# Simulating Smoothing Caps in TINA

I'm trying to simulate the workings of smoothing capacitors in TINA-TI. I have a varying voltage source (sine wave varying from 4.5-5.5V) and 2 capacitors (1uF and 100uF) connected. The load is a 1000 ohm resistor, and VF1 is the test point. After testing with various capacitor values, I'm finding that there's no change at all to the output waveform compared to the input. See the image below:

Basically, why would this occur? Is it because these capacitor values are not suitable for smoothing, or is it just a problem with my simulation?

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You also find that your capacitor models actually need to be more involved, there will be series resistance and inductance. –  placeholder Jul 14 '14 at 21:38

The problem is that you simulated the voltage source as an ideal source.

No matter what you connect to an ideal source, the voltage across it will be whatever you set the source output to be.

If you want to see the smoothing caps have some effect, you need to include some resistance and/or inductance in the connection between the source and the rest of the circuit.

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Strangely, when I add resistance in series between the source and circuit, the voltage goes into the negative range. I'm confused. –  d.free Jul 14 '14 at 23:49
I'm not sure what's going on with that simulation --- maybe because there's no ground node defined in your circuit? But that's just speculation because I don't know the tool you're using. Also, you probably want to look at frequencies in the range of 1 kHz to 1 MHz, not 2 Hz like you're doing now (or you need much bigger capacitor values to filter such low frequencies). –  The Photon Jul 15 '14 at 0:29
Yep, it was the ground node. I added it and everything works fine. :) –  d.free Jul 15 '14 at 1:50
I think in TINA you can set the internal voltage source resistance for the same effect. –  PlasmaHH Jul 15 '14 at 9:56

That's a problem with your circuit.

Look closely: you are probing a node driven by an ideal voltage source: the voltage at that node is decided by VG1, whatever you connect to that node VG1 will provide as much current as it's needed to keep the voltage as it wants. That's precisley its job actually.

Real world generators have something called internal resistance, i.e. they are not ideal but they are modeled by an ideal voltage source with a resistor in series. That's because a real life generator can not provide arbitrarily high amounts of current. Try adding a $10\Omega$ resistor between VG1 positive terminal and C2, and see what happens.

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Step back slowly from the simulator, then actually think about the circuit, not the simulation.

Remember what a perfect voltage source does. It always presents its voltage accross its terminals, regardless of what impedance you connect to it, or equivalently, what kind of current you draw from it.

The capacitors and resistor are just loads on the voltage source. They will cause it to produce some current to maintain the voltage, but by definition, it will maintain the voltage. Another way of looking at this is that a ideal voltage source has 0 impedance. Therefore paralleling it with any other non-zero impedance isn't going to change the voltage.

What you are probably confused by is that in the real world there are no such things as ideal voltage sources. They all have some equivalent series impedance, and can only supply up to some maximum current. A battery, for example, may have 1.5 V accross it open-circuit, but go down to 1.2 V when you draw one amp. Ideal voltage sources in simulators don't do that. You can model that to some extent by putting a resistance in series with a ideal voltage source. However, perfect ideal voltage sources only exist in the land of simulators and unicorns.

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It might be easier to understand what is happening if you think of "smoothing caps" as a low pass filter using an RC network. In this case, there is no R so the time constant will be zero. This practically means the ideal caps will charge and discharge instantly from the ideal voltage source. As everybody else suggested, there needs to be another resistor in series with Vs. Also, select a trace as the circuit ground so you have a 0V reference.

A more intuitive way would be to think as the ideal voltage source drawing and providing an infinite amount of current in order to instantly drain or raise the charge in the capacitors. This way, the capacitors will always follow the voltage source with a zero phase shift.

As everybody else suggested, you need to model either the ESR of the capacitors or that of the power supply. If it is a battery, its datasheet should have a resistance vs current curve.

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