• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Omit the filament windings -- LTM8033 from Linear

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#6
tomchr said:
Yeah..... Good luck soldering that one.

~Tom
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Yea I know what your mean. And you have a good point. However I could do it. Hot air is the trick.
I think that Jackinni's idea for a fix when filament windings are not avaiable, or not able to produce enough current is the obvious first idea. Anywhere that you need a higher voltage DC from a lower one. I wonder if you could put two back to back and get + and - for a solid state amp? If someone wants to lay out a board and have it built, I will install the parts properly no cost to anyone on the forum. Send me 10 I'll knock it then in one sitting.
 
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#7
Yes, if you take a look at some of SY's designs using the LM317 as a filament regulator, then lifting the filament well above ground for a cathode follower, use this module instead. The switching frequency is in the 100's of kHz.

Bas, if you think it's appropriate for a GB, any number of folks could have them burned and assembled at 6.3VDC, 12.6VDC etc for mounting on a couple of molex 0.100" headers.
 
#9
firechief said:
Yea I know what your mean. And you have a good point. However I could do it. Hot air is the trick.
Click to expand...

I've soldered SMDs with exposed DAP and such with a heatgun. It works alright. I also have access to a METCAL hot air reflow station at work. I *can* do the work. But why? The LMZ-series of SimpleSwitchers from National are way easier to work with. A couple of passives and you're set. Check out the LMZ12003 for example. Or LMZ14203. Easy to solder too. What's not to like?

~Tom
 
#11
I like the LTC3789. It can do buck or boost and seamlessly switch between them. Yes the circuit is a bit more complicated but it can power any tube up to 36 volts from any filament winding from 6.3 volts to 30 volts. Just the thing for lighting up those odd ball sweep tubes. The SSOP package can be soldered with a soldering iron and cleaned up with solder wick.


http://cds.linear.com/docs/Datasheet/3789f.pdf
 
#13

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#16
Both part and EVM spec sheet mention only 5V to 1.22V? Or maybe I have misread something too quickly.
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There are output voltage limitations imposed by the maximum duty cycle of 87%. This means that the output voltage can never be greater than 87% of the input voltage. There a few more terms in the maximum output voltage equation. It is on page 13 of the data sheet.
 
#20
So what happens during startup/shutdown while the input might be above UVLO 5.5V, but less than VOUT?
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That part is a buck only regulator. The output voltage will ALWAYS be less than the input voltage. There are 2 possibilities and it is not clear in the data sheet which will happen. If the duty cycle is truly limited to 87%, then the output will never be more than roughly 87% of the input (within 1 or 2%). I believe this to be the case here. You will get about 4.3 volts out of this chip with 5.5 volts input.

Many older buck regulators simple saturate the fet if asked to put out more than the input. In this case the regulator bscomes a resistor equal to the RDSon of the fet plus the other resistive losses.

In any buck regulator the efficiency is highest when the input and output voltages are nearly equal. It degrades as the difference between them increases.

The Linear Tech LTC3789 is a buck or boost part. As long as the input is within the allowable range the output will be equal to the programmed value. The output voltage can be below, equal to, or above the input voltage. The efficiency is lowest with low input voltages, and depends on the mode.

Voltage regulators can not create power. THe total power output (voltage times current) will ALWAYS be less than the input power. The difference between the input power and the output power is the efficiency. A linear regulator can have lousy efficiency if the voltage drop is high since it works like a continuously variable resistor.

EXAMPLES:

I have 12 volts and I want to light a 6 volt tube that draws 1 amp.

A linear regulator will drop 12 volts down to 6 volts but the total current will be 1 amp plus a few milliamps to feed the regulator. So the input will be 12 volts at 1.02 amps. the total power input is 12.24 watts to feed a 6 watt tube. 6.24 watts are burnt as heat in the regulator.

A BUCK regulator for this application will have an efficiency rating in the 90% range. This is given in the data sheet, but you will always get a few less percent than they state due to magnetic and switching losses. We need 6 watts for the tube. We assume 90% efficiency so we need to put 6.67 watts into the regulator to get 6 watts out. 6.67 watts at 12 volts is 555 mA. So we will need 555 mA at 12 volts to light up a 6 volt 1 amp tube. Only 0.67 watts will be lost as heat, most of it in the mosfet switch which is inside the regulator in the TI part.

The BUCK OR BOOST part will have a lower efficiency in this case since there are 4 mosfets in the path. You will loose about 1 watt in the above example, so the input current will be about 600 mA.

Lets say you have 12 volts but now want to light up a 24 volt sweep tube that draws 1/2 an amp. You can't get there from here with a linear or buck regulator. You need a BOOST regulator. Again these can be found in the 90% efficiency range at this power level. So a 24 volt 1/2 amp tube eats 12 watts. To get 12 watts out of a 90% regulator you need to put in 13.33 watts. So we need 12 volts at 1.11 amps to light up a 24 volt 1/2 amp tube. 1.33 watts are lost as heat.

Again the BUCK OR BOOST part will have a lower efficiency. so you need to feed it 1.2 amps or so.

I hope this explanation clears up a few things. Briefly you need a BUCK regulator when the output is lower than the input. You need a BOOST regulator when the output voltage is above the input voltage. Neither type of regulator works well when the input voltage is equal to the output voltage, and should not be used whan that possibility may exist over possible operating conditions. The BUCK OR BOST regulator works here and is the universal solution at the expense of efficiency and complexity.
 
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