Emitter diode protection in a regulated controll grid PSU - Page 1
Forum section: electronics / projects

author: 4cx10000
author_rank: Regular Contributor
timestamp: April 15, 2021, 11:52:07 am
content: Hi!Whenever I cross a regulated power supply with negative voltages I always get a bit confused. This is a regulated power supply for the control grid in a tube rf-amplifier, but could someone please explain why D3 is in the same direction as the emitter diode. It is for protection, but how? Maybe I should have posted this in "beginners".Thanks in advance

author: mjs
author_rank: Regular Contributor
timestamp: April 15, 2021, 12:43:50 pm
content: BE junction typically has a breakdown voltage of 5-7V and is easy to damage. Diode protects for that.

author: 4cx10000
author_rank: Regular Contributor
timestamp: April 15, 2021, 12:56:50 pm
content: Hi mjs and thank you for reply!Yes I have read that it is like that, but why is the protection diode in the same direction as the emitter diode? I try ... and I try, but I do not get it!

author: mikerj
author_rank: Super Contributor
timestamp: April 15, 2021, 01:39:56 pm
content: That looks drawn incorrectly to me.

author: Gyro
author_rank: Super Contributor
timestamp: April 15, 2021, 03:31:31 pm
content: Quote from: 4cx10000 on April 15, 2021, 12:56:50 pmYes, the diode is the wrong polarity to protect against reverse breakdown of the BE junction. As drawn, it will turn on at the same time as the transistor BE junction, 'stealing' base current. If both devices are silicon, the degree of current sharing will depend on their relative junction size and temperature. It might stop the transistor going into saturation maybe (though it seems unlikely in that context).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 15, 2021, 05:37:18 pm
content: Incidentally, the parallel diode, if matched, would give a circuit something like a current mirror, but since it's floating, it would act more like a transistor with really low (but fairly stable) hFE, equal to the ratio of areas; when done with matching transistors (one diode being a "diode strapped" transistor, B+C tied), hFE = 1.This obviously doesn't help any, and yeah it seems likely it's supposed to be backwards.There's also no bias current through the transistor (or diode), unless there's a negative supply elsewhere, which seems unlikely given the trouble of a whole transformer and rectifier here, with no supply symbol distributing that voltage elsewhere.The diode would at least constrain the output voltage to within +/-0.6V of the setting, ugly but better than nothing I guess.And the collector load RC is... irrelevant? Maybe it extends SOA by soaking up some of the voltage drop; it also serves as a current limiter, but a slow acting one given the capacitor.For grid bias, typically a fixed voltage at high impedance (10k+) is adequate, and this suffices for class 1 amplifiers; class 2, you have to source some grid current, which might be the intent here, but it's really sketchy doing it without any idle current through the transistor at all.Tim

author: 4cx10000
author_rank: Regular Contributor
timestamp: April 15, 2021, 06:04:45 pm
content: Thanks for all the replies!I made some changes in the schematic and added one more diode in series (D3 & D4) across the emitter diode. Would that change it to the better if you still use same diode direction?T3sl4co1l, you are right, this is an AB amplifier and will probably run into AB2 with some current draw on the control grid, so the bias voltage has to be stiff. I have not figured out why RC is there ... current limiting for some reason?

author: Gyro
author_rank: Super Contributor
timestamp: April 15, 2021, 06:15:21 pm
content: The BU508 seems a very strange transistor choice.It's a 1500V 125W transistor intended for CRT TV horizontal deflection. It has very low current gain and I can't see how you would ever get anywhere near that sort of voltage in the circuit shown.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 15, 2021, 06:23:47 pm
content: No, better with the diode reversed. Or if you're using two diodes, make one a 5V zener antiseries, but that just makes it worse anyway (if my reading is correct, see edit above).The voltage isn't much, a MJE340 would probably be fine, unless it needs to handle more current/power in which case I'd have to shop around but there are plenty to choose from with better specs than the original.Tim

author: floobydust
author_rank: Super Contributor
timestamp: April 15, 2021, 06:35:45 pm
content: OP where did you get that original circuit from, it's not great.Common failures are an open pot wiper and grid-cathode (pentode) or grid-heater short, neither of which are well covered.What current is needed, AB2 is from a cathode follower I imagine?D3/D4 are still backwards, although T1 can at least turn on now. BU508 is 5A 1,500V TO-3P is so brute force but low hFE of ~10-30 which makes the circuit not perform well. BU508D has a built-in C-E diode.I think it needs to tolerate the bias voltage going up, perhaps +ve yet have a low output impedance, which is the 47uF across R2 1k.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 15, 2021, 07:41:04 pm
content: For fault conditions, a diode from output to GND would do. The transistor being mildly reversed under that condition, is no problem.Tim

author: 4cx10000
author_rank: Regular Contributor
timestamp: April 15, 2021, 09:22:57 pm
content: Sorry, It was not meant to fool anyone, but I think the following will clear things up. The schematic I got - which originally came from a friend of mine but not made by him - was pencil drawn and as you all will see, is NOT as it should be; I did the schematic in sPlan though.QuoteYes I know these BU508 from that time I was in the tv repair circus and these have very poor current gain and in this circuit overkill.Well, here is the hand drawn schematic, which I will ask my friend who did, as well as a link to the original control grid psu to a 4CX250B amplifier. If you go through the PDF, there are a few things that does not match the attached schematic.https://www.theskywaves.net/technical/PA/4CX250B_PA_Regulated_Supplies.pdf

author: floobydust
author_rank: Super Contributor
timestamp: April 15, 2021, 09:43:32 pm
content: The sketch did draw the E-B diode in backwards. There is a 100uF cap across the output, not the 1k resistor only. 80VAC transformer gives 111VDC.From G8MNY explanations:"... In this design the sudden peak current from the grid RF drive is safely conducted to ground without upsetting the bias pot voltage. So the optimum no crossover distortion valve bias for about 100mA anode current can be accurately maintained.""... A diode from the pot to the grid (base-emitter) routes -ve volts through to the grid when there is no grid current. No sudden AC step change from the transistor to diode voltage is allowed due to the 100uF on the grid output."

author: floobydust
author_rank: Super Contributor
timestamp: April 16, 2021, 12:25:44 am
content: As a teenager I designed and built a grid C- regulator and remember using a PNP pass transistor...G8MNY has the zeners drawn backwards, OP has the E-B diode backwards and output capacitor in the wrong spot.With that fixed, I tried a simulation and sourcing current, it works OK. That is, current flowing out of the output to GND.Sinking current, it all flows backwards through a (proper) E-B diode (or avalanched transistor) back to the pot/zener, so the output somewhat gets pulled to that. The transistor remains off.Either I'm backwards or the design is, grid current (conventional) is out of the grid.

author: David Hess
author_rank: Super Contributor
timestamp: April 16, 2021, 03:55:10 am
content: The diode was probably drawn backwards and really should be anti-parallel in which case it protects the transistor from reverse breakdown when power is removed.Sometimes a diode is used as shown in parallel with the base-emitter junction for temperature compensation with the diode thermally coupled to some other part but I doubt that was intended here.

author: 4cx10000
author_rank: Regular Contributor
timestamp: April 16, 2021, 11:19:42 am
content: QuoteThanks floobydust!Yes, that hand drawn schematic really fooled me and besides, when it comes to circuit design it is sometimes hard to figure out how the designer was thinking, but luckily we all got it on print now.What transistor did you use in your simulation?When I think of this story I smile; my friend told me, when he tested the circuit, that the regulation was very very poor. As soon as he connected a light load to the regulator with a couple of mA current draw the voltage started to drop drastically.Well well ... thank you all for helping me out with this, but rightly, I should have asked my friend for the original circuit before I posted.

author: floobydust
author_rank: Super Contributor
timestamp: April 16, 2021, 05:53:37 pm
content: You can fix the circuit by changing to a PNP transistor, that makes it regulate while sinking current. Adding a reverse-diode across the output, and series base resistor to Q1 so base current is limited with the pot at the ends, and another resistor say 100k to cope with the pot wiper going open.But there is no real short-circuit protection. With the output at 0V, the 1k resistor will smoke trying to dissipate 10W (if the transformer keeps up) at ~90mA output. The 1k seems good for start-up inrush but the output capacitor is tiny. Either you live with that or add another transistor for current-limiting.For a PNP transistor, I'd suggest MJE15031, MJE350, 2N5415, 2N5401, 2N5551, MPSA92 etc. - it all depends on your output current. You don't need a lot, even 5mA is plenty.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 16, 2021, 06:04:05 pm
content: Well, sinking current isn't the problem, sourcing is. The grids don't source any significant current in normal operation (just ~uA), and they only sink in class 2 operation.So the transistor is appropriate for that condition.I wouldn't mind seeing a complementary pair, with diode bias to make a class B or AB follower; but this is a lot of work for just an additional 1.2V of stability, and it doesn't feel worth.A 10W resistor I think is within the realm of possibility, and I do think that's for fault limiting (or overdriving if you get the excitation level and tank tuning completely wrong).Tim

author: floobydust
author_rank: Super Contributor
timestamp: April 16, 2021, 06:20:32 pm
content: The grid's usually get a -ve voltage through a fixed bias resistor in AB1, so I have (conventional) current flowing into the bias supply transistor's emitter and I call that sinking output current.Ideally it's a push-pull regulator. But for AB2, the grid current makes it a non-linear load for the driver stage so I thought the bias supply can just take a holiday on those peaks.Using an NPN pass transistor alaLM337it's confusing Vout and Vin are flipped compared to a LM317, and you need more transistors.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 16, 2021, 06:28:10 pm
content: Yeah that's the trick, class 1 is supplied through a resistor because the current is SFA, who cares. So there's no current to bias a transistor, either -- of either polarity.In class 2, think about the equivalent circuit between drive transformer (there must be a transformer, RC coupling ain't gonna cut it!) and grid. The grid acts as a diode to ground (cathode, give or take bias sense resistors perhaps, but not usually much voltage away from GND). So the signal gets rectified, pushing the transformer's center-tap down. It's a transformer and rectifier, a power supply (just a poor one, ultimately limited by what grid current is available). So we need to pull the CT up, to keep grid voltage constant.Tim