Need help understanding this filter.  - Page 1
Forum section: electronics / beginners

author: chinoy
author_rank: Regular Contributor
timestamp: October 25, 2023, 09:04:51 am
content: Im trying to simulate this filter. Have tried freq sweep and voltage sweep. But still cant figure out what this filter does.Somebody told me its a shelf filter. But could not tell what Freq it impacts.The input to the circuit is a coil. Voltage range 1 to 30 volts.The top filter works on the top half of a wave form and the bottom filter works on the bottom half of the wave form.

author: MrAl
author_rank: Super Contributor
timestamp: October 25, 2023, 03:57:56 pm
content: Quote from: chinoy on October 25, 2023, 09:04:51 amHi,Without seeing the entire circuit this goes into it is hard to tell what it is supposed to do. Without that all we can do is a pure analysis and then we are stuck thinking about what that outcome can be used for.For this circuit without anything else, it looks like it may cut off the base of the half cycles, leaving mostly the top sections, and as the frequency increases we only see the top sections of the half cycles that are meant to be passed.Do a simulation and post the waveforms you get on the outputs and we can take a closer look.

author: chinoy
author_rank: Regular Contributor
timestamp: October 25, 2023, 06:29:12 pm
content: My interest lies only in what the filter circuit is doing.Is it a low pass / high pass / shelving filter. Once I know that. I can use one of the online calculators to see what it does.Somebody told me it looks like a Parallel RC = first order high pass shelving filter.Trying to simulate or do ffreq domain analysis to see how changing the cap and resistor values impact the signal.

author: MrAl
author_rank: Super Contributor
timestamp: October 26, 2023, 11:15:23 am
content: Quote from: chinoy on October 25, 2023, 06:29:12 pmHello,The two circuits are high pass but the diodes will force some residual voltage across the capacitors.However, it is not possible to know what they will actually do without knowing what the load is, and that goes especially for the lower circuit with no load at all.For example, in the lower circuit the capacitor will never charge and thus will look like a short circuit indefinitely. There has to be a load of some type before you can see the lower circuit do anything remotely resembling what it will do when a load is applied. The output of that lower circuit will follow the input even though the diode is there. If you apply even a 100k load you will see the behavior change radically. As is, it's a very theoretical thing where there are behaviors that are so theoretical that you would never see it behave that way in real life because you would always have a load of some type, either resistive, capacitive, inductive, a combination of those, or even a transistor or something. Each type of load could cause a completely different type of behavior. In these cases though it would be typical to apply a resistive load, but the resistor value then affects the bandwidth and waveshape.It would be best to know where these came from or else it may be impossible to know what they are supposed to do, especially the lower circuit.A simple way to figure out what the lower circuit does with no load is to short out the capacitor and measure the output. If you do the typical and measure with reference to ground, all you will see is a half wave rectified waveform and it will look like that regardless of the frequency, again in theory, and again that is because the capacitor will never charge up at all due to no current flow ever. If you decide to assume some current flow anyway, then you have a high pass filter with some DC also.If you have difficulty setting up a simulation I can do that for you and show some waveforms using different load values.

author: Terry Bites
author_rank: Super Contributor
timestamp: October 26, 2023, 03:52:09 pm
content: Ignoring the diodes (why such hv?), this is a pre-emphasis or eq filter.It is a very primitive highpass shelving filter. It reduces the attenuation the voltage divider at higher frequencies.It will not simulate easily with the diodes in place. Removing them will not affect the frequency response, just the DC voltages (if any)If your test waveform is a symmetrical sine you can ignore the diodes.I'm guessing that the lower section is terminated in a resistance as well.When Xc is << R2 or R3 these resistors are bypassed and the circuit functions a simple voltage divider.Attached simulation in Microcap12 (MC12)Where did this idea come from, it doesn’t look like a practical circuit.The diodes won't conduct until you have a signal>600mV, is that desirable?

author: wasedadoc
author_rank: Super Contributor
timestamp: October 26, 2023, 04:21:23 pm
content: In the first post one of the capacitors is reversed.

author: chinoy
author_rank: Regular Contributor
timestamp: October 27, 2023, 04:01:51 pm
content: When working alone you tend to make such stupid mistakes. And land up going crazy when something does not work.You sir are spot on I had mixed up the direction of the cap. Which is why my simulation results were making no sense.Hopefully once I correct this my sims should start working.The load is C106 and a BC transistor.Here is the rest of the schematic. The voltage source is a Sine wave from a coil.For it to work the voltage has to be at least 3 volts. I'm trying to get it to work at lower voltage without changing the character of the filter.The working range is 10 Hz to 400 Hz. That's it.If I could understand what the filter does I can then modify it to work with lower voltages. It would be nice if I could get it down to 1 volt.Max output of the coil is unknown but its a small coil so cant be more than say 50-60v.Ps: Not sure what R11 does but the simulation work even without it in place.

author: MrAl
author_rank: Super Contributor
timestamp: October 27, 2023, 09:34:53 pm
content: Quote from: chinoy on October 27, 2023, 04:01:51 pmOh so the more complete circuit finally shows upIt may be hard to get it to work at a lower voltage but you can try first changing the two diodes to Schottky type, see if that helps.If not, stick an op amp amplifier on the front end with gain whatever you need to get 3v on the output. Probably need a negative supply also though.We might try some transformations to keep the bandwidth the same.For C2 and R2 the transform would be to lower the value of R2 and increase the value of C2 by the same ratio.

author: Terry Bites
author_rank: Super Contributor
timestamp: November 01, 2023, 12:11:24 am
content: What information are you trying to exctract from the coil?Could there be a better way?In what mode exactly is the BC557 used in? C106?Is a BC like a 2N or an LM?More circuit secrets to be revealed from the right hand side I hope.Get yourslef a free sim, eg Qspice