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I am trying to impose an AC ripple of 0.9V peak to peak (1kHz) on 36 volts DC supply for one of the test using following circuit.

Without any load connected, I get the proper AC imposed DC voltage but as soon as I connect a resistive or capcitive load to the output, the AC ripple of 0.9V just vanishes. Only 36 volts is left behind. The AC current source is capable of sourcing 400mA which is much higher than my actual current requirement of 280mA DC input current of load.

What could be the reason of vanishing of AC ripple when the load is connected ?

  • Do I need to use higher current sourcing AC source?

  • Does the value of the coupling capacitor need to be increased to reduce the output the impedance?

Any other means to achieve the functionality are also welcome!!

enter image description here

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    \$\begingroup\$ Is this only for simulation purposes or are you trying to construct a physical circuit? \$\endgroup\$ – Elliot Alderson Jul 23 at 18:38
  • \$\begingroup\$ If only for simulation purposes, you can just put V1 and V2 in series (and delete D1 and C1). \$\endgroup\$ – Huisman Jul 23 at 18:45
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    \$\begingroup\$ Increase your 10µ or lower it's ESR. Electrolytic not a good idea for C1 as they behave strangely with reverse bias. Note that the 0.9vAC will ride ABOVE the 36v minus the loss across D1, so output is something like 35.5v to 36.4v. D1 will become a light-emitting diode at 280mA (150mA avg max.) \$\endgroup\$ – rdtsc Jul 23 at 20:17
  • \$\begingroup\$ What is the nature of the load that is being applied? Is it just a resistor? Is it capacitive? etc. \$\endgroup\$ – Aaron Jul 23 at 20:40
  • \$\begingroup\$ @ElliotAlderson I am building this circuit physically to test the Module. \$\endgroup\$ – Rohan Jul 24 at 14:38
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For simulation, just connect the two voltage sources in series. You can do the same with the AC source directly, as vangelo says, but it's more obvious on the schematic if you do it with two voltage sources.

For a real circuit you can do the same if at least one of your voltage sources is not grounded.

If the voltage sources are grounded, then you could add a series inductance (a few, say 3H) in place of the diode and something like 1000uF/50V electrolytic low-impedance blocking capacitor in series with the AC source, however the AC source needs to be protected against the 36V surge at power-on and capable of supplying the load current plus reactive current through the 3H inductor.

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  • \$\begingroup\$ i will increase the the voltage of DC source gradually . I will try to connect the two AC and DC sources in series . Current sinking capability of either source needs to be checked ? my AC souce is High Power Opamp based Current amplifier which has AC input signal from Function generator . \$\endgroup\$ – Rohan Jul 24 at 14:46
  • \$\begingroup\$ All the load current passes through both. Consider protecting your op-amp output with some back-to-back Zener diodes (more than the desired peak AC voltage but less than the supply rails). At 0.9V you could use a few silicon diodes in series instead of a zener. \$\endgroup\$ – Spehro Pefhany Jul 24 at 14:55
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For simulation, just put the DC and AC sources in series. Done.

For the real circuit, do you have access to the 36V supply feedback network? You could inject the ripple there and let the power supply do the heavy lifting for you.

Another approach would be to put an LDO in line with 36V and inject ripple in the LDO's feedback network. You would have the LDO overhead to contend with, but probably this would also give the desired result.

Here's another approach using a transformer inserted in-line: Adding small AC signal to a DC current using inductive coupling

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  • \$\begingroup\$ Actually injecting ripple at the reference of Error amplifier was the first idea but it is not allowed to do any altercation in test instruments of the LAB. I will try to figure out the option with the usage of LDO . \$\endgroup\$ – Rohan Jul 24 at 14:52
  • \$\begingroup\$ Some lab supplies have remote sensing so they can compensate for the IR drop in the power leads. Otherwise the LDO should work, and since you’re only dropping a volt or so the losses should be manageable. \$\endgroup\$ – hacktastical Jul 24 at 19:06

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