I have a 4Hz 8KV DC sinewave, I want to make it an AC sinewave. - Page 1
Forum section: electronics / projects

author: Qmavam
author_rank: Regular Contributor
timestamp: June 27, 2023, 10:30:58 pm
content: [ Specified attachment is not available ] I have a circuit that will output a squarewave from about 100 Volts up to 8000 Volts. It was designed to output an 8000 volt squarewave and does, but I changed the input drive from a square wave to a sinewave and it has the resulting 100 Volts up to 8000 Volt DC sinewave. I want to make the output AC and would like it to operate from 4 Hz to 50Hz. Is this as simple as putting in a blocking cap? The circuit consists of a 12kV DC PS, a 6MΩ resistor then a 6BK4 tube the is pulsed on and off. The load is a 12MΩ resistor. So tube open, I have a 6MΩ and 12MΩ resistive divider with 8kV out, when the tube conducts, it drops 12kV across the 6MΩ resistor and I have zero* volts on the 12MΩ resistor load. *(tube is not a short so I have about 100 volts across it.) If a blocking cap will work, what size do I need with a 12MΩ load? Attached is a simplified schematic.

author: TimFox
author_rank: Super Contributor
timestamp: June 27, 2023, 10:39:12 pm
content: A blocking capacitor will remove the DC component (mean voltage averaged over a long time) from the waveform, but will not (by itself) turn the "square" shape into a "sinusoidal" shape.The capacitor value is chosen to obtain the "tilt" on the horizontal edges of the square wave after the capacitor.To compute the RC time constant relevant to the "bass response" of the signal, in the high state (tube off) the series resistance is the sum of your 6 and 12 megohm resistors, but in the low state (tube on) it is the sum of the plate resistance of the tube and the 12 megohm load resistor.If you are driving the grid of the tube with a sinusoid, biased to avoid clipping in the negative direction, then the R value for RC is (the parallel combination of the plate resistance and 6 megohms) in series with your 12 megohm load.Of course, the plate resistance varies over the full voltage swing on the plate.A conservative value for R in RC is just 12 megohms.For AC, your output voltage will be down by 1/1.414 (-3 dB) at a frequencyfc= 1/(2pi x RC) and approximately - 1 dB at twice that frequency.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 27, 2023, 10:58:48 pm
content: Sorry, I didn't explain well. I have a sinewave input and nice sinewave output, but it is a DC. I want it to be AC.More explanation, If I drive the tube grid with a squarewave, I get a squarewave on the output. I changed the drive to the tube grid to a sinewave and have a nice sinewave on the output. 100V to 8kVThanks, Mikek

author: Stray Electron
author_rank: Super Contributor
timestamp: June 27, 2023, 11:05:13 pm
content: If it is a4Hzsine wave then it is AC. Anything above 0 Hz is AC. Do you mean that you want to make a 60 Hz sinewave or a 1000 Hz sinewave or something else?

author: TimFox
author_rank: Super Contributor
timestamp: June 27, 2023, 11:31:39 pm
content: Then you wantC> 1/(2pi xfx 12 M\$\Omega\$)Iff> 4 Hz, thenC> 3.3 nF.Better to haveC> 20 nF, at a suitably high voltage.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 27, 2023, 11:53:50 pm
content: Quote from: Stray Electron on June 27, 2023, 11:05:13 pmYa, I'll try to explain again. Trust me, I did a poor job.I have a beautiful 8kV sinewave at 4Hz, but it rides on top of about 100 Volts DC.Another way to say it, I have 100V DC signal with 7,900 volts of ripple at 4Hz. :-)Mikek

author: TimFox
author_rank: Super Contributor
timestamp: June 28, 2023, 12:37:09 am
content: Your sine wave (without adding the series capacitor) is "riding on" 3,950 V (the mean value of a sine wave from +100 V to +8000 V), not 100 V.After inserting a series capacitor, the mean value of the waveform across the following resistor (12 M\$\Omega\$ in your example) must be zero.If the capacitor is very large (compared to the frequency, see my discussion above), the peak-to-peak amplitude will be conserved, therefore you would get a sine wave from -3,950 V to +3,950 V, with a DC value of zero.(Exact values, except for DC = 0, depend on how badly distorted your sine wave is.)If you need to research this, look in any textbook from the vacuum-tube era about "resistance-coupled amplifiers" or seehttps://www.angelfire.com/electronic/funwithtubes/Amp-RC.htmlIf you reduce the amplitude of the signal (reducing the AC voltage into the grid), the voltage after the capacitor will still have a mean value of 0, and the positive and negative peak levels will shrink symmetrically around it.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 28, 2023, 01:05:25 am
content: Quote from: TimFox on June 28, 2023, 12:37:09 amYes, that correct, the so called ripple is subtracting from my DC voltage. Thanks for making me understand.QuoteI have two adjustments, I can vary my DC supply, 4kV to 30kV , or I can adjust the grid drive.I don't expect to go above 12kV.Thanks, Mikek

author: james_s
author_rank: Super Contributor
timestamp: June 28, 2023, 02:18:41 am
content: Quote from: Stray Electron on June 27, 2023, 11:05:13 pmHe wants to remove the DC offset from the sine wave so that the peaks are symmetrically spaced across 0V.

author: Stray Electron
author_rank: Super Contributor
timestamp: June 28, 2023, 04:49:56 am
content: Got it. And I sort of thought that might be the case from reading the description in his first post but the 4Hz bit had me wondering.

author: Qmavam
author_rank: Regular Contributor
timestamp: June 28, 2023, 12:38:24 pm
content: Quote from: Stray Electron on June 28, 2023, 04:49:56 amThe 4 Hz is mostly likely the lowest frequency that will be used, although we may test as high as 50 Hz.50nF cap, 15kV cap on order.Thanks, Mikek

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 28, 2023, 02:48:17 pm
content: Also consider:If you have a sine wave like voltage that varies between 0V and 8kV, and you put a DC blocking cap in series, then you will get a -4kV to +4kV signal. and this is 4/1.41 = 2.83 kV rms.Using a simple DC blocking cap also has some other limits / disadvantages. Whenever the signal form or frequency is changed, you will get a temporary DC offset until your capacitor is charged to the DC difference of the new waveform. And of course it depends on the application whether that is acceptable.