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Hoffman Amps Forum image Author Topic: Cap voltage question  (Read 3278 times)

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Offline rzenc

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Cap voltage question
« on: July 27, 2016, 07:13:44 am »
Hello there!


I have decided to build a bass preamp and would like to try the following power supply schem I found on Merlin Power Supply for Valve Amplifiers, pg.83. I also draw what I intend to use and attached it. However, I'm kind puzzled as to which DC voltage rating C02 - 100uF - should be rated at.


Can you guys gimme a help?


Thanks in advance!


Best Regards


With Respect


R.

Offline sluckey

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Re: Cap voltage question
« Reply #1 on: July 27, 2016, 08:21:00 am »
That zener diode will keep the voltage at or below 12V so a 50v or 100v cap should be fine.
A schematic, layout, and hi-rez pics are very useful for troubleshooting your amp. Don't wait to be asked. JUST DO IT!

Offline rzenc

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Re: Cap voltage question
« Reply #2 on: August 02, 2016, 12:28:34 am »
Many thanks Sluckey!!!


I have another question regarding the gyrator schem above:


Would it make sense to cascade 2 or more gyrator stages in order to have more (better) filtering?


Best Regards


With Respect


R.

Offline kagliostro

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Re: Cap voltage question
« Reply #3 on: August 02, 2016, 01:54:03 am »
Quote
Would it make sense to cascade 2 or more gyrator stages in order to have more (better) filtering?

I think yes

also there are formula to establish the H value of the Gyrator



Franco
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Offline rzenc

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Re: Cap voltage question
« Reply #4 on: August 02, 2016, 03:42:56 am »
Thanks Franco!!!!




Could you explain me what does Leff & Reff means??


Unfortunately I don't have the book on my hands...How could I calculate the attenuation provided by the gyrator?




Best Regards


R.

Offline kagliostro

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Re: Cap voltage question
« Reply #5 on: August 02, 2016, 08:18:52 am »
Have you find the info ?  :wink:


Franco
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Offline HotBluePlates

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Re: Cap voltage question
« Reply #6 on: August 02, 2016, 10:41:13 am »
Could you explain me what does Leff & Reff means??

Leff -> Effective Inductance (L)
Reff -> Effective (Series) Resistance (R)

...How could I calculate the attenuation provided by the gyrator?

Consider how to calculate the ripple reduction in a regular L-C (inductor-capacitor) filter:
 - Ripple voltage at the capacitor is the result of voltage division by the L and C
 - Find the reactance (X) of both the L and C for use in the formula for a voltage divider, at the ripple frequency (f):
   - XL = 2 * Π * f * L
   - XC = 1/(2 * Π * f * C); C is in Farads
 - Ripple Out = Ripple In * [XC/(XC+XL)]

In the steps above, XL is calculated by using Leff of the gyrator.

Voltage drop due to current by the load is provided by using Ohm's Law along with Reff of the gyrator.

Offline PRR

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Re: Cap voltage question
« Reply #7 on: August 02, 2016, 11:14:33 am »
I see HBP types faster...

> what does Leff & Reff means??

A real choke has L (inductance) and R (resistance).

The "fake choke" has effective values of L and R.

It is interesting that it says "47H 60 Ohms" but the formula for Reff, R3 [1+(R1/R2)] seems to give 20 Ohms.

> How could I calculate the attenuation provided by the gyrator?

Pretend it is a real choke and use the standard method for choke-capacitor filtering.

> Would it make sense to cascade 2 or more gyrator stages in order to have more (better) filtering?

Any time you ask if you need more than one somewhat complicated circuit, you should re-examine the real problem. Not stack-up solutions which may not be great solutions for your specific problem.

HOW much filtering do you need?

How much allowance for varying supply and load?

> to build a bass preamp

A preamp??

Preamp stages often suck 2mA each. Even a 10-stage preamplifier is only 20mA. A 250mA "solution" may be solving a different problem.

Common-sense. A zillion tube preamps were built without fake-choke filters. Half of them have no audible buzz from B+ ripple. Simple excess DC voltage and simple R-C filtering is usually ample for even complicated preamps.

The choke approach is "non-optimum" for several reasons. A choke's low DC resistance means delivered voltage hardly sags as load current varies. However typical preamp stage current does NOT vary. The low resistance promises low waste heat, but how much heat can you get from 20mA? (100V resistor drop at 20mA is 2 Watts, which is tiny next to the 10 Watt heater power of a 10-stage preamp.) The choke AC impedance should be "high" compared to load impedance, but a preamp's B+ load is a high resistance (300V @ 20mA is 15K) so we need a very-very high inductance choke to get any improvement.

Since electrolytic caps became cheaper than iron, THE way to power low-level stages is a C-R-C-R-C-R-C filter, with 1K-5K for each R and a bulk-pack (price break) of 40uFd caps for the C. Taking 2K for R, this gives 6K total resistance. At 20mA this is a 120V drop. If you start from 420V of raw B+ you get near 300V at the end, which ought to be ample for preamps.

Most preamps have low-low-level, low-level, and medium-level stages. The pickup input stages need the cleanest B+. The output stage can usually tolerate a bit more ripple on its B+. We also need to prevent signal "sneaking back" through the B+ from higher level to lower level stages. So we tap each stage or two from a different point on the C-R-C-R-C-R-C filter. This means the last R may only carry 2mA-4mA, so it can be a larger resistor for the same B+ drop. Maybe 10K-22K, which gives this one filter stage 10X the filtering of a 1K-2K+40uFd filter.

There's other rules of thumb. However IMHO the "best" way (if you want to build instead of think hard) is to steal a plan from a commercial product of similar purpose. Then, because you do not have to watch every penny like a profit-making factory, double the cap size. A 4-stage filter with 2X the caps is _16X_ the filtering (2*2*2*2). Since caps are often cheaper each when you buy 10, you can parallel 2 caps at each node, get great filtering, and have 2 spare caps for repair or other projects.
« Last Edit: August 02, 2016, 11:16:47 am by PRR »

 


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