Analog electronics riddle

[@victorman2227]

The diode bridge with a zener inside makes a bipolar limiter. R6 isolates U2 inverting input from C2. C2 will get charged extremely slowly through R3 R5.. So probably the circuit is an almost DC low pass filter + bipolar clamp Analog is fun!

[@sambenyaakov]

Getting very very close.

[@stjiu]

1. the circuit is somekind of DC low pass filter for signal within +/-6.5V with cut off combination of (R3+R5+R6) and C2 (~0.008Hz). However, it behaves as a soft clipper when it exceeded +/-6.5V. I have a feeling this circuit is used to detect the average voltage of the non DC input signal limited output at +/-6.5V. The U2 is there to provide additional precision buffer as well as low cost isolation to protect downstream components because of no high voltage involvement (except static). The signal is then buffered by U1 to the other circuit. The lack of resistance at the non inverted input at U2 shows that the signal is quite sensitive (possibly high impedance input) hence to lower the parasitic noise. The R4 is there to serve as protections for U2 by providing path to ground and also as grounding input to provide correct signal when the input is floated during connection/disconnection of IN. 2. the function of the diode bridge is to limit the output of U2 to within +/-6.5V. The instantaneous behaviour of the DC filtering properties changes slightly when the voltage exceeded +/-6.5V as the R3 become dormant. thus become low pass with a combination of (R5+R6) and C2 (0.016Hz).

[@sambenyaakov]

Thanks for participating. Very close to a perfect answer. Hold on to my answer.

[@stjiu]

@@sambenyaakov Thanks for your reply Prof. Ben-Yaakov. I look forward to reading your answer. Sugi

[@sergepetrov8598]

1. Clip signal softly to some 40uV peak, and frequency below R6C2. 2. Clip symmetrically wrt polarity.

[@sambenyaakov]

Getting very close. Good thinking.

[@vaks2l]

It is a kind of a polarity detector of the DC component of the input voltage. If the DC component of the input signal is Vdc, Zener diode voltage Vz and the capacitor voltage Vc, then if Vdc > Vz, Vc = Vz, If Vdc < -Vz, Vc = -Vz, otherwise Vc is somewhere between, but always Vc > 0 for Vdc > 0, and Vc < 0 for Vdc < 0.

[@sambenyaakov]

Not really. Hold on to my answer.

[@hardrocklobsterroll395]

Not sure about the whole circuit, but I think the diode bridge and zener would serve to clamp the excursion of the voltage from mid to ground. The bridge rectifies the AC and places it across the Zener. Once the Zener crosses its standoff voltage it will conduct and start pulling mid to roughly Vz+2Vd relative to ground. This should work regardless of input polarity. R6 isolates the inputs of U1 and U2 and creates an RC filter lowering the output bandwidth of U1. To be honest, I’m not entirely certain of the operation of the whole system, but can’t wait to learn. I love these riddle style videos it tests my knowledge and helps me think about deciphering circuits

[@sambenyaakov]

🙏Getting a bit close

[@toiganmalot8576]

r6 and c2 is a very low load for 20 meg output so it's works almost as a dc filter. Diode bridge is for voltage clipping

[@sambenyaakov]

Not really 🤗

[@Blanke94]

A Duty Cycle sensor to be used in the feedback loop.of a Duty Cycle Correction Circuit? U2 and the bridge serve as a clamp to keep the integrator saturated so that the feedback path has enough gain. Possibly an identical circuit is used to process the NOT(clk) signal and they are fed differentiallly to the corrector not shown here

[@sambenyaakov]

Not really. Hold on to my answer.

[@electronicaindia]

The circuit with zener is use to limit positive as well as negative voltage Peak limiter upto (6.1 +.8)=+/-7v , and the output of the buffer give a exponential output but very slow changing if input is high or low. I suggest the 4 diode and one zener can be replaced by back to back zener having same value .

[@sambenyaakov]

Close. Hold on to my answer.

[@32bits-of-a-bus59]

Seems to me like a first order filter with a cutoff frequency of 1.6 kHz equipped with input voltage clipping done by the Zener diode in the bridge. If no clipping takes place, the opamp maintains negligible voltage between its + and - terminals because of the feedback. The feedback mechanism hides the high resistance of 10M resistors, effectively mirroring the input voltage to the - terminal where it is sensed by the input of the RC filter. This hiding of high resistance is only possible for voltages low enough so that the current flowing thru the R5 is high enough to supply relatively low R6. The current is given by the potential in the MID point, which is clamped to Vclamp=2*0.65+Vzener. The thing is complicated by the derivative nature of that current caused by C2. So for Vin such that Vin + I*R5 = Vin + R5*C2*d/dt(Vout) < Vclamp it behaves as a low pass filter, for higher voltages it clamps the input. The peculiar thing is that the clamping level depends on the rate of change of the output. I am quite unsure what practical purpose this circuit could serve. Looking forwards to seeing its explanation!

[@sambenyaakov]

Getting close, keep going.

[@michaelk.1108]

With this explanation it looks to me like a circuit which reduces the harmonics by use of the low pass filter in case of voltage clipping. So a kind of protector circuit with signal reconstruction.

[@edenkidiahman]

The diodes with the high resistance turn The signal to a square current source (Depends on the Zener Rating and the resisters value). C2 turn this current to a function of voltage versus time so as lowere the frequency the higher the peak voltage of the output it seems that this circuit is kind of frequency to voltage transformer

[@sambenyaakov]

Thanks for participating. Not really. Hold on to my answer

[@Stelios.Posantzis]

The function of the diode bridge is to clip symmetrically anything above/below */- (Vzener+1.2V). These problems are impossible! I'm glad I'm not taking an exam on them!

[@sambenyaakov]

👍🙂. Hold on to my answer.

[@Stelios.Posantzis]

@@sambenyaakov My only guess is that, for a simple sinusoidal signal, this circuit works as a low pass filter for signals below the clamping voltage (as defined by the bridge) or as a fixed amplitude (close to the rail voltage) for signals above the clamping voltage and with a frequency lesser than that of the low pass filter. I have no idea what it does with complex AC signals or with DC signals.

[@sambenyaakov]

@@Stelios.Posantzis as explained in the answer video the parameters of this circuit are for very low frequency and small amplitude.

[@zaikindenis1775]

R3 and R5 are included in the feedback loop when no clipping. So output impedance is relatively small (~20MEG/AOL) for small signals and it is a low pass filter. But when the signal is big, the time constant increases a lot, and on the output we have still a sinus wave...Not sure here. Thank you for the interesting video.

[@sambenyaakov]

Getting close, keep going.

[@davidsecades6633]

My guess: D1 to D5 is essentially a bi-directional voltage clamp. If it is disconnected, and R3 & R5 shorted, U2 output voltage tracks IN, while R6 & C2 provide lowpass filtering, with U1 meauring the filtered voltage. So this simplified circuit is a low-pass measurement of IN. R4 works as a pull-down resistor for IN, when it is disconnected. Reintroducing the diode bridge, with Vclamp, there are no changes when IN < Vclamp. When Vclamp > IN, MID = Vclamp, and there would be a short between U2 output and ground, so R3 is added to stop this. However, this R3 is not in series with the bridge, but in series with RC, which also limits the maximum current draw of the RC circuit to (+V-MID)/R3. However, this design does not allow to control the allowed current without affecting the clamping lost current. With R5 added, the RC circuit voltage is still clamped, but now moreover the current drawn by the circuit is limited by R5, gaining this control. Now, assumptions aside, the allowed currents before clamping seem to be very small, so Idk how this affects the RC circuit. Same with IN < 0, just invert the inequalities. All in all, I belive this circuit takes an input voltage and procudes the same voltage low-passed and clamped in power (seeing that the lower IN is, the more current is allowed to pass to the RC). As per the purpose, perhaps it can be used for treatment before an ADC measurement, seeing that it provides low-pass filtering, voltage clamping for out of range measurements and max power drag. What do you tihnk?

[@sambenyaakov]

Not really. Hold on to my answer.

[@peplegal8253]

With that tiny current trough the Zener I would guess the ciruit takes advantage of the V/I curve of the Zener. Maybe a voltage squarer...or any other exponencial function (wich one I'm not sure). R6+C2 is just a filter to remove thermal noise...I can only imagine a diode working with such low current must be noisy as hell.

[@sambenyaakov]

Not really. Hold on to my answer.

[@sambenyaakov]

Not really. Hold on to my answer.

[@Korni0816]

1. High input impedance low pass filter with unity gain, slew rate limitation, output voltage limit and low output impedance. The Output is generally the buffered voltage across C2. The input voltage at IN is amplified by U2 so that a voltage equivalent to IN is at the feedback point to the inverting input across R6 and C2. I will divide the analysis in DC and AC operation. For a DC Input the voltage across C2 equals the input voltage. Since no current is flowing through C2 there is no voltage drop across R6, nor R5 and R2 (assuming ideal OpAmp and neglecting leakage current through D1-D5). Therefore the voltage at AMP and MID are also equal to IN. If IN is rising further, the voltage at MID will also rise and eventually D5 will reach breakdown, clamping the voltage at MID. U2 is trying to keep up and will eventually reach its output voltage limit of just below the supply voltage. Current supplied by the Output of U2 is flowing through the diodes to ground and can no longer charge C2. The current is limited by R3, so it will be fairly small (at or below 1uA). Since the voltage at MID is now limited and no other current can flow through R5 and R6, the voltage at C2 will not rise further and therefore OUT is also limited to this voltage. For an AC Signal C2 will be charged and discharged by a current supplied by The Output of U2. This current creates a significant voltage drop across R5 and R3. Since the voltage at MID is clamped, the voltage drop across R5 is limited and therefore the current to charge C2 is limited,.The voltage across C2 and the output is the integral of this current, therefore leading to a limited voltage rise rate at C2 and the output. This current limit is effective for discharging of C2 as well. This Voltage limit at MID therefore leads to a slew rate limit of C2 and therefore the Output voltage. Both DC Voltage limit and AC slew rate limit are active simultaneously. It makes a good filter for DC or very slow changing input voltages, limiting anything that is high frequency and and high amplitude AC noise. 2. bidirectional output voltage limitation and slew rate limitation. The rectification with D1-4 ensure that The TVS Zener D5 will limit the absolute value of MID, regardless of polarity. The activation threshold is symmetrical. Furthermore the diodes significantly help reducing the inevitable leakage current of the Zener.

[@sambenyaakov]

Very close to a perfect answer. hold on to my explanation video. Thanks for the input.

[@raja4521]

Voltage clipper which clips voltage above 7.5V on either directions

[@sambenyaakov]

Not really🤗

[@elektorix]

The diode bridge acts as classic full wave rectifier that is loaded by Zener diode D5 so D1...D5 are bipolar and symmetrical limiter. Limit voltage is about Zener voltage + 2*diode forward voltage. Here it would be roughly +/- (6.5 V + 2*0.7) V. It tends to static V(OUT)=V(IN) curve limitation. When voltage in the MID node is within range that I mentioned previously it's buffer. When the MID point voltage is lower than - (6.5 V + 2*0.7) V or higher then + (6.5 V + 2*0.7) V there’s high current flowing from the MID point. It tends to voltage drop on R3 and then U2 saturate at about +/- 14 V for +/- 15 V power supply. Obviously voltage at inverting input of U2 would clamped at about +/- (6.5 V + 2*0.7) V. I omitted diodes reverse current. Another issue would be during AC analysis. Here diodes capitaines and opamp capitaines can even couse oscillations but there’s also low pass filter R6-C2 with cutoff frequency about 1.6 kHz. R6 has small resistance and C2 large capitaine so node with inverting input of U2 would be loaded by R6-C2. I think it would be stable anyway but only with R6-C2. I guess that it's voltage limiter for DC and small frequency. With higher frequency it can be issue with opamp SR. I geuss that for higher frequency at OUT it would be something close to sine wave due to R6-C2 low pass filter. I think it's circuit for creating something like synchronization signal for further purposes and this signal would be independent of input voltage amplitude. Am I correct Sir?

[@sambenyaakov]

Close. Hold on to my answer.

[@pabloortiz7739]

Low pass filter, but the voltage clipping is acting as an anti windup or something like that, in such a way that the dynamic range of the operational in not limited until Vc is high (in absolute value) and opamp wants to continue charging above that value.

[@sambenyaakov]

Not really. Hold on to my answer.

[@Eldon_Dice]

All I can make out from the circuit is unity gain and bipolar voltage clamping -- the exact function and values are above me at the moment. You're getting am analog signal and buffering it -- inverting-- voltage clamping -- inverting signal to positive value. Edit: Didn't see the cap and res -- active filter not inverter.

[@sambenyaakov]

Thanks for participating. Hold on to my answer

[@TheRevenant-pn2xi]

R3 and R4 are large and within the feedback loop. This basically forces U2 to initially hit the rails on whatever input level. The diodes seem to be clipping the output to about half of the supply (5.6V zener + 2*0.6V). So at net MID we should always have ±7V. The voltage drop across R5 is Vin-Dependent. This voltage then charges C2 through R6 and the output is buffered by U1 and given out. Maybe this is a some funny looking voltage-controlled RC filter? 🤔

[@sambenyaakov]

You are very close. What is the consequence of +/_7V?

[@NishanthSalahudeen]

1st order lowpass filter with input clipping in both polarities. Thats the first impression. But, not so simple. The 10megs screwup everything. Gotta think more

[@sambenyaakov]

Indeed.

[@raphaelcoti692]

It's a slew rate limitor + clamp that need to reset input signal (diodes ref) to go back to normal, fast changes in voltage leads to clamping same if you go above diodes setting Input follower with hysteresis system Diodes to clamp the signal if the variation of input voltage is too violent (slew rate) once clamped you have to get down below diodes reference voltage (that's why I said hysteresis or Schmidt Trigger as you wish) 1rst order RC filter to set the maximum slew rate then follower Globally : This system is clamping on magnitude of voltage but in first place it reacts to how fast the input voltage changes, indeed if you change input very slowly the output of U2 won't make a voltage rush to match the input, so you don't go above diodes ref R3 R5 R6 and C2 set the slew rate and make a filter Diodes are clamping depending voltage You need to get down to diodes voltage to "reset" i'm not sure about that maybe you need to go down to 0V because in transient when you reach diodes voltage to reset you will oscillate between clamp and current rush that will creat enough voltage to go through the diodes ... To simulate !!

[@sambenyaakov]

Thanks for participating. Very close to a perfect answer. Hold on to my answer

[@NNNILabs]

The least I can say is that the diode bridge is a "precision clamp" a la Bob Pease and Jim Williams...

[@robson6285]

I think this is an audio compressor the 4 diodes rectifie and so there are always 3 diodejunctions in series who act like non linear resistanses. The same way for the positive and negative halves of the signal. Or so i think, but the 1μF capacitor seems too big for most of the audio frequencies so a lin to log for a VU meter? O no that should work more log to lin.. Only sure on one thing and thats that i look out for the answervideolesson, even more than normal because this is the way we made things work in "our time" so with good old opamps and a clever analog circuitdesign! Supergreat riddle, supergreat topic for a riddle!

[@sambenyaakov]

Not exactly

[@Stelios.Posantzis]

By the way, it is not obvious that the diode in the centre of the bridge is a Zener. It took me a while to see that. ROHM says RSA6.1J4 is an ESD protection Device (TVS) with Typ/Min 6.1V and Max 7.2V Zener voltage.

[@sambenyaakov]

👍thanks.

[@lapserdak24]

Creepy circuit. So as long as Vin is within +/- 6V, it's a buffer (not sure the bandwidth). But if it gets out of this range, the output remains the at the clipped voltage for a long time. Bandwidth is so low, like 0.015Hz. It's a detector of some kind.

[@sambenyaakov]

Some kind is correct😊

[@lapserdak24]

@@sambenyaakov is it something like DC level measurement? can't guess 😁

[@BkR090]

Looks like a noise filter and amp for AC signals. Audio maybe!? The bridge makes the zener bi-directional, and that will short AMP output above or below +/- 2*(Vzd5 + 2*Vd) referenced to Vc2.

[@BkR090]

But why the big cap? This is driving me crazy. Now I'm thinking it's just a noise filter for a DC or a low frequency signal!!

[@sambenyaakov]

😊Here s hint. Try to figure out how the circuit works when the 10Meg resistors are much smaller and without the diodes, what happens when you increase the resistors back and then the diodes. Good luck.

[@sambenyaakov]

Getting close.

[@BkR090]

​@@sambenyaakovI thought it was an attenuator that removes anything above 1V, but then I gave up and did a sim. It attenuated anything above+/-0.7V. So I still don't get it😊. It works well as an amp/buffer for AC in the mV/uV range. And I don't know of an application for attenuation to that range.

[@BkR090]

@@sambenyaakov the "switching" diodes are there to improve the zener's response time. So this circuit is expecting really fast AC!!!

[@k7iq]

This is a GOOD one, Sam !I would enter this circuit into LTspice but I don't want to cheat. That might give a better clue though. It definitely runs on some form of electricity !! :) I don't think that R6 does much ? Zener clamp within the center of feedback loop plus the very high impedance driving that Zener diode, stops the feedback voltage at whatever the Z clamp voltage... But possibly with a wide knee due to hi R. And also stops the charging of C2 any further. But C2 and its history is also affecting the response of the feedback or lack of feedback when the Zener diode is off, on, partly or full. When negative feedback is removed by clamping, I would imagine that C2 can charge/discharge faster since the op-amp output will be railed then. So speed of IN makes a bit difference. I don't know what the circuit's function is but it must be some kind of DC control voltage maker. Audio VCAs need control voltages. Maybe that is a usage. But usually, VCAs need a log amp for their control. Unless distortion is required. Many good answers here but I think I will have to wait for your follow-up. Thank you for making us think !

[@sambenyaakov]

Getting warm. Hold on to my answer.

[@armangevorkyan1975]

It might be some EFT filter, which clamps high voltage transient above some voltage(Vz+2Vd), in case input voltage higher than >( Vz+2Vd) or lower than

[@sambenyaakov]

Not too bad but not exactly. Good thinking.

[@santiall73]

I'd say some sort of triangle wave generator. I'd guess the output of the op amp would be a square wave that is integrated by the cap, only working at low frequencies as the amplitide of the output wave is frequency dependent? Not sure about the clipper, the 082 will have a max output of approx 12-13V, the clipper is in between the 2x10M divider agains whatever voltage is in the cap. So I'd say it isn't doing much haha but I am commuting after a work day so who knows what I am missing... Fun, looking forward to the explanation

[@sambenyaakov]

Not really. Hold on to my answer.

[@christianbuzzio1002]

That feedback after the 20Meg resistance, along the clamping circuit, gives me an idea of an oscillator, and the RC serves as an integrator. Is it something like a sigma-delta digital to analog converter?

[@sambenyaakov]

Thanks for participating. Not really. Hold on to my answer.

[@dr.dharmrajghodke974]

+ve and -ve clipper circuit to limit the peak amplitude.

[@sambenyaakov]

Not really🤗

[@sonoilmario]

It a limited slew rate circuit, like the One needed to adjust the Speed set of a motor in inverter drive

[@sambenyaakov]

Good. What a out the diodes?

[@sonoilmario]

The bridge allows the positive and negative swing to be of the same amplitude due to the only one zener used to clamp the errore voltage

[@sambenyaakov]

@@sonoilmario Correct but for what purpose?

[@monsieurLDN]

As someone with no analog experience i think 2. Is protecting something

[@sambenyaakov]

Good guess!

[@Blanke94]

Is it a regulator, clamped to +/-(Vz+2Vf), with the bridge providing bipolar functionality? |Vout| is max(|Vin|dc, |Vz|). If Vin

[@sambenyaakov]

Not really. Hold on to my answer.

[@dfgsdja]

Reverse leakage current in the diodes could be an issue.

[@sambenyaakov]

OK, but what is the purpose of the circuit?

[@codures]

Simplified bass guitar overdrive. Convert sine to square (clipping).

[@sambenyaakov]

Close but missed the point. Suppose it is the guitar clipper, what would be the BW with no clipping?

[@codures]

@@sambenyaakov indeed professor, given the R-C (20 Megs and 1μ) the fc it's closer to 0 Hz than to 1 Hz. So, it's outputting the Vin's DC bias up to around +/-7V because it's kinda hard to put a real value from the top of the head at such low currents. Opinion: 2 zeners in series back to back would yield the same results.

[@michaelk.1108]

OP AMP U2 can have a max. output voltage swing of +-13.5V. The diode bridge limits the voltage to the Zener Voltage of D5 (6.1V...7.2V) plus twice the forward Voltage of 1N4148 (2x1V). So the protection is for +/- 8.1V... +/- 9.2V. Maximum Voltage range at point MID can be -9.2V...+9.2V. Is the cathode of D4 connected properly?

[@sambenyaakov]

Getting close, but not exactly. You are correct about the diode D4 but this is just a drawing.

[@sonoilmario]

The current Is so low in the bridge that suggest me that diodes + zener are also with the purpose to compensate each other a temperature drift...May be?

[@sambenyaakov]

Thanks for participating. Hold on to my answer

[@BeMuslimOnly]

I am sorry if i am wrong, The diode network is clipping the positive and negative input voltage to some limit, depending upon zener diode voltage, and thus limiting output voltage at 2nd op-amp. R and C is acting like a low pass filter. So the circuit is low pass filter with limited output voltage swing.

[@sambenyaakov]

Not really. Hold on to my answer.

[@SuperHal2k]

Frequency to Voltage Converter ?

[@sambenyaakov]

Not really.

[@michalmichal2495]

Hmm... meaby it is a part of inductor/transformer saturation detection?

[@sambenyaakov]

Not really 🤗

[@kashifkhan5565]

Some type of peak detector?

[@sambenyaakov]

Not really🤗

[@smartups1]

Triangle to sinewave shaper

[@sambenyaakov]

See the posted answer.

[@smartups1]

Where the post ?

[@sambenyaakov]

@@smartups1 In 'sam ben yaakov" KZbin channel

[@electromatic2014]

Passive Voltage divider

[@sambenyaakov]

Thanks for participating. Not really. Hold on to my answer.

[@electromatic2014]

@@sambenyaakov Omg I didnt think you would answer, it was a joke 😭

[@sambenyaakov]

@@electromatic2014 No harm done. I got worse answers😂😂

[@byronwatkins2565]

It is an exponential 'triangle wave' generator. The Zener at the heart of the diode full bridge sets the charging voltage for the RC timer.

[@sambenyaakov]

Not really. Hold on to my answer.

[@pkumar_381]

The diode circuit is there to limit the swing at the MID node. VMID can swing b/w +/- (2*VD + Vzener_reverse). Beyond that, diode circuit will not allow any swing and feedback would be broken.

[@toiganmalot8576]

D5 requires 1mA to work as zener. With 10meg resistor that wpuld requires 10kV

[@sambenyaakov]

OK. And the purpose of the circuit is?

[@toiganmalot8576]

@@sambenyaakov i was wrong, it works actually😳. Looked for iz(vz) characteristic. Thanks professor, i just learnd something 🙏.

[@biswajit681]

Answerrrr plsssss!!

[@sambenyaakov]

Some say: 'the journey is worth more than the destination

[@tamaseduard5145]

👍🙏❤️

[@sambenyaakov]

#1 😊👍🙏

[@electrowizard2000]

Looks like the output of diode bridge is RC lowpass filtered with cutoff 1.6kHz and buffered. The bridge is sampling opamp output current and if the produced voltage is above zener+1.2V the current is diverted to ground. Diverting opamp output current seems act as a slew rate limiter: With a 1V input step, 10mA would charge C2, so MID would be 1+10M*10m... nope, opamp output would hit the rails at +15, Mid would be 7.5V, but since zener is certainly conducting, it will be held to zener+1.2, allowing a current (zener+1.2-V(C2))/10M to charge C2. Diode bridge makes this true regardless of input polarity. I belive R4 is just to allow input leakage currents. Since I'm not clear why you would do this I must be at least partially wrong. The opamp would often be saturated and struggle to recover if polarity switches often, and at slow rate of change the zener won't activate at all, with just the 20M impedance limiting the drive.

[@sambenyaakov]

A bit off.

[@eduardinification]

Its a follower (Vout=V- = V+) until VMID reaches Vz+2Vd, after this point, the OPAMP (Saturated) and the capacitor, provide the rectifier bias current until Vout = Vin returning to linear region. So, is it an oscillator?

[@sambenyaakov]

Very close, but the conclusion is wrong.