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For executing a (stress) test scenario, I need to simulate the following behavior on my circuit:

  • for 2 minutes, deliver 11V to the circuit
  • for 2 minutes, deliver 14V to the circuit
  • for 2 minutes, deliver 11V to the circuit
  • for 2 minutes, deliver 14V to the circuit
  • ...

I have available :

  • a lab bench capable of supplying the 14V
  • a micro-controller + relay setup that I can program to switch something on/off every 2 minutes.
  • a stepdown converter that I can use to bring 14V to 11V.

What would be the easiest / most convenient circuitry be to handle this ?

I was thinking about using the microcontroller to switch a relay on / off very 2 minutes. When the relay is on it would just supply 14V to the circuit. When the relay is off it would somehow "divert" the flow into a stepdown converter that brought down the 14V to 11V.

But I'm not 100% sure if this is correct and if it is how I should draw the circuit.

I was thinking about something like this. When the relay is on it just passes the 14V to the circuit. When the relay is off current goes through the stepdown and brings it down to 11V. Does this make sense ?

CircuitLab Schematic y7a477

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    \$\begingroup\$ I think I'd revers the set-up. Use the two poles of the relay as 'inputs', connected to the 14V or 11V stepdown. Then choose between those, so the single 'output' Connects to the circuit. Put a diode across the reactive load (solenoid) to bypass spikes generated by its inductance. \$\endgroup\$ – gbulmer Sep 3 '14 at 16:11
  • \$\begingroup\$ Yes .................................. \$\endgroup\$ – Russell McMahon Sep 3 '14 at 16:33
  • \$\begingroup\$ @gbulmer why do you suggest he reverse the setup? Curious here... \$\endgroup\$ – ACD Sep 3 '14 at 19:11
  • \$\begingroup\$ @ACD - have a look at my answer. There are two different voltages, and one load. So it makes more sense to me to select one of the two voltages with the relay. \$\endgroup\$ – gbulmer Sep 3 '14 at 20:49
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    \$\begingroup\$ @ACD - I would avoid putting 14V on the output of U1, the 11V stepdown. There is no information about U1, and it may survive, but I would avoid doing that. I would expect the stepdown to be happier with an open-output. (I might put a resistor on its output.) Because of that, it makes more sense to me. \$\endgroup\$ – gbulmer Sep 3 '14 at 23:00
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Use the solenoid to select between the two input voltages (11V and 14V).

Edit3: I would avoid putting 14V on the output of U1, the 11V stepdown, as shown in the questions schematic. There is no information about U1, and it may survive, but I would avoid doing that. I would expect the stepdown to be happier with an open-output. If that is still a concern, add a resistor between the output of "U1 STEPDOWN" and ground to keep it within spec. At a guess it might need a few mA, so a 1K resistor may be enough.

schematic

simulate this circuit – Schematic created using CircuitLab

The transistor switch could be a N-MOSFET, or a BJT with current limiting resistor on the base.

Edit:
The N-MOSFET (or NPN BJT) could be driven by any square wave, it doesn't need to be an MCU.

In favour of a low-speed oscillator is simplicity and cost.
In favour of an MCU is flexibility. For example a sub-£3 Arduino could monitor some aspect of the circuit under test, and capture data, or be reprogrammed to have more complex timing.

Edit2: tidied up schematic, hopefully addressing concerns.

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  • \$\begingroup\$ +1 gblumer, for OP, if you want to avoid MCU you can use a 55 timer for switching the power supply every 2 mins. \$\endgroup\$ – AKR Sep 4 '14 at 4:14
  • \$\begingroup\$ I was also thinking of removing the MCU. I assume the contol could be managed by a flip/flop circuit ? depending on the resistor/caps a 2 min on/off should be possible right ? \$\endgroup\$ – ddewaele Sep 5 '14 at 9:38
  • \$\begingroup\$ gbulmer: The output of a 7805 is 5V, not 11V. \$\endgroup\$ – EM Fields Sep 5 '14 at 11:24
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    \$\begingroup\$ @EMFields - Agreed. I should have done a better job on the schematic. It should have said 11V. I don't have an account on circuit labs, so AFAIK, there is no way to go back and edit the schematic. Circuit labs is sufficiently clumsy that I don't want to to do it all again from scratch. \$\endgroup\$ – gbulmer Sep 5 '14 at 11:50
  • \$\begingroup\$ gbulmer: <G>, it does say 11V, which is what it can't be since the regulator's a 7805. The LM317 fairchildsemi.com/datasheets/LM/LM317.pdf is an adjustable regulator you could easily substitute, and all it takes is two extra resistors. \$\endgroup\$ – EM Fields Sep 5 '14 at 12:26

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