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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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27th Jul 2018, 4:09 pm | #1 |
Dekatron
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Amplifier circuit problem.
Whilst trying to find out why the big amp I built had such a dire frequency response and after going down a few rabbit holes I found the cause, see pic.
With the OP of the anode of the valve on the left connected to the following stage all is well, but with the 0.1u coupling cap connected frequency is attenuated, roughly it starts to roll off at 1500hz. I would say this is caused by the 100k preset. The circuit on the RHS is part of a clipping indicator (680k/330k and transistor etc) and VU meter driver. I've only drawn the relavant bits. I put the 100k pot in as without it and the signal just going into an opamp input via a 10k R is series, the signal amplitude dropped to much. So effectively the 100k preset is there to increase Z in, into the VU driver circuitry. So as this 100k pot is causing problems, what can I replace it with. I did try shunting it with a 10M R in parallel, no joy. I do have a spare OP amp, 1/4 of a TL074, use it as a buffer maybe? TFL, Andy.
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27th Jul 2018, 4:46 pm | #2 |
Dekatron
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Re: Circuit problem.
100K seems a rather low value to be used in such a situation: it's significantly lower than the anode-load resistor you've used for the triode.
I'd sugggest something more like 500K or 1MOhm as being 'usual' values in valve amplifiers. Equally, if your op-amp is presenting a low resistance (via the 100K pot and the 10K? resistor you mention as being part of the input circuit, and unspecified input-capacitance) this will act as a low-pass filter and introduce treble-cut. If you're interfacing like this you need to do something to ensure your op-amp presents near-as-makes-no-difference an infinite impedance at the frequencies you're interested in. |
27th Jul 2018, 8:11 pm | #3 |
Hexode
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Re: Circuit problem.
Just to add if the op amp has the 10K series with the feedback to that input then it will be operating in virtual earth mode. That is the presented impedance will be the 10K.
Pete |
27th Jul 2018, 8:34 pm | #4 |
Pentode
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Re: Circuit problem.
hmmm. ....
what about capacitor after potentiometer? |
27th Jul 2018, 8:35 pm | #5 |
Dekatron
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Re: Circuit problem.
Surely Cseries/Rshunt is a high pass not a low pass so you will lose bass, not treble.
Whatever, a high Z buffer would help and that 100K could go up to something more usual.
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27th Jul 2018, 8:51 pm | #6 |
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Re: Amplifier circuit problem.
To create a roll-off with increasing frequency, you need shunt capacitance. There is none shown in that circuit.
The 100k load resistance will decrease the gain and as Chris has said, push up the frequency below which bass rolls off. By reducing the anode load impedance, by effectively being in parallel with the anode resistor it also reduces the effective source impedance driving any stray C... this actually increases the frequency above which the high frequency roll-off begins. The usual scheme is to pick an anode resistor to get the target amount of gain, and then to make sure that load impedances downstream, whether capacitor DC-blocked or not, all combine to an impedance much larger than the anode resistor. The effective anode load applied to the valve is the 220k resistor, in parallel with the 100k, and also in parallel with the chain with the transistor in it. You don't know what the latter is like unless we know what else is going on around the transistor. 220k//100k//(680k+330k) = 64.36k so you get 64.368/220 = only 29.4% of the gain you may have thought you had. A shortfall of 10.7dB David
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28th Jul 2018, 5:48 am | #7 |
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Re: Amplifier circuit problem.
Thanks chaps. I'll try a bigger pot or buffer which as the TL074 is a JFET op amp should have high Z in.
As you say David the 100k etc is in parallel with 220k anode R. Now I have a better idea of what's what I'll re do the circuit. Andy.
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28th Jul 2018, 10:33 am | #8 |
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Re: Amplifier circuit problem.
The TL074 buffer circuit will only have a high-z in if it's used non inverting with signal direct to + input and no extraneous resistors connected there. If it's used in classic inverting mode, the z will be determined by the input resistor.
Having said that, you would be able to use megohm range resistors with a JFET amp so you could get away with it that way.
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28th Jul 2018, 11:34 am | #9 |
Hexode
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Re: Amplifier circuit problem.
Hi also as you are in a design stage and using op amps you might like to consider tapping a lower impedance signal off of the first or 2nd stage cathode stages depending on available signal levels. Perhaps with two cathode resistors with a lower value one unbypassed or one resistor in series with the bypass capacitor and tapping off that. This will depending on values give a stage gain reduction but you will have a low impedance signal to feed an op amp buffer to get the required levels with minimum impact on the amp anodes.
The other advantage here will be that you are less likely to spike the op amp with turn on or plug in clicks that could be routed from the anode via the coupling capacitor. Just some thoughts off the top of my head as I am not an amp designer. Pete |
28th Jul 2018, 10:44 pm | #10 |
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Re: Amplifier circuit problem.
Ignoring any frequency response considerations (as these can be easily fixed with the suggestions in the posts above) there is another issue.
If you are coupling into an OP amp from a high voltage valve circuit like this, you need to make sure you have added a diode pair on the signal feeding the OP amp's input resistors. These are connected from the signal line (or OP amp inputs) to the OP amp's power supply rail and from the signal line to the OP amp's earth, to make sure the signal swing cannot exceed the OP amp's supply rail and earth voltages.These can transiently go very high or low with power cycling. If the OP amp's negative input is being used and that is acting as a virtual earth it can help, but that relies on the OP amp's output voltage being able to dynamically keep it there and with power cycling this can not work sometimes, and you might be using the + input anyway. In any case, when the power cycles on and off the voltage transients, without the two clamp diodes, can kill the OP amp or less severely, with time, degrade the input transistors, if it is a type with BJT's, as they get zenered with power cycling, without the clamp diodes. |
28th Jul 2018, 11:16 pm | #11 |
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Re: Amplifier circuit problem.
It's something to watch out for whenever you have things connected together which are running from different power rails.
Daid
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29th Jul 2018, 7:08 am | #12 |
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Re: Amplifier circuit problem.
Thanks all. Used the unused 1/4" of the TL074 yesterday as a non inverting buffer with a 0.01u 630v rated coupling cap connecting the anode to the + input. Clamping diodes are a very good idea Argus.
With this altered circuit and the 100k pot off all is well, I altered the valve amp and I now have a flat frequency response from 10hz to 40khz and it's stable with NFB. I'll post a schematic later. Pete, I'm retro fitting to an already built circuit unfortunately, so some bodging is inevitable though I'll take note of your ideas. Later, A.
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29th Jul 2018, 9:27 am | #13 |
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Re: Amplifier circuit problem.
Here's the proposed schematic. As you see I have to supply two separate circuits, a VU meter driver and clipping indicator. The input is 6v RMS, so I can't really put clamping diodes on the IP - zeners?
One other thing I've used a tantalum as the cathode bypass cap, this saves using an electrolytic, which is a problem as the only suitable ones I have are 25v. I've read that using an electrolytic that won't have enough polarising voltage on it can cause distortion at low frequencies. Andy.
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30th Jul 2018, 10:23 am | #14 |
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Re: Amplifier circuit problem.
Clamping diodes on the input to an opamp are simply a reverse biassed series pair between the opamp's rails with the centre a-k point connected to the input. Any input signal can swing between +rail (plus a diode drop) and -rail (-a diode drop) so no zeners or other exotica are needed.
To ensure opamp input protection, there needs to be sufficient source impedance from the output of whatever drives the opamp input to prevent the diodes from going pop if the voltage swings sufficiently outside the opamp rails.
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30th Jul 2018, 11:07 am | #15 |
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Re: Amplifier circuit problem.
Thanks Chris, thought I had it wrong. So if I put a pair on 1N1007's as shown ont schematic that'll be ok.
The circuit as is doesn't work, with the non inverting IP floating like that, the OP goes high with the result that there's about 8v DC on the meter. I spent a day trying to elliminate shorts on the PCB. Tying the + IP to ground with a 1M R sorts the problem, but in that case I may as well not use the buffer and just use a 1M R. Any thoughts? Perhaps the coupling cap needs a resistor in series after it? Andy.
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30th Jul 2018, 7:32 pm | #16 |
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Re: Amplifier circuit problem.
I wouldn't use 1N4007s for opamp input protection! 1N4148, 1N914 signal diodes would be more suitable. Or for high impedance inputs PAD50 perhaps. Ensure that whatever happens, they can't get more than a few 10s of mA through them.
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30th Jul 2018, 10:21 pm | #17 |
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Re: Amplifier circuit problem.
1N4007 are so slow you'll have lost the opamp before they've even begun to think of turning on. With back-back diodes you don't need much voltage rating! The 1n4007 have a fair bit of capacitance too. They really aren't worth even keeping.
David
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30th Jul 2018, 11:49 pm | #18 |
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Re: Amplifier circuit problem.
Ah, the floating +input..... yes it does need a dc 0V reference. If it's a FET input opamp, 10 megs should be OK and will load things less.
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31st Jul 2018, 12:23 am | #19 | |
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Re: Amplifier circuit problem.
Quote:
You will find that the actual music peaks, that would correspond with clipping, will be far too brief and the LED will not light up long enough on those to be of much use a a clipping indicator with your current circuit. It is easily fixed though, if you rearrange the base circuit of the transistor to be fed by a diode and put a capacitor in the the base of the transistor to store charge on peaks. This effectively makes a charge pump & pulse stretcher to keep the LED on long enough to give a useful indication. I have a transistor circuit for this with values that works very well, which makes a good peak clipper indicator that I designed for my Pantry Tx's, I can post it, if you want it. Hugo. |
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31st Jul 2018, 5:50 am | #20 |
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Re: Amplifier circuit problem.
"Ensure that whatever happens, they can't get more than a few 10s of mA through them." For a full HT fault that'd mean 20k R's in series with the diodes I presume. Might be a problem, room on PCB is very tight.
1N4007's, chuck em David? Have they no use whatsoever? "10 megs should be OK and will load things less." I've used a 1M5 = 10hz HPF with 10n coupling cap. This doesn't load the valve down or so a quick test tells me, I thought 10M would be a bit mad. Your right hugo, it does come on like a glow worm on it's last legs. I presume the diode is cathode to positive end of small ish value of electrolytic, 10u maybe? Could you post the circuit please? Andy.
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