Question CD4053 input voltage

[Voltron]

Hello.

I'm confused. What happens ifyou feed a 100v pp square into one of the 4053 inputs?.

[Voltron]

EDit: An image to better clarify my question.

[armdnrdy]

What happens ifyou feed a 100v pp square into one of the 4053 inputs?.

[R.G.]

It depends on the source impedance. There are clamp diodes on the inputs that route signals bigger than the power supplies to the power supplies. The amount of current that flows depends how much current the source of the signal and the power supply on the IC will let flow.

A'n'R's comment is correct for very low impedances.       

[PRR]

> to better clarify my question.

"Show the whole circuit!"

What is the power for the 4053??

Another notorious quip: "Read the datasheet!"

CD4051B, CD4052B, CD4053B
Absolute Maximum Ratings
DC Input Voltage Range .  . . . -0.5V to VDD +0.5V
DC Input Current, Any One Input . . . +/-10mA

I will ass-ume the 4053 power is a nice round small number like + and - 5V.

The "100Vpp" is 50V peak.

At the point of +50V at top of R2 and +5V at 4053 input, we have 45V across 330K or 0.14mA available to kill the 4053 input.

The 680K will divert some of this, less than 0.01mA.

Some research (or reading R.G.) will suggest that when you take a CMOS input "outside its supply rails", you run into a diode which attempts to clamp the input pin to 0.6V beyond the supply rail.

Changing 4053 power to 0/+9V etc makes very little difference in face of 100V swings.

Since 0.14mA is less than the Ab MAX 10mA, on paper it is "safe". (But really playing too close to unexpected disaster.)

However you have carefully and painfully built-up a 50V peak tube-flavor signal, and then whacked it off with a chip's dumb diode to a low voltage. This is a very costly diode clipper, NOT a tube-color circuit.

Cut your 100V swing down to a Volt or so, so that the 4053 can pass it without added damage. Your R2 R6 net is about 1meg. You want 100:1 signal drop. 990K+10K will do that. But of course 1% values are excess precision here. Also I know the cathode follower can drive much less then 1Meg; in fact the original app (5F6A) drove under 60K. High impedance invites buzz and treble-loss. Personally I like 270K with 2.7K or 3.3K.

[nick d]

       I think you mean 100mV .......... surely ??

[GibsonGM]

       I think you mean 100mV .......... surely ??

More like 67 volts, with the voltage divider following the point shown as 100V.  But that's still 67VAC P-P, WAYYY more than anyone would want to feed that chip, as described above.  The instinct sure is to look for mV!

Tube circuits are made to be run at high voltages (160V is actually not that high), and very low currents.  "At the end", you cut it down to a reasonable level for guitar amp input, if doing a preamp, or it feeds the power tubes in a full amp. 

It is common to need to bring a signal of this level down via resistive voltage divider, which is what I suspect the OP will be doing...

[PRR]

> I think you mean 100mV .......... surely HuhHuh??

It is "reasonable" for a happy tube amp to deliver a peak output of 20% of its supply voltage. 160V*0.20= 32V peak, or 64V peak-peak. The "20%" is linear; if we SLAM the amplifier we can get more, very possibly Voltron's 100Vp-p number.

The 330K:680K divider, un-loaded, knocks this down to about 2/3rd or 44Vpp to 66vpp.

Yes, 66,000 milli-Volts.

In the original (5F6A, which ran a higher B+), the sweet-spot out of this stage was around 20 volts RMS (60Vpp), because it drove a tone-stack with loss like 10:1, then a power-amp with about 2Vrms sensitivity. That's to get "just clipping" in the 40 Watt output; obviously many-many players SLAMMED far past the just-clipped point to 40Vrms or 60Vrms out of this stage.

So switching with 15V micro-devices is probably folly. If he wants the true tube sound, he should use relays, or drop the level WAY down to fit through the CMOS without diode-damage, then boost-up if he needs to drive lossy tone-stacks with hot power-amp level output.

And sadly Larry's excellent illustration is wrong. CMOS dies very quietly, without flames or shrapnel. (The old-old relay-driver chip in Apple flop-drives would sometimes at least burst-off its lid, quietly but some satisfactory damage.)

[Voltron]

Sorry didn't had time to reply this week. Thanks for all the comments and answers.

"Show the whole circuit!"

Another notorious quip: "Read the datasheet!"

This is a fragment of the schematic, original has several pages, hope all relevant is on this image. Also the measure waveforms I found but I dont know if either of them are legit.





Seems like datasheet specs varies slightly from brand, there's no brand specified on schematic, i think best would be to find a more contemporary print according to date production unit. I don't know if that would be any relevant thou. Anyways.

http://www.ti.com/lit/ds/symlink/cd4053b.pdf


"When a switch is connected, it looks like a moderately nonlinear resistor between pins. The resistance varies with the signal voltage from pin to pin, which really means that for low distortion, you have to keep the current through the switch low." -- R.G.
http://www.geofex.com/article_folders/cd4053/cd4053.htm

So, please correct me if i'm wrong, but this isn't an inverter so no CMOS distortion, right?. So basically this goes down to:



plus a 80 Ohm resistor connected between in and out pins... a bridge rectifier for crossover distortion wtf?

According to datasheet a 1 kHz sinewave at 100 Vpp would show a typ distortion of 0.8% with a Vdd of 10v (anyways, it would work the same as long as the range keeps above 2v and below 15v).

I don't know how a 4 Vpp signal at input of tube stage would get you a 100 Vpp at output. But if 64 Vpp should be the linear range, it would mean lowest signals still got a 0.512% distortion added for whatever "typ distortion" means. Waveform chart is too inaccurate to see any crossover distortion on 4053 output.

[anotherjim]

What worries me, is that a fault condition, could put 96v on IC12b  , at least that's how I see it.

[R.G.]

What worries me, is that a fault condition, could put 96v on IC12b  , at least that's how I see it.

As my friend Jim would say, "... then all you need is a rotating magazine of spare IC12's". 
Press the button and rotate a new one in after a fault.

[PRR]

> I don't know how a 4 Vpp signal at input of tube stage would get you a 100 Vpp at output.

Why do w have tubes? For GAIN. Done right, output is bigger than input.

You can compute this. But the lazy way: 12AX7 has Mu of 100, we can never get more than half of that in a practical stage, and this stage has no cathode bypass, so half of that. 100/2/2 is 25. 4V in should give 4V*25= 100V output.

Why 4V? Because of the diode-bridge INto the tube. Going around the loop we find three junctions (either way). Peak current is no more than 15V/5.6K or under 3mA. Taking 1N4007 diode drop at few-mA as 0.6V-0.7V we have 1.8V-2.1V peak, or 4V p-p.

> there's no brand specified on schematic

Do you mean the brand of the CMOS?? Should not matter. I bet "any" CD4053 will "work".

Now the key question: YES they are putting 100Vpp through 330K into the CMOS input. The other side of the switch is 680K in a 9:1 pad, or a fixed "tonestack" of similar properties. i do not see the power supply for IC12b, but ass-uming it is simple (not actively driven to follow the signal), that CMOS input is getting hit with >0.1mA of signal current forcing its input protection diodes into conduction. That is the "clipping action", not the tube slamming (which happens when level is maybe 5X higher than CMOS-input slam).

So what is your question? Should you copy this? Sometimes that goes well, but this contraption is very complicated with too many parts. You should be able to make music simpler. Also there are some very rare chips (TC9176).

Do you have a dead whatsit and are wondering if that CMOS chip has been beat to death? It "seems" to be within spec limits, but stuff does happen, especially if you slam the spec 440 times a second over the long haul. Yank the chip, install a socket, and put in a new chip. (If it died once, it may die again.)

Personally I would not do it that way, but anybody who can draw like that has sold a LOT more stuff than me.

> a 80 Ohm resistor connected between in and out pins

When switched "on", the CMOS switch is like a resistor of 70 to 140 Ohms. When switched "off" it is like a many-Megs resistor. In all cases you have those 0.6V-drop diodes to Vdd and Vss. If powered on 0V and +15V, you can swing 1V to 14V without gross distortion. In this case I think they *want* distortion.

> 1 kHz sinewave at 100 Vpp would show a typ distortion of 0.8% with a Vdd of 10v (anyways, it would work the same as long as the range keeps above 2v and below 15v).

Well, but any way you bias it, "100Vpp" is "mostly" outside the 2V-15V range, no?

Also that part-percent figure is very dependent on Load. If you keep signals well inside the rails and use a high load impedance the THD is 0.01%. For a 50 Ohm load you may observe 10%. And if you throw 100V at 15V CMOS it is going to clip all to heck, much-much-THD. However when not CMOS-clipped, the 70-140 Ohm switch in series with 330K will give THD down near 0.02%.

One thing CMOS switches don't do is crossover distort. "In the middle" both P and N devices conduct good.


> a fault condition, could put 96v on IC12b

But through 330K, right? (It is fairly unlikely that a resistor on an inspected PCB will go short.) 100V in 300K is 0.3mA, far-far less than the several-mA spec of a faulted CMOS input.

[anotherjim]

Paul has a lot more faith in CMOS input protection networks than me. The current may be safe, but there is still excessive potential. That's all before we see what's making the HT supply. Is it protected against 240v when expecting 120v? Is the AC transformer double insulated? Is there overvoltage crowbar protection?

CD5053
Absolute Maximum Ratings
Supply Voltage (V+ to V-)
Voltages Referenced to VSS Terminal . . . . . . . . . . . -0.5V to 20V
DC Input Voltage Range . . . . . . . . . . . . . . . . . . -0.5V to VDD +0.5V
DC Input Current, Any One Input. . . . . . . . . . . . . . . . . . . . . . ±10mA

So, current max 10mA -  but voltage max 0.5v outside of supply - this is a split rail +/- 5.6Volt in the scheme.

Call me nervous and over-cautious, but I'd want to see proper galvanic isolation (audio transformers) between all the LT and HT circuits.
At least nose-to-nose zeners to ground on the LT inputs. Would 5.1volt zeners be about right?

[merlinb]

Paul has a lot more faith in CMOS input protection networks than me. The current may be safe, but there is still excessive potential.

There is no excess potential, because it is limited by the input protection. Catch 22 (aka Ohm's law). Assuming the input protection diodes can handle the 10mA quoted on the data sheet, the with nothing but that 330k resistor in series with the input you could apply 10mA*330k = 3300 volts before the chip would complain. Since it's part of a potential divider, the voltage could in fact be even higher. Hence no further protection is required with a measly 160V HT.

[bool]

Zeners

[PRR]

> Is it protected against 240v when expecting 120v? Is the AC transformer double insulated? Is there overvoltage crowbar protection?

Valid points but usually not the CMOS's problem. Ample line-to-chassis isolation is done for the USER's benefit (and to reduce wrongful death law-suits). Is my 120V light-bulb protected against a 240V socket? Well, no, that's why it says "120V" on the box. Also why we use different plugs for 120 and 240 juice. Plug adapters exist, true, but nothing is fool-proof, and well-paid lawyers put fine-print in manuals.(*)

> proper galvanic isolation (audio transformers) between all the LT and HT circuits.

That will not protect against the signal we want, the 60V audio, which will shoot-through any transformer we would want to listen to. (This would also be a very costly transformer, and liable to "change the tone".)

To preserve the tube-stage tone, and get well away from reliance from protection features, all we need is a 50:1 attenuator, perhaps 330K:6.8K. The worst-case AC or DC voltage at the CMOS would be knocked-down to 5V. The heavy loss could easily be made-up with a 19 cent opamp. Surely they need loss anyway, since everything else is +/-15V chips.

> Zeners

Are Zeners more trustworthy than simple diodes to +/-5V rails?

There is one "gotcha". CMOS switch are SO low-power that we sometimes use incredibly lame power supplies. We "could" get the 5V from 15V with a 220K:100K divider. The CMOS worst-case idle current would not sag that, and switching demand is not an issue with foot-control. That's cool if there's no strong external current INTO the CMOS. In this case there could be a part-mA, and we must design the CMOS power to absorb that. Certainly a 2K:1K divider would do. Zeners on the rails are lovely also. (Note that 7805-like 3-pin regulators often *won't* absorb pull-up, though part-mA may be tolerated by being absorbed in the regulator bias loss.)

(*) I have a weed-wacker strong-trimmer, electric, plug-in. Many-many page manual full of DON'Ts. Nowhere in there does it tell me not to cut my power-cord with the wacker. The thing is lame, so it would be hard work, but I am sure eventually it would flay my cord open and leave live wires on wet grass. Someone slipped-up.

[Voltron]

Hello. Its me again. Sorry, I couldnt get enough time to chew all great anwsers and info you guys provided. Thanks.

So I settle a few days in and I just made a simulation of the CD4053, just for kicks.



On my cd4053 sim seems the signal attenuates to the rails but is not clipping? WTF?

I also build a diode clamp as suppose to be found in CMOS protection to compare. As expected the signal gets heavily clipped to 5.6V plus 1n4148.

I was expecting the CD4053 clipping the same as the diode clamp. Maybe this is material for the other simulation section of the forum ut Is theres somethign wrong with my sim?. Is my diode clamp wrong?

[PRR]

> On my cd4053 sim seems the signal attenuates to the rails but is not clipping?

I'm not sure what is happening on either sim.

The "Input Protection" sim appears to be drawn right, however the nodes are not numbered so I can not know(?) what node "V(n005)" is. But none of the nodes should show +/-0.6V clipping... they should show (for dual 5.6V rails) 6.2V clipping.

The "4053VeeTest" sim drawing is so tangled that I do not know what is connected, what the node names are, or what the plot is showing. (My sim would not assume that nodes of the same name are connected together, as you seem to have with "Vss". Mine would call that an error, other sims may do what may be reasonable.

The internal diodes "have" to be in a CMOS model. In real CMOS use you may drive a long wire, which at 5MHz will "ring" outside the supply rails. You may instead rig an R-C network around inverters, which again will kick beyond the rails. SPICE is supposed to figure this stuff for us (if we hadda do nonlinear analysis by pencil, then why bother with a computer?). Adding diodes costs the model maker nothing. It may cost the chip-maker even less (diodes pop-up everywhere on a chip).

(However I have seen vacuum-tube models which appear to omit any form of grid-cathode diode, so you have to check for yourself.)

I did a hasty sim. I don't have a 4053 so I used a 4016. The amp-plan in Reply #8 does not show the CMOS power rails, and I (mistakenly??) assumed zero and +12V.

Clips cleanly 0.6V either side of the rails.



I re-did it with +/-5V rails, and now it clips at +5.58V to -5.58V. Image omitted (fingers tired).

The "blow-up current".... I get 0.127mA peak into the chip, 0.117mA peak from the power supply, 0.008mA to the load (switch on). While this does not quite add-up (0.002mA has vanished in numeric fog), the maximum current is less than 2% of the spec-sheet rated MAX current. If I build my dog-porch for a 100 pound dog, and only put a 1.27-pound dog on it, I would not worry about the porch falling-down. (I might worry about my friends laughing at my teeny terror.)

[Voltron]

OK. I got that, the chip will not blow cause we are way under the max.

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SOrry my 4053 sim is mess. I used a ready made circuit example (hence the CD4024B at the left) I connected everything as in Reply #8 schematic. Powered by +- 5.6V.

The input bxoby gets the sine at 100Vp thru 330k / 680k-ground (70 Vpp?), it enters the 4053 and then comes out from pin "by" to the 680k / 82k divider and goes in at pin "cy".

 the plot shows output from [cxoxy_pin4_out] which is pin4 on the 4053. The lines out of the image go to the fixed tonestack (which is disconnected as the chip isnt switched on) and to the 100Vpp Sine gen +  47n and 330k/680k.
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The CD4053 model is inside a library of a lot of CD4xxx devices most of them specifies their own diode model. However, the 4053 I used, doesnt specify a diode, it implements .subckt by Helmut Sennewald: I/O Translators adapters subcircuits. Which I think was written for general CD40xx I/O. Here the diode model is:

.MODEL CD40DIO1 D(Is=1e-12 Rs=100)
.MODEL DIO2 D(Is=1e-12 Rs=10)

I know nothing about spice parameters but quick look at the big SPICE paper: Is = sat current and Rs = resistance. I don't know how accurate that sounds thou. I was expecting at least depletion capacitance and reverse breakdown as important data here.

I think it's the only 4053 spice model available online. ANyways, now that the exploding chip issue was tacked down, I am more concerned about what is happening with the audio signal. Here I settle to sim the input protection diode scheme (powered by +-5.6V).



Here is what I got when I connect to 680k/82k



ANd here is what I got with the fixed tonestack connected



Doesn't make sense to me that the tonestack signal doesnt even look slightly squared. I was also expecting the signal thru 680k/82k should be equal or even higher than the tonestack. It also does not match at all to waveform chart on reply #8.

OH, now that I look at the waves again, I also forgot completely that the signal was already squared BEFORE the tube stage!.

So, again, it was a 1.9 Vpp sine into a TL072 (with almost 200 gain?) powered by +-15V so I guess thats why 30 Vpp at output.
Then that goes thru 5.6k+220n and the bridge rectifier arrengement (0.6V-0.7V x 3 each side) which should slightly round the square and tame it down to 4Vpp.
Then that goes into the tube at 1Meg and I remember "12AX7 has Mu of 100, we can never get more than half of that in a practical stage, and this stage has no cathode bypass, so half of that. 100/2/2 is 25. 4V in should give 4V*25= 100V output."
WHich goes to the 47n+330k/680k divider (70 Vpp?) and that goes into a switching chip with diode protection powered by +-5.6V, which should clamp the signal from 70Vpp down to a 6.2 Vpp and put it thru a fixed TS or 680/82k which should output a squarey 1.4Vpp or 1.5Vpp, according to wave chart. Not happening  

[Ravi]

Hello everyone!

I came accross this thread since i was looking for a simulation model for CD4053, I was wondering if it is possible to get it here?

Thanks in advance!