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I have the following circuit which is designed to take a longer pulse (on the order of miliseconds) from a microcontroller (source v3 below) and use two 555 timers in cascade to output two consecutive smaller width pulses (ranging from 0.5us to 10us and individually adjustable through a pot [not shown in the schematic below]). These two consecutive pulses are triggered from this longer pulse. The output pulse from the second 555 timer should follow after, when the output from the first timer goes low. I've used RC differentiators between the first BJT inverter and between the two 555 timers to avoid multiple triggering of the timers.

555 timer circuit

The simulated circuit below shows from top to bottom the input pulse, the output after the BJT inverter, the output after the first edge detector (555 trigger input number 1), the output of the first 555 timer, the output after the second edge detector (555 trigger input 2) and the second 555 timer output.

555 simulation

I've built this circuit and the output of the first 555 timer produces a satisfactory pulse however the input to the trigger of the second 555 timer with edge detection reaches a value just short of the 1/3VCC value required to trigger the second timer. In simulation it dips below this 1/3VCC value and the second 555 timer triggers correctly however in practice this does not happen as shown in the oscilliscope reading below. The lowest value it reaches is 2V.

Oscilliscope measurement at the input to the second 555 timer. (node titled in1 in the schematic above)

Oscilliscope reading

I have tried adjusting C4 and R9 (edge detector time constant) to get this value lower but it affects only the discharge rate rather than the amplitude. Connecting the top of the second edge detector to a lower voltage such as 3v rather than 5v results in a lower trigger value however I don't want to use a seperate lower source to solve this. I'm looking for a simple way to drop this trigger voltage by 500mV or so, so that the second timer will trigger.

edit

  • I'm actually using an NE556 IC instead of two NE555s.
  • The outputs of the 555 timers go into unity gain buffers with an op-amp.
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  • \$\begingroup\$ What else is the first 555 driving? Are they bipolar or CMOS 555s? \$\endgroup\$ – Spehro Pefhany Aug 13 '20 at 12:24
  • \$\begingroup\$ Both 555 outputs go into unity gain buffers. The output of the buffers are used as input signals to a high voltage H bridge type switching circuit. I'm using an NE556 timer so I assume they're the bipolar assortment. \$\endgroup\$ – Blargian Aug 13 '20 at 12:28
  • \$\begingroup\$ Cleanup: 1. R6 is in parallel with R5 and can be deleted. 2. A 555 has a full totem-pole output, so R8 can be deleted. 3. The third trace is not correct for the output of a differentiator. \$\endgroup\$ – AnalogKid Aug 13 '20 at 12:30
  • \$\begingroup\$ @AnalogKid Thanks, have amended the diagram to show the output of the differentiator. \$\endgroup\$ – Blargian Aug 13 '20 at 12:33
  • \$\begingroup\$ What if you add 10K from in1 to GND? \$\endgroup\$ – Spehro Pefhany Aug 13 '20 at 12:35
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enter image description here

Trigger activates at \$1\over3\$ Vcc there's no need to pull up all the way to VCC; pull to \$1\over2\$ VCC instead.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks! This solved my problem. \$\endgroup\$ – Blargian Aug 14 '20 at 9:28
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You could try a 10-100k pull-down resistor to ground on the output of the first and the trigger of the second. You may need to play with a few values to get the correct response. Depending on the 555 type the output can be 2 volts below vcc. Although most diagrams I have seen with one 555 triggering another are just directly connected between out and trigger.

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  • \$\begingroup\$ I took so long answering - many have commented in that time. \$\endgroup\$ – Steve Aug 13 '20 at 12:52
  • \$\begingroup\$ Hi, I've tried adding between 10 and 100k to both spots you suggested. Sadly not the slightest difference. I would have thought that would work too. \$\endgroup\$ – Blargian Aug 13 '20 at 13:17
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I would suggest replacing the capacitors, same resistors and diodes at the Trigger inputs of LM556. 555trig

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  • \$\begingroup\$ Many thanks for your answer, I ended up going with another user's response just because I've built this circuit using a home PCB method and only had surface mount schottkys available so it would have been harder to solder. Would you mind elaborating on why you went for the schottky diodes, it's quite interesting for me. \$\endgroup\$ – Blargian Aug 14 '20 at 9:30
  • \$\begingroup\$ There is a problem with the first input, which is controlled by a 12V signal. With a normal diode connected to ground, it will be -0.9V at the Trigger input. This is enough to open the diode B-C of the internal transistor. This can only be protected with a Schottky diode. The rest could be normal too, I just thought it was easier to work with one type. \$\endgroup\$ – csabahu Aug 14 '20 at 11:02

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