Відео

With the volume control all the way up, about 500K ohms with the balance pot centered. With the volume turned all the way up and the balance pot all the way to one side, it would be close to 0 ohms for the nulled out channel and 1M ohm for the other. This design is actually not good for an input stage because it really needs to be buffered. If you take out the balance pot, the input impedance simply becomes 1M ohm for each channel as usual.

The CD4053 is not newer, it's a 3-throw version of the CD4051. The CD4052 shown here is the 2-throw version. The CD4066 is basically just four separate single-pole single-throw switches. For your pedal, likely any of these would work, but I know some pedal designs need multiple poles for the switch. In that case the CD4052 or CD4053 would be needed. Do put some zeners or some kind of protection on the input and output of the circuit though, these CMOS switches are not immune to static.

@@thetuberoaster8321 Thx a lot for your explanations and helpfull advices to point me in the right direction!

Yes, it should

​​@@thetuberoaster8321Great.... also...works at any frequency from 0-20khz?

The input and output are approximately equal and are something like 550mV peak-to-peak in this video. It doesn't really matter though, as long as your peak-to-peak signal is within ~4.7V, it shouldn't clip.

There is an equation for re that is a function of emitter current and temperature. That number should be taken with a grain of salt since it can be as high as 42 ohms or as low as 21 ohms for a given current depending on what you take to be the ambient temperature. For 42 ohms the gain would be something like 0.96 instead of 0.97. The capacitor value should be set to pass all frequencies above 1Hz at the input impedance of the stage before bootstrapping. This is the impedance of the voltage divider which is about 13K, in parallel with the input impedance of the base, which is ~150K for an Hfe of 150. The 150K base impedance swamps the 4.7K resistor because it can change +/- 30% or more depending on the transistor Hfe and temperature. For a 10uF cap, this gives a rolloff of just under 1Hz.

@@thetuberoaster8321 bootstrapping the input to have greater Rin ( input impedance) . Your circuit on the first page. Your out is comming what is your collector right. So what is your gain ? With high impedance. What is your input Ac voltage : 20mv Vpp ? I am really trying to understand what's going on your video. What is the benefit of having high impedance if I am not doing audio. I am building a 3 stage amplifier CE- CE-CC. What's your thoughts on that.

​@@Tankzim.z The gain is approximately Rc over Re. The intrinsic emitter resistance changes that slightly but in this case it is almost 10. The output impedance is roughly Rc which is 10K. The input impedance without bootstrapping would be dominated by the base bias network, making it hard to get anything above 10s of Kohms. Bootstrapping allows this input impedance to be at least and order of magnitude higher. This is sometimes useful for when you can't afford to load down the preceding circuit. If you have one CE stage driving another, the first will be loaded by the input impedance of the second. This will cause the gain to drop and the distortion to increase somewhat. The more obvious solution is to use a FET, but for low-noise applications this may not be feasible. This technique allows you to use a regular BJT with its benefits but without its usual trade-off of low input impedance. The input sensitivity of this amp is something like 400mV before clipping. You can tweak the biasing and the gain to change the operating point to make the input sensitivity whatever you like. The three stage amp you are talking about - the CE-CE-CC amp - allows you to split the gain between two stages, which is better in terms of distortion. The CC output will make the output much lower impedance than a normal CE amp, allowing the amp to drive heavier loads without increased distortion or loss of gain.

I should have said in the video that this series only represents one part of the cycle of the waveform. For multiple cycles the Fourier series more accurate since it shows sine components cancelling. The (infinite) series itself does not converge but 1/n does converge to zero as n goes to infinity. As n becomes infinitely large the amplitude of the nth component becomes infinitely small. As the number of terms approach infinity the wavy saw slope becomes smooth. The waves in the middle gradually cancel out as many terms add together. This is very similar to the behavior of a square wave. In real circuits, the high frequency components cannot be infinite in number because the circuit cannot respond at infinitely high frequencies. Those components would then have amplitudes of zero and therefore not matter. It is in some ways akin to integrating 1/n. Technically, the slope only approaches zero so therefore the integral can be infinitely large. Generally, the integral is taken over some convenient bound so that this integral is finite and predictable. (Like computing radioactive decay or heat transfer.) If the circuit is part of a digital system where there is a DAC, there will be aliasing and low-pass filters to prevent the aliasing. If you look at a digitally generated square or saw tooth wave on an oscilloscope you will see little waves instead of a flat line. The closer the filter rolloff is to the fundamental of the wave being generated, the more pronounced this effect is. This is due to those high frequency components being filtered out.

TBH I don't actually know where that comes from. It has also been a long time since I've looked at this circuit. The book shows the expected Vcc²/2xRl but then later it also shows Vcc²/8xRl for the same output stage calling it maximum average output power.

@@thetuberoaster8321 i can see in the book that we assume that Vout = Vcc /2 and Vout rms = Vout/sq2 so Pout avg= Vout²/R = Vcc²/(2²xsq2²xR)

The book is called: Transistor Circuit Techniques by G. J. Ritchie

It's been a long time, but I remember getting the datasheet with the characteristic curves for the 12AX7 and plotting loadlines for the anode load to find the cathode resistor. I wanted to keep the gain relatively the same so I kept the 100k anode load resistor. I wanted the anode voltage to be just above half of the available HT voltage. Ohms law gave the required anode current in order to make that happen. I settled on a desired grid voltage of something like -1.5V. Tying the grid to ground and putting a resistor of 1.2K between the cathode and ground yielded ~1.6V. The second stage is biased colder for some reason at something like -3V near cutoff. I do not remember the reason for this nor the reason for why the two input stages were so different. I do remember trying to emulate the real superlead circuit and keep the bias voltage and anode current ratios as similar as possible. The bypass caps are different between channels because one is the bright channel with less bass and the other is the normal channel. I think I did recalculate those values as well using the standard -3dB RC filter equation. The stage after the volume control is biased the same as the normal channel input stage. (The bright channel is the one with the bright cap on the volume control). I do not remember changing anything in the cathode follower driving the tone stack. I may have left it the same. The tone stack is literally cut and paste from the original. I'm pretty sure the phase splitter long-tailed pair stage was also original. The only thing that I changed was the feedback resistor in the presence knob circuit since the output stage had a different amount of gain. I did this experimentally and settled on 47K but I might have been able to go lower. The output stage I had to change the grid leak resistors to something lower to drop the gain a little and added big grid stopper resistors to hopefully knock down RF and prevent oscillation. I'm pretty sure the 12AU7s in the output make way more voltage gain than El-34s do. This causes the amp to compress the sound a bit too much at full volume. The sweet spot is about 2/3 of the way up. You could probably decrease the gain somewhere and make it more manageable. I could not tell if this excessive compression was due to the power transformer sagging, the output transformer saturating, or the gain issue.

The negative terminals of the power supplies are not directly connected to ground. There are separate ground and negative terminals. I connected the two power supplies in series and the connection point of the + and - I took as ground for the device. It's no different than connecting two batteries in series.

@@thetuberoaster8321 Thx. Didn't know that meanwell PSU-s come "isolated" out of factory.

The original one that inspired this build came from the oatley electronics kit. The link is in the description. The schematic is listed there as a pdf download which you can look at for reference. Or you can just use the schematic I provided and change the front-end volume control to your liking.

If you make the emitter resistor lower than 4.7k, it will amplify the signal, but it will also phase shift it more since. The signal from the emitter will be lower than that of the collector, therefore it will act upon it less. As you change the 50k pot, the signal will change phase but also amplitude, arguably making it less useful.

An aluminum case would help noise. However, PCB has power planes on both sides that shield the traces to some degree. Also, the gain of the circuit is rather low, which helps. The shot noise of the tube and the thermal noise of the resistors are nearly as loud as the mains hum anyway.

I replied to another comment similar to this which you can also look at. You can solve for RC in the equation but the trick is making the phase shift range large enough but useful. A 180 degree phase shift is not practically possible because it can only occur when the term for arc tangent is infinity. The best you can do is something like 176 or 178 degrees. If you solve for a phase shift much higher than 176, the range of the pot becomes really nonlinear and it does not sweep the phase range in a way that is useful. I'd recommend plugging in 176 degrees for the phase shift and the desired frequency to then solve for RC. This will correspond to the max value of the potentiometer. The 0 degrees phase shift condition is of course simple. When the pot is turned down it will go to 0 ohms or near enough making the phase shift nearly 0 degrees.

You need a decoder circuit to turn the binary coded decimal logic back into normal counting numbers again, unless you want your LEDs counting in binary. For two inputs this is easy, all you need is an inverter. For four inputs, you need a 2 to 4 decoder chip like the SN74LVC1G139 which will give an output for each LED. The outputs are active low (inverted logic) so you will need to switch the low side of the LEDs to make one light up for each input selection.

@@thetuberoaster8321 I am using decoder circuit of 2:4 Decoder using 7404 and 7408 ics

@the tube roaster - can you plz help how to select stereo and dvd which has 6 channels like front, left, surround left, surround right and centre. I can using relay but it would be. Much expensive, need 5 relays so thought once I select dvd, output should give 6 channel instead of stereo output(right, left)

@@svem1972 That should work

You can use two CD4053s together to get six channels that can switch between two sources. The 4053 has three separate single-pole double-throw switches inside. You can tie the input select pins together so they all switch at the same time. To accommodate more sources, you can add another pair in parallel and use the inhibit pins to select which pair passes the signal. The logic circuit might need to be modified however.

There are RCA connectors on the back for input and output left and right channels.