LTspice modeling of EH Scott Bass boost Circuit.  - Page 1
Forum section: electronics / projects

author: polaraligned
author_rank: Newbie
timestamp: June 07, 2015, 11:31:26 pm
content: Hi,First post here.I am trying to model the bass boost circuit of an EH Scott Philharmonic Radio. It uses push/pull arrangement thru the filter.I used FET's to simulate the tubes in the push/pull circuit. I am just looking for the frequency response of this filter, nothing else.Problem is that I can't get LTspice to show the output in the frequency domain. If I do a transient analysis, I see a sinewave approx 70v p-p at the output in the time domain. Why no output when I do an AC analysis? I am new to LTspice and this really has me scratching my head. The 250v DC supply on the center tapped choke is just like the original circuit.Anyone know where I may be going wrong?Also attached is the area of the original circuit in question.

author: codeboy2k
author_rank: Super Contributor
timestamp: June 07, 2015, 11:48:34 pm
content: as a test, change your frequency range.. change the end of the frequency sweep to.ac oct 50 20 200Megand see if you get something.. maybe it's there, but not where you expect it.

author: c4757p
author_rank: Super Contributor
timestamp: June 08, 2015, 12:03:47 am
content: Those MOSFETs aren't working - you have them biased deep in cutoff. Bias them up.

author: polaraligned
author_rank: Newbie
timestamp: June 08, 2015, 12:19:07 am
content: Changing freq range did nothing.If the problem was the FET's being deep in cutoff, which it would seem is the problem, why do I get this on the output? (see screen capture below) Each FET should be running class B.

author: c4757p
author_rank: Super Contributor
timestamp: June 08, 2015, 12:28:03 am
content: Because you can't do class B in an AC analysis. AC analysis assumes the bias point never changes - it doesn't notice that your signal is large enough to swing the FETs in and out of conduction, it just assumes they stay at the same bias current all the time and the input signal is relatively small.

author: polaraligned
author_rank: Newbie
timestamp: June 08, 2015, 12:28:34 am
content: Here is the FET drains plotted along with the output.How the heck do we get 450v peak on the FET drains? The DC supply is 250v.

author: polaraligned
author_rank: Newbie
timestamp: June 08, 2015, 12:30:39 am
content: Quote from: c4757p on June 08, 2015, 12:28:03 amThanks. I had a feeling the answer was along these lines.I guess I need to figure out how to model this circuit in a different way to find the response of this filter.

author: Andy Watson
author_rank: Super Contributor
timestamp: June 08, 2015, 01:09:43 am
content: Quote from: polaraligned on June 08, 2015, 12:28:34 amIsn't that the intended function of the tank circuits - i.e. they are being operated near resonance?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 08, 2015, 04:03:10 am
content: The choke halves should be 150 H / 4 = 37.5, not 87.5. (Where did 87.5 even come from..?)With choke supply load, the "plate" voltage is naturally capable of swinging above 250V, by about the same amount that it is pulled below (on an equal-area basis -- "area under the curve" being flux, the magic thing that makes inductors work).10 ohms is a poor choice for source resistance. 1k would be more representative of the transconductance you'd see from something like a 6L7. A few mA of bias should do. Keep the input swing under 1V for linear conditions.Tim

author: MK
author_rank: Regular Contributor
timestamp: June 08, 2015, 05:43:11 am
content: There are some tube spice models available on the web, I came across some pages that had them when I was looking for some other device models, didnt need them so I have no copy to upload.

author: polaraligned
author_rank: Newbie
timestamp: June 08, 2015, 11:33:10 am
content: Quote from: T3sl4co1l on June 08, 2015, 04:03:10 amHi Tim.The inductors are each 175 Hy and center tapped = 87.5 hy each "section" which I used a "K" directive to tie together.I lowered the input signal to 1v. I changed the source resistors to 1k.Attached is what happened. The lower traces are the output, and the top traces are the filter side of the 15k drain resistor. My original intent was to have the FET's operate class B because it is push-pull circuit.It looks like the FET's are working in the linear region now, but still no AC analysis possible with this circuit. The waveform window shows no output when I do AC analysis.I don't see the point in modeling using tubes as this filter should work whether driven by solid state devices or hollow state devices.Any ideas how to model the frequency response of this filter?Thank you all.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 08, 2015, 11:12:42 pm
content: Right, 175.Perfectly coupled inductors divide by 4. The end-to-end inductance of two 87.5H in series, coupled (in phase), is 87.5 + 87.5 + 2*k*sqrt(87.5 * 87.5) = 350.This circuit is not class B. It's class A push-pull. It's actually worse than that, because the 6L7 heptodes appear to be in use as volume control (post detection AVC??), which will vary plate current and signal current proportionally. The DC level will change slightly as the AVC control is varied. (A good analogy is a single balanced differential modulator using BJTs. Or an operational transconductance amplifier, but with a complementary pair of single ended outputs.)A few volts DC offset on top of those AC sources is required to get AC steady state conditions.Tim

author: polaraligned
author_rank: Newbie
timestamp: June 09, 2015, 09:54:09 pm
content: Many thanks to you Tim. I really appreciate your help. Somehow I got caught up thinking about the 6L7's being in class B.Attached is the response of the bass boost circuit that I was after.I measured the values of the inductors to arrive at the values used.Thanks again,Scott

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 09, 2015, 10:10:07 pm
content: Doesn't look like much of a boost...

author: polaraligned
author_rank: Newbie
timestamp: June 09, 2015, 10:22:24 pm
content: The Scott's are known to have a weak bottom end. The Zenith's always packed more punch on the bottom.

author: c4757p
author_rank: Super Contributor
timestamp: June 09, 2015, 10:24:36 pm
content: Quote from: polaraligned on June 09, 2015, 10:22:24 pm*Beavis and Butthead laugh*