Carvin X100 Schematic Verification and Ultra Hot Cement Power Resistors

[R.G.]

I apologize for being testy. I was having a grumpy day.

[Ten Over]

The only time that the rectified and filtered voltage is 1.414 times the peak AC voltage is when there is no load on the DC voltage and you have a perfect rectifier. Once a DC load is introduced, the sine wave that the transformer puts out becomes flattened. Taken to the extreme, the transformer would put out a square wave so that the peak voltage would be the same as the RMS voltage. In actual practice, the peak voltage will be somewhere between the extremes of the RMS voltage and 1.414 times the RMS voltage.

Different voltage meters have different opinions about what the RMS voltage is for a flattened sine wave. A typical voltage meter might read 50Vac when the peak AC voltage is 61V, so the meter is not just spitting out 0.707 times the peak voltage. It would pretty much require an oscilloscope to determine what the peak AC voltage is when your voltage meter reads 50Vac.

[Ten Over]

The AC voltage source on at least some of the simulators puts out a sine wave of the same magnitude regardless of the load that it is driving. This causes the simulator to come up with voltages that are higher than those obtained by driving the same load with a real world transformer. Sometimes the difference between reality and simulation can be quite large as in the case of this Carvin bipolar power supply. With no zener diode, The simulator came up with 26.7Vdc presumably at the reservoir capacitor while I came up with 16.0Vdc at the reservoir capacitor when both cases were running at 50mA. The difference is because the simulator put out a sine wave and my transformer put out a flattened wave. My load was 300 ohms for 50mA and the simulator was 500 ohms for 50mA because the simulator had a higher DC voltage.

[Ten Over]

For the Carvin at hand, the two 4136's will typically draw 12mA just sitting there with no signal and no load. The 5532 will typically draw 8mA under the same conditions. The two 2N3391 circuits will draw 2.8mA when the switches are closed. This leaves the 2N5550/2N5400 push-pull amplifier that drives the reverb. I know from experience that these little amplifiers can draw an astonishing amount of current, so a total 50mA load on the bipolar power supply is actually plausible.

[Ten Over]

The zener diode does not conduct for most of the AC cycle. When the rectifier diode first becomes forward biased, the zener diode still doesn't break over because the reservoir capacitor has discharged to a voltage that is below the zener voltage. Once the reservoir capacitor has charged-up enough, current starts to flow through the zener diode, so the duration of the current pulse through the zener diode is even smaller than the duration of the charging current pulse. It is difficult to ascertain the overall heat dissipation of the zener diode when the "on" time is so small and the "off" time is so large.

[R.G.]

The AC voltage source on at least some of the simulators puts out a sine wave of the same magnitude regardless of the load that it is driving. This causes the simulator to come up with voltages that are higher than those obtained by driving the same load with a real world transformer. Sometimes the difference between reality and simulation can be quite large as in the case of this Carvin bipolar power supply. With no zener diode, The simulator came up with 26.7Vdc presumably at the reservoir capacitor while I came up with 16.0Vdc at the reservoir capacitor when both cases were running at 50mA. The difference is because the simulator put out a sine wave and my transformer put out a flattened wave. My load was 300 ohms for 50mA and the simulator was 500 ohms for 50mA because the simulator had a higher DC voltage.

The only time that the rectified and filtered voltage is 1.414 times the peak AC voltage is when there is no load on the DC voltage and you have a perfect rectifier. Once a DC load is introduced, the sine wave that the transformer puts out becomes flattened.

This is true. A rectifier will drop the peak voltage by 0.6V to 1.0V for silicon diodes, depending on the diode's internal characteristics. The sine wave from the transformer becomes flattened by the fact that the diode only flows current into the filter capacitor for a short time before the peak of the half wave. The degree of flattening depends on the equivalent resistance in the diode windings, the internal resistance of the wiring and diodes, and a small amount of voltage loss in any parasitic inductance. By far the biggest flattening is from the transformer winding resistances. The exact amount of flattening depends on the transformer, and each transformer may well be different, especially in the case of a low-voltage tap on a high voltage winding, as this case seems to be.

Another source of flattening is the peak distortion on the AC mains themselves. There are so many rectifier/filter loads on the AC mains that they can cause flattening on the voltage of the AC mains. It's a recognized problem. I can provide references if you would like to dig deeper.

In this case, apparently the transformer resistances did not flatten the peak voltage on that tap enough; that's why the design includes that 200 ohm resistor. It flattens the AC wave as seen at the rectifier by an amount of the charging current times the 200 ohm resistance. This probably dramatically overshadows any tranformer winding resistance voltage losses, but if we want to be precise, we ought to consider them both.

Taken to the extreme, the transformer would put out a square wave so that the peak voltage would be the same as the RMS voltage. In actual practice, the peak voltage will be somewhere between the extremes of the RMS voltage and 1.414 times the RMS voltage.

I'm having a little trouble understanding that. Can you elaborate? I get that with a rectifier you would see increased flattening, especially with a huge capacitance or extremely low load, and that eventually the AC voltage would get clamped to a square-ish wave, representing the part of the sine wave around the zero crossing where the rectifier(s) are not conducting, but I don't understand why this would be limited to dropping to the AC RMS voltage. A big enough load would drag the AC voltage down to +/- one diode drop - the voltage would all be dropped inside the transformer windings. This is probably not going to be good for the transfomer.
Um, did you mean in the case where there was only one half wave rectifier, not two as in the circuit under question? In that case, a big enough load would clamp the conducting half wave down to a diode drop, and leave the other half wave untouched, so that the transformer output voltage would be a half-wave sine on the non-conducting half-waves and clamped nearly to ground on the conducting ones.
Can you elaborate? I don't understand what you mean there. I don't see why the flattening should stop when the external transformer voltage drops to the nominal RMS.

Different voltage meters have different opinions about what the RMS voltage is for a flattened sine wave. A typical voltage meter might read 50Vac when the peak AC voltage is 61V, so the meter is not just spitting out 0.707 times the peak voltage. It would pretty much require an oscilloscope to determine what the peak AC voltage is when your voltage meter reads 50Vac.

That's true! I loves me some oscilloscope! In fact what most voltmeters do is to to an internal half- or full-wave rectifier/peak detector and just scale whatever that voltage happens to be down to indicate what a sine waveform of that peak voltage "ought to be" as an RMS number. There are true RMS voltmeters, and thankfully the prices on these have come down. But the ordinary analog or digital meter probably relies on peak-detect-and-scale.
This was a big point in my instruments class back in the 1970s - learning all the ways that meters can misrepresent the truth of the matter and fool you.

Beyond that, I took the "50Vac" as an indication, not a precise measurement. It's a good place to start, but in my analysis, I allowed for that voltage being between 45Vac and 60Vac. I was expecting it to be imprecise. But it's a useful indicator as a starting point. If the one, true, known-only-to-&deity voltage was 50.000000Vac RMS, the peak would be 1.414 times that. But in the real world, nothing is ever that precise. I haven't yet mentioned AC line tolerance, but I did allow some for it in the alternate circuits. Even if the transformer designer did try to make it 50.0000Vac, the AC mains can be lower or higher. So 50 is a starting point. Could be higher, could be lower. It's a good, conservative design practice to allow for reasonable variations in the power inputs.

[R.G.]

The AC voltage source on at least some of the simulators puts out a sine wave of the same magnitude regardless of the load that it is driving. This causes the simulator to come up with voltages that are higher than those obtained by driving the same load with a real world transformer.

Believing everything a simulator tells you is a beginner's mistake, all right.
A good simulator will offer you both ideal components and real world models. The ideal parts let you add in your own "imperfections" to circuits to see what happens when you to model for parts you don't have. So yes, simulators usually offer "perfect" sine wave power sources as an option. Of course the perfect ideal parts don't reflect the real world - they rely on the user to be informed enough to put in their own "imperfections". I got into the habit of doing that back in the 1970s.
In the case of this simulation, I modeled the transformer output as a "perfect" sine wave followed by a resistor to represent the transformer winding resistances. I could have modeled in the primary inductance, primary and secondary leakages, primary and secondary resistances, core nonlinearity, and non-ideal coupling factors. I didn't do that because I have enough decades designing power supplies and transformers to have a feel for when that is needed. But the "perfect" sines let me do that if needed.

Sometimes the difference between reality and simulation can be quite large as in the case of this Carvin bipolar power supply. With no zener diode, The simulator came up with 26.7Vdc presumably at the reservoir capacitor while I came up with 16.0Vdc at the reservoir capacitor when both cases were running at 50mA. The difference is because the simulator put out a sine wave and my transformer put out a flattened wave. My load was 300 ohms for 50mA and the simulator was 500 ohms for 50mA because the simulator had a higher DC voltage.

I disagree with your reasoning on exactly why there was a difference, but I can see how you might get there. The fact is, I suspect that I would have had a sim answer matching your results within millivolts if I had the actual transformer at hand to measure the actual transformer's winding resistances and so on. I did not have the benefit of having a physical device to measure. I had to estimate what it might be, based on experience.
In particular, the simulator put out a true sine wave, which was then flattened by the estimated resistances and not affected by the other loads on the other windings of the transformer. That is - the simulator case was flattened by an estimate for internal resistances, the transformer output flattened inside the trannie where you can't get at it to measure.

I think I mentioned that I had to estimate the loads. That does happen.

[R.G.]

The zener diode does not conduct for most of the AC cycle.

It conducts from before the peak of the half-wave that charges the cap and tails off to zero for the rest of the half-wave. In the picture, the mountain-looking trace is the diode current, sensed at 1ma/1mV. The peak is about 243ma. So it conducts for the last 7.5mS out of the 8.66mS of the half-wave. See picture below. If I had a working head here, I could instrument it. I'm aware of lots of places where simulators give erroneous results, but in this case, I think it's very close. After all, if simulators were always wrong, no one would use them. They have to be right sometimes.

When the rectifier diode first becomes forward biased, the zener diode still doesn't break over because the reservoir capacitor has discharged to a voltage that is below the zener voltage. Once the reservoir capacitor has charged-up enough, current starts to flow through the zener diode, so the duration of the current pulse through the zener diode is even smaller than the duration of the charging current pulse.

Referring to the picture, The mountain-peak trace is the zener current, sensed by a 1mV/1mA sensor to convert it to a scope trace. The zener current turns on at T1 and off at T2, total of about 7.5mS. The capacitor voltage ripple is show in the middle sawtooth waveform, amplified up to 500mV per division. The cap stops charging at the peak of the sawtooth and starts discharging. The zener's current peaks when the cap voltage peaks, but for the rest of the zener current time, the incoming voltage, flattened by the 200R resistor and the transformer winding resistances, continue to pour current into the zener. So the zener continues to conduct for a long time because the transformer half-wave voltage is reduced by the resistances and tails off, with the zener eating the difference in the rectifier current and the capacitor and load current.

It is difficult to ascertain the overall heat dissipation of the zener diode when the "on" time is so small and the "off" time is so large.

Not in the sim, or with an RMS reading current sensor. The yellow area is a simulator probe report. The RMS current in the zener is 89.7ma in this one run, and the RMS voltage is 14.6. The power must then be 14.6 * 0.089 = 1.309W.