Rectifier and max electrolytic Cap

[Guus]

Hi guys,

I have a question about the max electrolytic Cap you can put direct after a rectifier tube.

Why question?
I built a guitar amp with 5Y3 rectifier(JJ) followed by a 47uF cap.(no choke)
Some time in use and so far no problems, but later on, when reading a spec list of the 5Y3, I have doubts comming up.

I once read, if using a choke, the max C after the choke isn't so critical anymore? but, in case of using NO choke you have to be carefully with the first C.
So,
- why are there limitations on the max Cap, put direct after the rectifier?

- what happens if not?

Let's say: a 5Y3GT can (formal)be followed up by 20uF or 33 uF max?
What problems can, over time, occur when using a bigger Cap, like 47uF?

- What is the max C placed after a 5AR4/GZ34 with and/or without choke?

regards
Guus

[Milkmansound]

I was just talking about this with a tech friend last night.

He said 40uF after a 5Y3 is common practice in many amps. A lot of people ignore the specs - you should be fine. It will strain the tube a bit more during power on

[Structo]

It's because of the high current draw to charge the first cap.

If the cap is too large, it can exceed the rectifiers rating.

[Firestorm]

If you look in your tube manual, you will not see a "maximum capacitance" listed (sometimes there is a capacitance value shown in the "typical operation" section). But there is a "maximum hot switching transient plate current" and a series of charts showing the safe area of operation for given voltages and currents. And these all change with different levels of "effective plate supply impedance." This why the issue is more complex than simply giving a capacitance number. What we see published and repeated are sort of guidelines.

When you first throw the switch, the capacitor looks like a dead short to the rectifier and during the first few cycles the rectifier will be pumping a lot of current to "top up" the filter. Fortunately, the tubes haven't warmed up yet and so aren't passing current at first, so this happens quickly. On the other hand, if you had a standby switch right after the rectifier and had the tubes all warmed up when you switched to "play," the stress on the rectifier would be much greater. That's hot switching.

You must limit rectifier current to keep it within safe operating levels: smaller capacitance does this, a choke does this (if the choke is first in line, the maximum plate current rating at a given voltage goes up) and plate supply impedance does this. A "cheap" power transformer with poor regulation has a highish impedance to begin with, but resistance can be added to either side of the PT to increase it. Nobody seems to do this.

As long as you don't hot switch and you don't try to get more output current than the rectifier can provide (you can't run a JTM45 on a single 5Y3), you will be fine with 40uF on a 5Y3 (some people say 60) and 60uF on a GZ34.

There are also fairly complex ways to split the load between a tube and a solid state rectifier to maintain some tube feel without stressing it. Gerald Weber published one somewhere.

[JazzGuitarGimp]

I think stressing the rectifier tube goes beyond initial power up: every time you hit a huge chord with the amp dimed, the first cap is discharged. Maybe not all the way, but it does need to be "topped off" again every time this happens.

[Firestorm]

I think stressing the rectifier tube goes beyond initial power up: every time you hit a huge chord with the amp dimed, the first cap is discharged. Maybe not all the way, but it does need to be "topped off" again every time this happens.

Which is why you can't run a JTM45 on one 5Y3. The cap recharges 120 times per second and a 40uF cap at 400V can store enough energy to provide 384W for 1/120th of a second so you never really pull that cap down as far as you think.

[JazzGuitarGimp]

Which is why you can't run a JTM45 on one 5Y3. The cap recharges 120 times per second and a 40uF cap at 400V can store enough energy to provide 384W for 1/120th of a second so you never really pull that cap down as far as you think.

I don't think instantaneous current demand from the amplifier circuit is part of the original discussion. So, let's put that 5Y3 into a Tweed Deluxe for a minute. When you hit the amp with a huge signal, the reservoir cap gets pulled down, and at this point, the rectifier is completely warmed up. The larger the reservoir cap, the more the stress that is placed on the rectifier during top off. Think about how many times this can happen in just one gig. Or am I not fully understanding this?

[Firestorm]

Rectifier tubes are designed so that they can run into a continuous load (MI amps don't typically present that, but radio transmitters and other gear did). So one maximum rating of consequence is the maximum DC output current per plate from the rating chart. It varies with voltage. This is the average current over each half cycle. For a 5Y3 at 350VAC with a cap filter, this is 62.5mA per plate. It can do this all day everyday until it wears out. The other rating of consequence is the peak current per plate, which for the 5Y3 is 440mA. This value can be hit (but not exceeded) recurrently.

So to your point, if the input to the power supply is a big choke, the load current and the peak current will be nearly the same (the peak being limited by the choke). But capacitors, being dischargeable, will have peak current demands that can be many times the load. So in our hypothetical Deluxe idling at, say 40mA per tube, the 5Y3 can deliver 125mA continuously and the load is going hang out around 80mA no matter how many power chords you hit (until the amp slips into Class B). So the question is, how likely is it that the peak current would ever get past 440mA?

If you tried to use 6L6s, I think the first line you'd cross would be the 125mA limit
even before the 440mA came into play.

[JazzGuitarGimp]

Hi Firestorm,

I'm still learning here, so please feel free to correct / add information.

As I understand it, the inrush current required to charge a cap is expressed by the equation

I=CV/T

Where:
I = Inrush Current in Amps
C = Capacitance in Farads
V = Supply Voltage in Volts
T = Time in Seconds

Assuming the voltage is applied to the cap at the zero crossing point (that's a crap shoot), the cap will charge in one half-cycle of the 120Hz rectified DC voltage, or 4.17 Milliseconds. Let's say we've got a 47uF cap and an HT of 450V.

Then,
I=(0.000047 x 450) / 0.00417 = 5.07A

But this assumes that the PT is ideal, and has no internal DC resistance. Which of course, is not the case. The DC resistance of the PT will slow the charging time, and in-turn lower the inrush current. Now, I have no idea what the DC resistance of typical HT winding is, but for sake of discussion, let's say it is 500R. Then, (I think) we can recalculate for T as follows:

T=RC
T=500 x 0.000047 = 23.5 mS

Then re-run the equation as follows:

I=(0.000047 x 450) / 0.0235 = 0.9A

Am I on the right track?

Cheers,
Lou

[tubeswell]

- why are there limitations on the max Cap, put direct after the rectifier?

- what happens if not?

Each positive 'pulse' of DC going into the reservoir cap causes the cap to charge at a certain part of the pulse cycle - which is not the complete 'pulse', just a part of it. This is referred to as the 'ripple current'. When the pulse from the rectifier reaches its maximum DC intensity, the ripple current stops flowing and the cap then gradually discharges until it is hit with the next pulse from the rectifier. The bigger the capacitance of the reservoir cap, then harder the rectifier has to work in order to supply it with ripple current (because there is 'more' capacitance to 'charge') and the rectifier anodes run hotter. Too much reservoir capacitance causes the rectifier to overcook itself.

Manufacturers give their rectifier tubes 'peak ripple current' ratings so that people buying and using their product will know that the product will operate reliably if used within the recommended parameters.

Let's say: a 5Y3GT can (formal)be followed up by 20uF or 33 uF max?
What problems can, over time, occur when using a bigger Cap, like 47uF?

It will wear out sooner

What is the max C placed after a 5AR4/GZ34 with and/or without choke?

A good learning exercise is to look at a few tube datasheets and see for yourself.

[Firestorm]

Hi Firestorm,

I'm still learning here, so please feel free to correct / add information.

As I understand it, the inrush current required to charge a cap is expressed by the equation

I=CV/T

Where:
I = Inrush Current in Amps
C = Capacitance in Farads
V = Supply Voltage in Volts
T = Time in Seconds

Assuming the voltage is applied to the cap at the zero crossing point (that's a crap shoot), the cap will charge in one half-cycle of the 120Hz rectified DC voltage, or 4.17 Milliseconds. Let's say we've got a 47uF cap and an HT of 450V.

Then,
I=(0.000047 x 450) / 0.00417 = 5.07A

But this assumes that the PT is ideal, and has no internal DC resistance. Which of course, is not the case. The DC resistance of the PT will slow the charging time, and in-turn lower the inrush current. Now, I have no idea what the DC resistance of typical HT winding is, but for sake of discussion, let's say it is 500R. Then, (I think) we can recalculate for T as follows:

T=RC
T=500 x 0.000047 = 23.5 mS

Then re-run the equation as follows:

I=(0.000047 x 450) / 0.0235 = 0.9A

Am I on the right track?

Cheers,
Lou

If you really want to understand all the math behind this stuff, you are better man than I. It's all phase angles and square roots of frequency calculations.

I think the key voltage is the ripple voltage since the capacitor shunts that, and yes the first charging pulse is a doozy, but only lasts a really tiny time. Fortunately, there no ideal transformers or rectifiers: current is limited by transformer impedance and by the impedance of the rectifier itself.

[martin manning]

Playing with a sim that varies the size of the reservoir from 20uF to 80uF, I find that it is all about the start-up. If you control that by putting the reservoir before the standby and letting it slow-start as the filament warms up it's all good, since the peak value of the current spikes in run mode is not affected in any significant way by the size of the reservoir.

[JazzGuitarGimp]

Playing with a sim that varies the size of the reservoir from 20uF to 80uF, I find that it is all about the start-up. If you control that by putting the reservoir before the standby and letting it slow-start as the filament warms up it's all good, since the peak value of the current spikes in run mode is not affected in any significant way by the size of the reservoir.

So the only mode that might be troublesome is when you turn the power switch off with the standy switch in the play mode - the caps discharge, then while the rectifier is still hot, you turn the power switch back on, yes?

[martin manning]

So the only mode that might be troublesome is when you turn the power switch off with the standy switch in the play mode - the caps discharge, then while the rectifier is still hot, you turn the power switch back on, yes?

Yes, exactly.

[Milkmansound]

Easy fix. Goodbye standby switch - or tweak it so its just a mute