I have simulated the circuit provided in this link:http://www.circuitdiagram.org/dc-voltage-doubler-multiplier-circuit.html. But, when I simulate the circuit, the output voltage is dropped and I get a simulation error after 0.56s saying A simulation error has occurred. Would you like to run the Convergence Assistant to attempt to resolve this problem automatically?. I ran the convergence assistant to fix this issue and it worked saying successfully resolved convergence issue, and then again if I simulate, it gives the same simulation error and drop at the output.

Details : when I give an input voltage Vin=5V, the output should be doubled to 10V, but there is a drop in the voltage giving an output voltage of 8.916V as shown in the figure below: enter image description here

//Multisim Convergence Assistant Summary Report

   Convergence Assistant Summary Report


           The simulation error was corrected successfully.

   Changes Made:

           Multisim Convergence Assistant Log 

     Step 1: Verifying Error Scenario

     Step 2: Setting parameter Integration Method (METHOD) to Gear 
     ...completed. Simulation error fixed.

     Step 3: Attempting rollback of Integration Method (METHOD) to Trapezoidal 
     ...Rollback successful.

      Convergence Assistant completed successfully.

I am not able to figure out the reason behind this output voltage drop and the simulation error. Any help and guidance regarding this would be appreciated. Thanks.

  • \$\begingroup\$ You will lose some voltage across the diodes. As for convergence, simulators usually have problems where there are no DC paths. Try a high value resistor (start with 100M) across the output (C4). Experiment with different values, and note that even small load currents (say into 1M or 100k) will reduce that output voltage. \$\endgroup\$ – Brian Drummond Jun 6 '15 at 11:47
  • \$\begingroup\$ By across the C4, you mean connecting a high value R in parallel to C4, from +ve terminal of the source to the ground?. \$\endgroup\$ – PsychedGuy Jun 7 '15 at 4:41
  • \$\begingroup\$ Yes, exactly... \$\endgroup\$ – Brian Drummond Jun 7 '15 at 8:24
  • \$\begingroup\$ For an output load resistance of 330 ohms, the load current is high and the voltage drop at the output is high. Any ways to overcome this issue? \$\endgroup\$ – PsychedGuy Jun 9 '15 at 6:07
  • \$\begingroup\$ Surprisingly high output impedance is a feature of doublers, and even more so of higher order multipliers. You can improve it by increasing C3,C4, and by driving it from a beefier output stage than the 555. Or learn the limitation, and limit your current demand. If you need high current, a voltage doubler is the wrong approach. I suggested this test because it's better to know this now than later on. \$\endgroup\$ – Brian Drummond Jun 9 '15 at 11:55

When the 555 output is low, the right side of C3 will be about 0.7 volt less than 5 volts, due to the drop across D2, and the output of the 555 will probably not go quite to Ground, so there will be about 4.2 volts across C3.

When the output of the 555 is high, it will be a bit below 5 volts, so the right side of C3 will be perhaps 9.0 volts (4.8V from the 555 plus 4.2 volts from C3), and a little more voltage will be lost across D1, so think your 8.916 volts is very good.

  • \$\begingroup\$ That makes sense. Can you also please tell me what is the purpose of the capacitors C4 and C5 in this circuit?. And is there anyway to compromise for this drop to get a full output voltage of 10V. I presume that the C4 and C5 are supposed to do it. \$\endgroup\$ – PsychedGuy Jun 7 '15 at 6:00
  • \$\begingroup\$ C5 is the 5 volt power supply bypass capacitor - it is intended to stabilize the 5 volt supply. C4 is the output filter capacitor - it will be charged to 9 volts or so through D1 when the 555 output is high. \$\endgroup\$ – Peter Bennett Jun 7 '15 at 6:59

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