I did not connect any load to the Voutput when first turned out to test it after assembled the board. I realized it after the fact and now it does not seem to work as I have anticipated and not really sure if that caused a problem. Output voltage measurement is about 46 volt which is 10 times lower than what I have in mind with 10K load connected to Vout. Output voltage is regulated at about 46V with input varied from 70V to a few hundreds volt. The weird thing is that when I increased or decreased the output load from 10K it changed output voltage a lot. In simulation using LTspice the Output voltage should be 457 volt with input of about 480 volt and more regardless of the output load. Any suggestions?
Your board are rock stable regardless input and load current variation. I have not tested out the overload current feature yet. One thing I noted to watch out for (just for reference purpose in case someone do the measurement) if one takes a voltage measurement at R11 the voltage will be impacted by the loading input impedance of the multimeter. At first I did not pay attention to it and wonder why the output dropped 28 volt (in my case) when making measurement there.
Just completed testing all features of the power supply today. Input is about 470-485V. Regulated Output is 455V with varying load from 100mA to 300mA. I also tested out the overload feature and the circuit shut downed at about 310mA as expected and turned back on after I decreased the load current and activated the momentary switch feature. It worked great.👍
a question:
the PCB assembly guide states the following: "To set the current limit you need to select a single resistor, R16. Current limiting occurs when the voltage across R16 reaches approximately 0.8 V. So, for example, if you want a current limit of 180 mA, your R6 will be 0.8/0.180 = 3.3 Ω."
I wonder if by any chance there is a printing error and the value of 3.3ohm refers to R16 instead of R6 as it is written?
the PCB assembly guide states the following: "To set the current limit you need to select a single resistor, R16. Current limiting occurs when the voltage across R16 reaches approximately 0.8 V. So, for example, if you want a current limit of 180 mA, your R6 will be 0.8/0.180 = 3.3 Ω."
I wonder if by any chance there is a printing error and the value of 3.3ohm refers to R16 instead of R6 as it is written?
Sto ora per assemblare 4 PCB T-Reg per i primi due stadi di amplificazione del mio preamplificatore phono mm (che ha un totale di tre stadi di amplificazione, l'ultimo dei quali utilizza un giratore Ale Moglia). Per quanto riguarda il T-reg capisco che è bene mantenere la differenza tra tensione in ingresso e tensione in uscita entro 20V, mi chiedo se è possibile aumentare leggermente la differenza di tensione tra ingresso e uscita? (cioè rilasciare più tensione sul T-reg). Lo chiedo per sapere se ho un minimo di manovra per poter regolare la tensione in uscita dal t-reg mantenendo fissa una certa tensione in ingresso.
Good day Jan and all,
I have a practical question please.
I am developing an Push-pull class A amplifier. Required Uht=465V.
I calculated that before T-reg i need 535V (ripple only 4V because CRC filter). When mains voltage decrease -10%, before T-reg is 482V (4V ripple). So T-reg has enought voltage across for his work.
The opposite is the case when the mains voltage increase +10%. Before T-reg is 588V. This is still fine for a power transistor.
But what if there is a error in the amplifier and the amplifier stops drawing current (only 3mA drawing remains necessary for T-reg). Then the voltage before T-reg increases to 690V and this is too much for power transistor. I see it as a problem for a safe design. How do you see it?
PS: All this situations i simulate in PSUD2
I have a practical question please.
I am developing an Push-pull class A amplifier. Required Uht=465V.
I calculated that before T-reg i need 535V (ripple only 4V because CRC filter). When mains voltage decrease -10%, before T-reg is 482V (4V ripple). So T-reg has enought voltage across for his work.
The opposite is the case when the mains voltage increase +10%. Before T-reg is 588V. This is still fine for a power transistor.
But what if there is a error in the amplifier and the amplifier stops drawing current (only 3mA drawing remains necessary for T-reg). Then the voltage before T-reg increases to 690V and this is too much for power transistor. I see it as a problem for a safe design. How do you see it?
PS: All this situations i simulate in PSUD2
I would look into a MOSFET that rates for at least 750V or perhaps try IGBT transistor. I plan to test it using IGBT but have not done it yet. I believed that if IGBT can be used without a problem then it's easier to find than MOSFET for higher voltage/current.
It is what it is. You correctly identify the important points, max and min voltage. You either must decrease the voltages or increase the transistor/FET ratings (and don't forget the caps).
Also, check the max dissipation when the input voltage is max, (Vin(max) - Vout) * load current.
Jan
I was checking these out again today, did I find another typo? Sorry Jan not trying to bug you, I see someone a few posts back did this. I love your work. I was looking at the T-reg tonight. I didn't look, but obviously some voltage has to be dropped by the regulator. This screenshot is from the DIYaudio store, at the beginning. Vout is 500v, correct? Or is max input B+ 500v.
Thanks,
Loren
Thanks,
Loren
Yes that is not worded very clear. Both the max output voltage and the max input voltage with the given BOM are 500V, but as you mention not at the same time. If you read the article you will see that I mention that when you change one of the MOSFETs to another part the Vin can go to 600V, but that part is a bit hard to find. So you are correct, with the given BOM Vout max is limited to about 495V with Vin 500V, and to get to 500V you need more input voltage and the other part.
Looked at it another way, with the given BOM Vin max is 500V and that limits Vout max to slightly below that, especially when you figure in ripple. So if you want outputs at 500V you'd better get the other part and that gives you more breathing room at the input.
Sorry about the confusion, hope this helps.
Jan
Looked at it another way, with the given BOM Vin max is 500V and that limits Vout max to slightly below that, especially when you figure in ripple. So if you want outputs at 500V you'd better get the other part and that gives you more breathing room at the input.
Sorry about the confusion, hope this helps.
Jan
I am building a power supply for my KT88 PP amp that required 134mA/channel (just for KT88 tube). I could install them as a two chassis configuration (1 for power supply and 1 for KT88 tubes and its components). Or as a separate mono blocks consisting of four chassis.
Attached is the front end power supply of my T-reg. The purpose is to bring the voltage down to a safe level before the T-reg. The output from choke/capacitor C1 is over 730VDC with no load, 650VDC@35mA and still over 600VDC @140mmA load.
The bleeder network served two purposes. Since it's a choke input then it needs a minimum current at a certain voltage. With above network I can bring the voltage down around 610V@65mA first 45 seconds before applied to T-Reg board and waiting for the KT88 tubes to conduct. So the bleeder network provides the minimum load for the choke the first 45 seconds. After 45 seconds it acts as a normal bleeder resistor network after power turn off.
As I understand from searching through the web it said the power tubes similar to KT88 required about 12 seconds to conduct when first turn on.
If my C3 capacitor in T-Reg board is rated as 1KV, does it really necessary for installation of above front end network to bring the voltage down. If I install above network, should I decrease the single shot timer to about 12 seconds or less?
Inputs, comments and suggestion are greatly appreciated. Thank you.
Attached is the front end power supply of my T-reg. The purpose is to bring the voltage down to a safe level before the T-reg. The output from choke/capacitor C1 is over 730VDC with no load, 650VDC@35mA and still over 600VDC @140mmA load.
The bleeder network served two purposes. Since it's a choke input then it needs a minimum current at a certain voltage. With above network I can bring the voltage down around 610V@65mA first 45 seconds before applied to T-Reg board and waiting for the KT88 tubes to conduct. So the bleeder network provides the minimum load for the choke the first 45 seconds. After 45 seconds it acts as a normal bleeder resistor network after power turn off.
As I understand from searching through the web it said the power tubes similar to KT88 required about 12 seconds to conduct when first turn on.
If my C3 capacitor in T-Reg board is rated as 1KV, does it really necessary for installation of above front end network to bring the voltage down. If I install above network, should I decrease the single shot timer to about 12 seconds or less?
Inputs, comments and suggestion are greatly appreciated. Thank you.