Electronic – Slightly weird astable oscillator in a 555, where does this capacitor discharge to

555astablecapacitoroscillatortransistors

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.

Best Answer

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.