diverting current from the timing capacitors, and there are many reasons why we may not assume that this current is also different by a factor of two. This is seen on ammeters AM1 and AM2.

Take a look at the charge rates of capacitors C1 and C2:

enter image description here

From an initially discharged state, the voltage across C1 (blue) reaches 0.7V in 440μs. If this system contained no non-linear elements, you could predict that C2 would arrive there twice as quickly, after only 220μs, but that's not the case. The non-linear elements (LED, transistor) cause C2 to charge faster than that, arriving at 0.7V after only 130μs.

The transistors both switch on when their bases approach 0.7V, with collector potentials looking like this:

enter image description here

Even though you can see why increasing supply voltage would decrease the time it takes for the transistor to switch on, which might lead you to suspect that frequency of oscillation might double, for each doubling of the supply, it clearly isn't that simple.

\$\endgroup\$
1
\$\begingroup\$

As pointed out by @Simon Fitch, breakdown between B-E (power supply > 5 V) of the transistor change a little the frequency (around here 15%).

enter image description here

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.