# Resonant - Tank Circuits

Discussion in 'Homework Help' started by JDR04, Aug 10, 2012.

1. ### JDR04 Thread Starter Active Member

May 5, 2011
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I'm currently learning about resonant/tank circuits in my studies and would like to build a few practical circuits to measure out on my oscilloscope and check out behaviour etc. In other words being real curious

I'm told that I could very easily damage my oscilloscope due to high voltage spikes etc when using inductors. As I'm new to this, how would I go about protecting my oscilloscope from any damage and what other precautions should I take.

Your help will be appreciated - Thanks again JDR04

2. ### crutschow Expert

Mar 14, 2008
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3,236
Inductors generate spikes if they are conducting current and and you suddenly try to interrupt the current flow, such as by removing one of the connections or opening a switch. That will generally generate a large voltage spike and arc at the point of disconnection since the inductance will cause try it's best to keep the current flowing. If you don't do that you should be OK.

3. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks crutschow, so in short is there some sort of simple circuit I could build to protect my instruments, ie meter,scope etc from the voltage spike you mentioned or is it all way more complicated than I am understanding it? Thanks again JDR04

4. ### vk6zgo Active Member

Jul 21, 2012
677
85
Unless you are trying to make LC circuits which are resonant at 50Hz & run them off the mains,or run high DC currents through high inductance iron or ferrite cored chokes,you are unlikely to see any spikes which can hurt your Oscilloscope!!

If,like most people,you are going to make up some LC resonant circuits,& feed them from a signal generator,they will not have DC currents flowing through them,so there will not be any stored energy if you open the circuit.

In any case,an LC circuit does not look reactive at resonance,but,rather,resistive.
In other words,what you have been told is a load of old bollocks!

PS:Back in the late 1950s & early '60s when I was first learning about such things,the term "Tank" for a resonant LC circuit was already being spoken of as "archaic",& I haven't heard it used by real RF people since around the mid '60s,
It seems to have been revived recently,for some inane reason.
It annoys me!

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5. ### WBahn Moderator

Mar 31, 2012
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I've always been curious about the voltage across the inductor (and the capacitor) at resonance in a series LC circuit. Isn't it the applied voltage time the Q of the circuit (or something along those lines)?

I assume that as soon as you try to pull anything beyond a small current from the junction that the Q gets trashed and the voltage falls rapidly, but what about circuits that have high input impedance? Could they see high enough voltages to damage them?

Also, is this (part of) the mechanism by which resonant circuit are able to pull such weak signals out of the air? Namely, that signals at or near the carrier frequency are boosted by the Q of the circuit?

Oh, and I came up in the 80's and 90's and the use of the term 'tank circuit' was real common, though I can't claim to have ever been tight with the RF community.

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
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When I think of resonant Circuits I think of series RLC.

When I hear tank circuits, I'm reminded of "anti-resonant" circuits or parallel connections.

7. ### GetDeviceInfo Senior Member

Jun 7, 2009
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I grew up with the term 'tank'. We have large RF heating tunnels where we still use the term.

8. ### vk6zgo Active Member

Jul 21, 2012
677
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Ideal resonant circuits don't have any R,so are more properly LC circuits.
This is fairly legitimate,because the R part doesn't play any part in determining the resonant frequency,it just affects the Q of the circuit.

"Anti-resonant" circuits is technically correct,but like "tank",it smacks more of the 1930s than any more recent time.

W.Bahn: A series resonant circuit appears as a minimum impedance,because the voltages across the L & C reactances are 180 degrees out of phase,& of equal value,so if this type of circuit is placed across the two conductors leading from the output of a signal she source to the input of an Oscilloscope,the 'scope display will dip as the source frequency is moved across the resonance point.
A parallel resonant circuit appears as a high impedance at resonance,so the display reading will peak at resonance if connected in the same manner.

Yes,in a Radio receiver,such as a crystal set,Resonant circuit "magnification" increases the signal voltage enough to turn on the germanium diode & produce enough audio to drive headphones.
The point is,that the voltage is still very low,even when the parallel tuned circuit is driven by a signal generator.

You would have to be very inventive to produce enough voltage to do any damage to an Oscilloscope input,which is what the OP was worrying about.
Any Radio Tech or Ham with access to a signal generator & a 'scope will have done this test at one time or another.
It is a useful way to see if that inductor in the "junkbox" can be usedin a particular application.

Interestingly,it seems like the term "tank" circuit has survived longer outside the RF community where it was first coined.

9. ### WBahn Moderator

Mar 31, 2012
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But that's not the measurement I described.

A 10Vpp sinusoidal source (capable of delivering, say, 10A of current into a short) drives a series RLC circuit with R=10Ω, L=10mH, C=10nF

I look at the signal across the capacitor (not across the whole thing).

What would the oscilloscope display at resonance?

The scope I have has the following faceplate placard: 1MΩ, 13pF, 400V max

On a similar note, what voltage rating for the cap should I use in this circuit?

10. ### crutschow Expert

Mar 14, 2008
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3,236
An LC circuit does store energy with an AC input, even if there is no DC. If you open the connection to a series resonant circuit or between the capacitor and inductor of a parallel resonant circuit at the peak of the AC current through the inductor you can indeed generate a voltage spike, with peak voltage determined by the inductor size, the inductor peak resonant current, and the stray capacitance.

11. ### vk6zgo Active Member

Jul 21, 2012
677
85
Yes,I guess I oversimplified the matter,but the resonant frequency of any LC network which is likely to be used by a real person experimenting with LC resonant circuits will have a period so short,that any physical action to open the circuit will cover many cycles,the probability of opening the circuit at either peak is remote.

As you say,the peak voltage generated is determined by,amongst other things,the peak resonant current,which will almost always,in the practical case,be minimal.

In my first post,I suggested that circuits resonant at 50/60Hz,across the mains may well have a problem.

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
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It's used in the ebooks here so, you might want to take that case up with the author.

Should we drop Anode and Cathode too?

13. ### crutschow Expert

Mar 14, 2008
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Not necessarily. If you use the value of 95% to 100% of the peak current value as containing 90% of the maximum energy then you are looking at the time between 71.6° to 108.4° of the peak of the sinewave. This is ≈20% of the cycle giving you a 20% chance of opening the circuit within 90% of its peak energy value.

So, at least for lower frequency resonant circuits, the chances of opening the circuit near the peak of it's inductive energy is significant. Even though the physical action of opening the circuit may be relatively slow, such as physically removing a wire, the time it takes for the circuit to open can still be very short leading to a large voltage spike.

14. ### vk6zgo Active Member

Jul 21, 2012
677
85
A really valid point! For some reason,Ive always just placed the whole resonant circuit either in shunt or series with the line.

Lets have a look at the example:

(a) 10v p-p ------p-p is really just a convenient myth so we can easily compare waveforms with an Oscilloscope,so that becomes 5vpk.

(b) The above supply can deliver 10Apk to a short circuit,so that means its internal impedance is 0.5 ohms.

(c) The values of L&C given produce a resonance at approx 15.9kHz.

(d) Calculating xL & xC from the above, at resonance they are of equal value,at 1kOhm.

(e) To the external circuit,the series resonant circuit is zero ohms,so current,I(res) through the RLC network is: 5/10.5=0.476Apk.

As I(res) flows through both L&C in series,the voltages across them are equal ,but 180 degrees out of phase.

We only care about that across the C ,which is:
I(res)* xC= 1000 x 0.476= 476 vpk.

There are the big volts,just as predicted!!
If ,as seems probable the scope spec is vpk,it exceeds the rating of that device.

OK,now for a reality check!

Remember my comments about someone playing with LC networks with a signal generator?

For the frequency in this example,the most likely generators to be used are a Function Generator,with an output impedance of 50 ohms,or an Audio generator,with a 600 Ohms output.

Both of these devices,normally top out at +24dBm,which in the former ,is 5vpk in 50 ohms,or 10vpk open circuit.

For the latter,it is 17.36 vpk in 600 ohms,or 34.72vpk open circuit.

Lets look at the 50 ohm case, where I(res) now is 10/60=0.166Apk,& the volts across C = 1000 x 0.166,or 166vpkwell in spec!

For the 600 ohm case,we have I(res)= 34.72/610=0.057Apk.
Volts across C are now 1000 x0.057,or 57vpk!

Maybe we should suggest that people playing around keep the generator output fairly low,but from the above practical cases,there is,as I originally said, no likelihood of damage to the Oscilloscope.

I have never said that very high voltages were not possible in resonant circuits,they are a fact of life in the design of antenna tuning units & suchlike at similar power levels,but they are unlikely in the extreme during ordinary bench tests with normal generators.

Last edited: Aug 14, 2012
15. ### WBahn Moderator

Mar 31, 2012
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I've got a pretty cheap function generator that outputs 20V peak (open circuit) with a 50Ω output. Whether it is able to deliver 400mA to a short, I don't know. But let's assume we have one that is 10V peak and that can deliver 200mA to a short.

If we want to get the voltage across the cap to exceed 400V, we only need

$
\omega C < 0.5mA/V
$

Let's choose that same 10nF cap from the previous example, for lack of an obviously better choice.

That means we need the resononant frequency (assuming just LC)

$
$

which is basically 80kHz, requiring a 400uH inductor.

So I'm not at all sure that someone can't get voltages that endanger their equipment, though I am more than willing to admit that it is unlikely they will do so without having something happen that should have given them a decent heads-up first.

But here is what I'm really curious about.

Should it be possible to make a very simple circuit, say for capturing something in the AM broadcast band or perhaps in the 2MHz WWVB range, with a circuit that is basically an antenna feeding a resononant circuit and taking the voltage across the capacitor to an A2D converter that is sampling at, say, 10MSa/s. All processing would then be done digitally. I'm not looking for any kind of good performance, but if I can demonstrate that it is possible to have an SDR receiver that is this simple, then it would make a good starting point for then walking up to better circuits.

16. ### vk6zgo Active Member

Jul 21, 2012
677
85
I think that the SDR Rx idea should work,& no doubt be a lot of fun to build.
There are a few websites describing experiments with simple SDR,especially among "Hams"

In the meantime,you may care to revisit your calculations,as a capacitor which has an Xc of 1000Ω at 15.9kHz will not look like
2000Ω at 80kHz,which it would need to do to have 400V across it with a current of 200mA.

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17. ### WBahn Moderator

Mar 31, 2012
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I see your point. I'll revisit the calcs tomorrow as I have to leave town in a few minutes. I was making the computations for my generator (assuming it can deliver 400mA) and switched half way through. Probably failed to catch something when walking the change through.

18. ### WBahn Moderator

Mar 31, 2012
17,757
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Okay, so stupid math error. It should be w<50krad/s, or 7.96kHz (half of the prior frequency, which makes sense) and an inductor of 40mH.

Once again underscores the value of always asking if the answer makes sense. I'm just starting to work with this stuff this way, so my sense of what makes sense isn't too well developed yet, but having mistakes like this pointed out the way that you did helps to develop it, so thanks.

Of course, we can always choose a smaller cap to get the needed frequency up and inductor size down.

19. ### vk6zgo Active Member

Jul 21, 2012
677
85
With the correction,you have shown that it is still possible to get voltages above the max voltage spec of your Oscilloscope in some conditions.

Thinking about it,the likely mechanism for damage would be "zapping" a FET in the input circuitry.

If,as is common,a 10X probe is used,instead of directly connecting the LC network to the Oscilloscope input,the probe would protect the 'scope, even if the voltage across the C was in excess of the probe ratings,as it is a passive device.
Although we may have a lot of volts,there is very little energy present to damage the probe's internal components.

As the occurrence of such problems is almost unknown,it seems as if most people do not have available,(or use) a generator output level of this magnitude,or have used X10 probes,or both!

Perhaps it is fair to suggest that the OP keeps the generator signal to a fairly low level,& uses a 10X probe.

20. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
I tend to agree that the actual risk of damaging your test equipment is probably pretty low. But it still begs the question of component selection. Both the capacitor and the inductor are seeing 400V+ voltages across them and therefore, it would seem, would need to be rated for it. Even if we are working with input signals that are quite a bit smaller, say just a few volts, we would be seeing upward of 100V across each. Most of my caps that I have laying around are ~50V working voltage. So, once I start playing with this stuff, I should probably get higher voltage caps (in addition to making sure I have the right types of caps). Since I expect to be winding most of my own coils, I probably should spend some time making sure that I understand the major factors in determining the max voltage of such coils. I expect it will be the insulation breakdown voltage assuming two wires are in physical contact with each other (and does that make it nominally twice the rated breakdown voltage of the wire?). Even if it does (and assuming that I am right that this is the dominant factor), I would probably just use the rated voltage.