Right PT specs for rack tube preamp

[ACHIEVEIT]

I don't know how experienced you are with building amps, but if you are planning to build a SLO-like hi-gain rack preamp, perhaps a kit like this can save you a lot time and effort, especially for the chassis and the faceplate, and of course a proven layout/design.

No actually i want to build a modified preamp based on the clean and lead channels of the marshall 6100. I have attached the schematics with some voltage analysis i did in case anyone is interested

[jazbo8]

I see, then that particular kit is too different to be useful, good luck with your build.

[R.G.]

Unfortunately, i don't have access to a pc to use the PSUiiDesigner - i am on vacation, so i use a tablet to post. I am trying to understand how things work so as to order the pt asap. I was thinking about ordering it from a local experienced winder. I didn't know about the tube town pts - they sure look nice although i am not certain the 6v3 winding is center-tapped? The local winder gave me an even lower estimate than tt's, so i think i ll go with him since i will spare the shipping + i will get the center tap

Get the 6.3V winding center tapped. In all cases, the heater winding must be referenced to ground, and the simple way is to use a CT on it. It is possible to construct a "center tap" for grounding by using two low value (~100R) resistors in series across the winding, and that can even be quieter, but if you don't need it, a wire is simpler and cheaper than resistors.

So, with the target dc being ~310v, say that the pt ht secondary is 245v. After the rectification, the voltage should be (in loaded state and considering the 1v4 voltage drop of the diodes) around 330v.

By choosing a 10uF reservoir cap, the voltage ripple is Vripple=t*I/C = 0.01*0.02/(10*10^-6) = 20v, which, percentage-wise, is ~6% of 330v.
Then by placing an RC filter with R=750 Ω, the voltage drop on the resistor is Vdrop=750*0,02= 15v and the power consumption is P=I^2 *R = 0.3W (so even a 0.5W seems sufficient although i would go with a 1W). With that in mind, the voltage of the second capacitor would be ~315V. So to further reduce the ripple by 10:1, Xc=75Ω, so C=1/(2*π*f*Xc)= ~10.61uF. So by placing the RC filter, i further reduce the ripple by a factor of 10 while ending up with ~315v and keeping the components happy and relatively stress-free.
Regarding the current of 20mA, 20mA*1.8=36mA is the minimum pt rating, so let's say above 40mA.
Have i understood anything wrong?

You have the essence of it correct. You've busy learning as hard as you can, so I went off and did some more direct looking for you. I put a model of this into my circuit simulator and messed with it a bit. Here are some of the short answers.
1. I tried values of 245 and 255Vac at 50Hz for an input, 80, 100, 250 and 300 for equivalent transformer wiring resitances, and 10uf and 22uF in both cap positions.
2. 10uF in both cap positions gets you under 1% ripple in all case, 22uF makes it even lower, if that matters. It may not. 1% is pretty good.
3. The transformer equivalent wiring resistance makes a difference. This is the measured wire resistance of the primary times the secondary voltage divided by the primary voltage, added to the measured secondary wiring resistance. It happens that you're very near unity on that voltage to voltage thing, so it's approximately the sum of the two. The difference between this number being 80 and 250 (which are reasonable estimates for transformers of this size) makes about a 13V dc difference on the final DC. At 80, you get 313vdc out, at 250, you get 300.
4. The transformer RMS current changes with the transformer resistance, too. Plan for 50ma secondary output minimum to be sure. The lower the winding resistance, the higher this number.
5. Consider going to 260Vrms on the high voltage winding, and then adjusting that "750R" resistor to get the DC voltage you want. It drops off DC volts at the same time it reduces ripple. This is a lot easier than getting a transformer that's got too low a DC out, and then ordering another one. Or consider making 300Vdc your target.
6. This analysis ignored the effect of wall-socket voltage variation. Your DC voltage out will go up and down in direct proprtion to it's variation from the AC voltage the transformer was designed for.
7. If you're tight on space or height, consider using two transformers, one for HV and one for filaments.

Do i need to ask the pt winder what the ht winding resistance will be in order to get a correct psuii emulation?

Yes. It makes 10-20Vdc difference on the simulation I ran.

Remember the wall-socket voltage variation, and that you can always get a lower voltage by dropping it a bit with resistors, transistors, or zeners, but it takes a new transformer to get it a little higher.

[jazbo8]

No actually i want to build a modified preamp based on the clean and lead channels of the marshall 6100. I have attached the schematics with some voltage analysis i did in case anyone is interested

Just curious, where do you live or source the parts to get a custom wound PT cheaper (I am assuming) than one off-the-shelf? I'm always on the lookout for a good deal...

[ACHIEVEIT]

Again, thanks a lot for all your help R.G. I just sent an email to the transformer winder and i will update the thread accordingly once i get a reply. I asked for a 260vac secondary as it seems wiser as you said

On a different note, i am thinking about using a 5v dpdt latching relay to change channels (clean, lead) with the two poles switching both the input & output of each channel. I would use a momentary spst switch (and the same goes for the footswitch) to close the 5v connection to the relay (actually i would switch the ground of the 5v relay power supply). Does this sound like a good idea? If yes, can i use the same logic to calculate the power supply to the relay? (i would rectify and put a pi filter on the 6v3 winding - in parallel with the heaters - to end up with 5vdc). In that case, the current i need to calculate is only the current drawn from the relay right?

P.S. In case you are ok with that i would be really interested to see the simulation files you ran R G. - granted they are in a file i can use (with ltspice or psuiidesigner or something similarly available)!

Thanks again, i ve really learnt a great deal so far!

[Gregg]

In that case, the current i need to calculate is only the current drawn from the relay right?

For current draw check out your relays datasheet but 5V relays usually draw ~30mA each so multiply that by the number of relays you intend to use to get the total current draw.

[R.G.]

On a different note, i am thinking about using a 5v dpdt latching relay to change channels (clean, lead) with the two poles switching both the input & output of each channel. I would use a momentary spst switch (and the same goes for the footswitch) to close the 5v connection to the relay (actually i would switch the ground of the 5v relay power supply). Does this sound like a good idea?

Should be fine. Be sure to take note of the special timing and other requirements of the actual latching relays you get. There are two kinds of latching relays: single coil and dual coil. For single coil, you reverse the direction of voltage on the coil to make it flip. For dual coil, you put a voltage on the other-side latching coil to make it flip.

In both cases, the whole point of a latching relay is that you only need to blast the coil with voltage/current for a fraction of a second it takes the relay to flip to the other side, then you can let go of the coil. As such, the net DC current of the relay is nearly zero as long as you can store up enough juice to make it flip. Most smaller latchers only need a tenth of a second or so to flip. So you get the advantage of no standing DC current to hold it in a position, but pay for that in having to figure out how to pulse the coils, not hold them.

Also, the heater currents on your 12AX7s is about 300ma per tube. The relays pull about 30ma each during the time they're on.

If yes, can i use the same logic to calculate the power supply to the relay? (i would rectify and put a pi filter on the 6v3 winding - in parallel with the heaters - to end up with 5vdc). In that case, the current i need to calculate is only the current drawn from the relay right?

Same logic/process, only the numbers have changed. Except for the gotcha that the current only needs to flow for a rraction of a second, not continuously, and that mucks up your abiliity to say what your load current is. Pulsed loads tend to involve charging up a capacitor to enough energy stored to make the change happen, then re-charging them over time when the load isn't on. In this case, I happen to know that a 100-220uF cap charged to 5V will make a latching relay in the small DIP package flip. But if you leave the current on all the time, it will be bigger.

P.S. In case you are ok with that i would be really interested to see the simulation files you ran R G. - granted they are in a file i can use (with ltspice or psuiidesigner or something similarly available)!

I'll see if it will spit out a SPICE file. It's a graphics based simulator front end to a SPICE background. I only use it at the graphic front end level because I'm lazy about writing SPICE models.

[Dave H]

P.S. In case you are ok with that i would be really interested to see the simulation files you ran R G. - granted they are in a file i can use (with ltspice or psuiidesigner or something similarly available)!

Here's one you can run in LTSpice. I set the input voltage to 350V peak (about 250V RMS) and the transformer resistance to 250R.

Bridge PSU.zip

[ACHIEVEIT]

Thanks a lot for all the answers guys.
@Gregg - i checked the datasheets of a few relays and they are indeed ~30 ma each

@Dave H - thanks for the file - i ll have a chance to check it in a few days!

@R.G. - i received a reply from the pt winder and he said that the exact value will be known after the winding is complete, so i think it's best to go with a worst case scenario and ask for a >=260v secondary and mess with the dropping resistors if necessary. As to understand the theory behind that, was the resistance you asked the complex resistance of the pt as seen on the secondary side?

On a different note,i thought about what you said that latching relays are not as plug n play as i thought and they actually need a "pulse generating" circuit. With that in mind, do you think it is worth venturing into latching relays or should i stay with non-latching relays for a simple 2-channel switching? Is there a simple tried and tested circuit for simple 2 channel switching? From what i ve gathered so far from googling it seems there is a few dozen circuits for simple channel switching and i don't know which to choose

Thanks a lot!

[R.G.]

@R.G. - i received a reply from the pt winder and he said that the exact value will be known after the winding is complete, so i think it's best to go with a worst case scenario and ask for a >=260v secondary and mess with the dropping resistors if necessary.

Good choice.

As to understand the theory behind that, was the resistance you asked the complex resistance of the pt as seen on the secondary side?

Transformers have two real resistances and a complicated virtual one (or two). The real resistances are the wire resistance of the wire in the primary, and in the secondary. The complicated virtual one is the core loss, which is variable depending on the size of the magnetic field.

A real transformer can be modeled as an ideal transformer, no losses or imperfect behavior, hidden behind some imperfection adding components. In a perfect transformer with Np primary turns and Ns secondary turns, the secondary voltage is Ns/Np times the primary voltage, the primary current is Ns/Np times the secondary current. If the secondary load is a resistor, then the transformer acts like it "reflects" the secondary load resistor to the primary as (Np/Ns)². The AC current in the primary flows through the primary wire resistance, and then powers the parallel combination of the primary inductance, the nonlinear core losses, and the reflected secondary resistance.

Any current flowing through the primary wire resistance causes a voltage drop, so the ideal transformer hiding inside sees a lower primary voltage than the real AC voltage. This is what is transformed to the secondary, not the real AC voltage. And any current flowing through the secondary causes a drop in the AC voltage on the secondary wires, equal to the secondary current times the secondary wire resistance.

Generally the transformer core and windings are designed so the nonlinear core losses and primary inductance current are trivially small compared to the current caused by the secondary load, so these are often neglected for quick and dirty design work.

(honest, I'm getting to the point )

At no load, the voltage on the secondary is Ns/Np times the primary voltage, as there is negligibly little current to cause it to drop. At full load, you lose voltage to both wire resistances. It is common in transformer design work to transform the wire resistance to the side it's easiest to work with. So you can tranform the primary resistance to act like it's in series with the secondary resistance by multiplying by (Ns/Np)². This lets you calculate the wiring voltage losses more easily.

So to make a good estimate of transformer performance, you need the wire resistance of the primary, the wire resistance of the secondary, and the voltage ratio from primary to secondary (as that's identical to the turns ratio). You can then do the voltage ratio and impedance ratio, and calculate what happens with what load current on the secondary.

On your transformer, you're going to be asking for a primary voltage of 240, wasn't it?, and 260 on the secondary. So whatever the primary wire resistance is, it will act like it is multiplied by (260/240)² or 1.17 times as big as it really is. This will then add to the measured secondary resistance, which ought to be nearly the same as the primary resistance since the current ratios are very close. So you could model its behavior with a 260Vac source in series with 1.17 times the primary resistance plus 1.000 times the secondary resistance, and that's what goes into the rectifiers. And that's what I was modelling.

So ask your transformer guy for an estimate of the primary wire resistance and secondary wire resistance, then do the math. Since the voltage ratio is nearly 1:1, the errors will remain small.

On a different note,i thought about what you said that latching relays are not as plug n play as i thought and they actually need a "pulse generating" circuit. With that in mind, do you think it is worth venturing into latching relays or should i stay with non-latching relays for a simple 2-channel switching?

In going to latching relays, you pay more circuit complexity to save power. If you have a power source available all the time that will run the relays, it's really much simpler to use non-latching ("single side stable") relays and use a latching footswitch.

Is there a simple tried and tested circuit for simple 2 channel switching? From what i ve gathered so far from googling it seems there is a few dozen circuits for simple channel switching and i don't know which to choose

I'm not the best guy to answer this.

[ACHIEVEIT]

Thanks a lot for the detailed answer R.G. Your answers have truly helped me understand a lot!

By the way,i would like to ask you a detail that arose after revisiting the previous posts. Why is the lowest ripple frequency twice the lines frequency??

About the channel switching, i have been searching a lot the last days and unfortunately things are not as straightforward as i thought. At this point i am not interested whether it is a latching relay with momentary switches or latching switches with non-latching relays as long as i can have leds on both the panel and the footswitch. After spending a few hours messing with the el34world schematics, trying to incorporate switches and leds on both the panel and footswitch, into my despair i stumbled upon the marshall jubilee switching system. I have attached a schematic of it from steve (sdm) over at metroampforum.

The circuit can be seen on the left side and it is powered by the heater winding that is center-tapped, like my transformer will be. I have a jubilee and i know that it can power two leds (one on the panel and another on the footswitch). Do you think this circuit will work? Can you help me understand how it works??

P.S. It calls for a G5A-234P-DC5, which draws 40mA with a coil resistance of 125 ohm (so a voltage drop of 5v - just crosschecking) but it is discontinued and i can find the G5V-2-H1-DC5 which draws 30mA with a coil resistance of 166.7 ohm (so again 5v voltage drop of course). I guess they are interchangeable for my purpose

[ACHIEVEIT]

Channel switching questions

Well, after much thought i came up with a solution that in theory should work and i really would appreciate some feedback guys. So, with a 6v3 center-tapped winding, using a full wave rectifier wields a voltage after the diode drop of 8v21. For a reservoir cap of c=4700uf, Vripple=0,068v (so under 1%) and so the dc voltage after the reservoir cap should be 8v176. That voltage should be fine for a 5vdc relay with a low-current led (forward voltage of 1v9) connected in series, but should not be sufficient for two of them and this is why i chose to let the panel led function only when a footswitch is not connected. Once a footswitch is connected, the led of the footswitch would only function. By the way, in order for the footswitch to operate, the panel switch should be engaged. Do you think this would work??



P.S. Is there a way to change the thread title? In order to reflect reality i should add something like "+ channel switching questions"