In the 555 datasheet, the cap is usually connected to pin 7 (discharge), but in this circuit, it's not:

weird 555 astable oscillator

When the capacitor voltage is >2/3Vcc, output will turn off and current will have no where to go. Does it go through Vbe?

Also, I've constructed this circuit with the 'normal' astable configuration and it seems to draw >1A current total on 9V, and after a while will kill the 555. I was under the assumption that most of the current will go through the transistor since we are controlling current via the 1k resistor.

The normal configuration works but the output voltage gradually decreases overtime, doesn't seem stable.

Never tried this configuration in the complete circuit, so unsure if same thing will happen.


3 Answers 3


When the circuit is first turned on the voltage on pin 2/6 is low, this triggers the 555 which turns on pin 3. The 10n cap then charges through the 33k resistor. When the voltage on pin 2/6 reaches 2/3 Vcc the output turns off and the 10n capacitor starts to discharge through the 33k resistor into pin 3. When the voltage drops to below 1/3 Vcc, the process repeats.

When the 555 is triggered the NPN is turned on, the exact current allowed through the coil depends on the 1k resistor and which transistor was chosen. You can vary this by change the values of these two components or changing the timing values presented by the 10n cap or the 33k resistor.

Since the current through a coil cannot change immediately it slowly ramps up when the 555 is triggered. If the timing interval is too long the current can ramp up to an extremely high value. You thus have to make sure the current is limited either by component values or the timing.

Something else to take note of is that the voltage at the collector of the transistor can be extremely high (couple of hundred volts easy), it is pretty likely that you are blowing the NPN transistor after only a short operating time. I would recommend putting in some feedback at the output. Use a voltage divider on the output of the transformer, into an NPN transistor that pulls the reset line low (it should have a pullup). This will stop over-voltage killing your circuit.

So to summarize, there are three problems with this design:

  1. Timing values probably aren't calculated correctly.
  2. Current isn't limited, either by timing or components.
  3. No protection against over-voltage

As for the voltage decreasing over time, pulling 1A from a 9V will degrade its ability to supply power pretty quickly, this will drop the output voltage/current which affects the circuit. To test this you can monitor the voltage of the 9V battery while the circuit is operating, you should see it drop drastically.

  • \$\begingroup\$ The transistor is a BD679 and it hasn't blown, however, the normal astable configuration has blown many 555. Do you know why this is the case? Also, how does the cap discharge into pin 3? Pin 3 is the output from an SR latch (internal to 555). \$\endgroup\$
    – PGT
    Commented Jan 11, 2015 at 9:12
  • \$\begingroup\$ @PGT when pin 3 is low and the 10nF cap is charged above ground, the cap will discharge; when pin 3 is high and the 10nF cap is charged to below Vcc the cap will charge. \$\endgroup\$
    – markt
    Commented Jan 11, 2015 at 10:10

The 555 is likely being blown out by the high voltage flyback spikes produced by the primary side of the transformer. Place a medium current diode across the transformers primary, with the anode side on the transistor's collector. This shorts out the spikes.

If the oscillation is slow the current in the transformer could be very high.

The transistor's current is being determined more by the transformers primary resistance than the base resistor, if the oscillation is slow. If the oscillation is high the inductance of the primary may help limit the current to a lower level.

A parallel RC combination on the transistor's emitter can limit DC current while still allowing high AC currents.

In your circuit pin 3 will discharge the 10n cap. Pin 7 would also discharge the cap, as in the standard data sheet arrangement. Using pin 7 isolates the cap voltage from the changing output at pin 3.

  • \$\begingroup\$ Is there any other way to protect the 555 without shorting the transformer's primary? Also, do you have a link to some reading I can do on low/high oscillation versus current or inductance (or just terminology I can lookup)? I have vague memories of the theory from school but can't remember the specifics. If the current through the transformer is reduced, this will also reduce the peak voltage of the spikes when the 555 switches, right? \$\endgroup\$
    – PGT
    Commented Jan 16, 2015 at 8:20
  • \$\begingroup\$ @PGT: The diode will not "short circuit" the primary side in the traditional sense of "short circuiting". When the output from the 555 is high the diode does nothing, and when the output is low the diode will allow the same current to flow through the primary. (If you short circuit a charged inductor it will remain charged, ie. the opposite of how a capacitor behaves.) \$\endgroup\$
    – Oskar Skog
    Commented Feb 18, 2017 at 12:47

I suspect that what's happening to your 555's that they're being killed by the high-voltage transients generated by the transformer primary when the 555 switches a square wave through it.

You could either try to protect the 555 from the transients by filtering them out of the power rail, or you could just not create them in the first place by adding a low-pass filter between the output of the 555 and the base of the BJT so that the square wave is "rounded off".


simulate this circuit – Schematic created using CircuitLab

A 100nF cap should be effective, although you may need to experiment a little with values because it's going to attenuate your audio signal as well. The diode prevents the cap from discharging back into the 555 when the output goes low.

  • \$\begingroup\$ So from what you're saying, when the square wave switches, the primary will try to stabilize the current, causing a spike in voltage in the reverse direction, which will cause overvoltage into the VCC of the 555? Or is it going into any other pin of the 555? \$\endgroup\$
    – PGT
    Commented Jan 16, 2015 at 8:16
  • \$\begingroup\$ It's not about stabilization. The problem is that when your put a voltage step (a sudden change in voltage, positive or negative) across an inductor, e.g. a transformer coil, the coil responds by storing some of the energy from the step and then releasing it as a very short, very high voltage spike, in your case on Vcc. Those spikes are (probably) what is killing your 555's. You need to either damp the spikes, or prevent them from occurring in the first place. \$\endgroup\$
    – markt
    Commented Jan 16, 2015 at 11:12
  • \$\begingroup\$ @markt: Inductors don't mind instant changes in voltage. What they "object" to is instant changes in current. Though it should also be noted that transformers don't like being run off DC. Unless the secondary has an asymmetrical current draw, the transformer won't work well being driven from juts one side. \$\endgroup\$
    – supercat
    Commented Nov 3, 2015 at 22:21

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