Re: deja vu ?
Hmmm, not a deja vu. If using some kind of feedback in a op amp my limited knowledge would tell me that part of the eq must be in that loop. In the version before, both C4 and C5 goes to ground and then easy to adjust and understand, perhaps.......... at least very easy to build.
Zen Mod said:
Hmmm, not a deja vu. If using some kind of feedback in a op amp my limited knowledge would tell me that part of the eq must be in that loop. In the version before, both C4 and C5 goes to ground and then easy to adjust and understand, perhaps.......... at least very easy to build.
DC On Input?
Hi, Sorry if this has been asked before.
Can the B1 take 2v DC on its input? I want to put the B1 directly after a voltage output DAC chip that has 2v DC on it.
I have a blocking cap on the both the input and output of the B1 right now but would like to omit the input cap if it's possible to do so safely. Will keep the output cap in place to block DC from the buffer to my amplifier input.
One less cap in the signal path seems like a good idea 🙂 .
Anyone know if this is ok to do?
Thanks
Hi, Sorry if this has been asked before.
Can the B1 take 2v DC on its input? I want to put the B1 directly after a voltage output DAC chip that has 2v DC on it.
I have a blocking cap on the both the input and output of the B1 right now but would like to omit the input cap if it's possible to do so safely. Will keep the output cap in place to block DC from the buffer to my amplifier input.
One less cap in the signal path seems like a good idea 🙂 .
Anyone know if this is ok to do?
Thanks
Yes, it's single supply.
So we can't have any DC on the input of the buffer even if we have DC blockers on the output of the buffer to stop the DC from being passed to the component downstream. Is it a bias issue?
Thanks,
So we can't have any DC on the input of the buffer even if we have DC blockers on the output of the buffer to stop the DC from being passed to the component downstream. Is it a bias issue?
Thanks,
TV Man said:
exactly - bias issue
you can try that , but only if you use split PSU , counting that lower JFet will do biasing and compensate that 2V shift ;
Ok, thanks 🙂
I was looking at Mr. Pass' schematic on enjoythemusic.com thinking C100 and C200, the 1uf caps right after 1k ohm R102 and R202 would block DC at the input of the buffer making it unnecessary to have a DC blocker on the output of the component in front of the buffer.
I was looking at Mr. Pass' schematic on enjoythemusic.com thinking C100 and C200, the 1uf caps right after 1k ohm R102 and R202 would block DC at the input of the buffer making it unnecessary to have a DC blocker on the output of the component in front of the buffer.
Assuming that I'm understanding the question correctly, yes, the caps will block DC.
As long as the DC does not exceed the wattage rating of the pot--no problem at mid or low point, but possibly so at nearly full volume when the heat is dissipated across a very short length of resistive track--then you should be okay.
Grey
As long as the DC does not exceed the wattage rating of the pot--no problem at mid or low point, but possibly so at nearly full volume when the heat is dissipated across a very short length of resistive track--then you should be okay.
Grey
GRollins said:
The input impedance of the circuit is very high, so it would take a
scary amount of input DC.
You're right. I was trying to work from memory and obviously lost track of something, somewhere. I downloaded a copy of the schematic and scratched my head--can't reconstruct my line of reasoning. Oh, well...
Grey
Grey
To Zen Mod: Sorry about the confusion... my specialty is computer networks - I can describe those kinds of problems/details pretty well, electronics is a foreign language to me. Thanks for your comments 🙂
To GRollins: Something I would never have thought of, burning out a pot that way. Will add that good info to the things I've learned about electronics so far. Might not be applicable in this situation but its a good thing to know to avoid difficulties somewhere down the road. Thanks for your insight 🙂
To Mr Pass: Thanks very much for the clarification. 🙂 I have to say this little buffer is amazing. It sounds wonderful... I never knew how colored sounding my tube preamp is until now. My head is trying to wrap itself around this new sound paradigm.
To GRollins: Something I would never have thought of, burning out a pot that way. Will add that good info to the things I've learned about electronics so far. Might not be applicable in this situation but its a good thing to know to avoid difficulties somewhere down the road. Thanks for your insight 🙂
To Mr Pass: Thanks very much for the clarification. 🙂 I have to say this little buffer is amazing. It sounds wonderful... I never knew how colored sounding my tube preamp is until now. My head is trying to wrap itself around this new sound paradigm.
Potentiometer Impedance
Question for Papa.
I noted that you said the pot could be higher than 25K this interests me, I'd like to use a 100K pot, however in an article (by another author) he did say that series resistance greater than 10K will introduce distortion, which I must admit I found strange these are FETs after all! So whats the deal? is a 250K pot too high?
I guess this circuit is the basis of your active crossover, I for one cannot wait. this is truly a golden age for DIY Audio.
RC😕
Question for Papa.
I noted that you said the pot could be higher than 25K this interests me, I'd like to use a 100K pot, however in an article (by another author) he did say that series resistance greater than 10K will introduce distortion, which I must admit I found strange these are FETs after all! So whats the deal? is a 250K pot too high?
I guess this circuit is the basis of your active crossover, I for one cannot wait. this is truly a golden age for DIY Audio.
RC😕
Dear all,
I will power original B1 from a regulated supply (after C1).
Can I spare C2 and C3???
Thanks,
M
I will power original B1 from a regulated supply (after C1).
Can I spare C2 and C3???
Thanks,
M
Re: Potentiometer Impedance
Not papa, but lurking this thread, so I thought I would add some thoughts.
Why are you using 100K? Because you have it in hand? It will be fine, the pot value, as shown in the B1 schematic, essentially sets the load the upstream device will see. Maybe someone else can comment about the distortion, I can't see a problem for that happening in this circuit.
The DC input impedance is R100 + P100 in parallel, the AC input impedance adds R203 (1M) in parallel to the other 2(R100+P100+R203 in parallel). This ignores other effects too small or too big to be of significance. So for the values shown, the AC input impedance is 1M + 1M + 25k in parallel, or about 23.8K. If you use P100 = 100K, it becomes about 83.3K, and with P100=250K, it is about 166K.
If your upstream device has a large output impedance, say 10K, (vintage tube tuner, preamp, etc) you may *need* to have an overall high input impedance on the order of 100K to 200K, or else you will lose a lot of signal level. And likely you will also lose low end response, as the existing tube circuit will have a smallish coupling cap, expecting to see a very much higher input impedance than say 25K.
When using a large pot like 250K, you'll be sending the signal through a big resistance like say 200K at low volume. This big resistance adds noise, which you probably don't want for no good reason. For the math, see this example where a 100K resistor adds 40uV rms noise to a 1MHz bandwidth circuit.
http://www.ecircuitcenter.com/Circuits/Noise/Noise_Analysis/res_noise.htm
But, as many will see, the need for low noise at low volume settings is probably not a major concern.
You can make the large pot smaller by putting another resistor across the two non-wiper terminals. For instance, putting 100K across a 100K pot makes it look like 50K now. You may sacrifice some of the log taper action, though.
Bob
Not papa, but lurking this thread, so I thought I would add some thoughts.
calvert73 said:
Why are you using 100K? Because you have it in hand? It will be fine, the pot value, as shown in the B1 schematic, essentially sets the load the upstream device will see. Maybe someone else can comment about the distortion, I can't see a problem for that happening in this circuit.
The DC input impedance is R100 + P100 in parallel, the AC input impedance adds R203 (1M) in parallel to the other 2(R100+P100+R203 in parallel). This ignores other effects too small or too big to be of significance. So for the values shown, the AC input impedance is 1M + 1M + 25k in parallel, or about 23.8K. If you use P100 = 100K, it becomes about 83.3K, and with P100=250K, it is about 166K.
If your upstream device has a large output impedance, say 10K, (vintage tube tuner, preamp, etc) you may *need* to have an overall high input impedance on the order of 100K to 200K, or else you will lose a lot of signal level. And likely you will also lose low end response, as the existing tube circuit will have a smallish coupling cap, expecting to see a very much higher input impedance than say 25K.
When using a large pot like 250K, you'll be sending the signal through a big resistance like say 200K at low volume. This big resistance adds noise, which you probably don't want for no good reason. For the math, see this example where a 100K resistor adds 40uV rms noise to a 1MHz bandwidth circuit.
http://www.ecircuitcenter.com/Circuits/Noise/Noise_Analysis/res_noise.htm
But, as many will see, the need for low noise at low volume settings is probably not a major concern.
You can make the large pot smaller by putting another resistor across the two non-wiper terminals. For instance, putting 100K across a 100K pot makes it look like 50K now. You may sacrifice some of the log taper action, though.
Bob
Re: Potentiometer Impedance
Go ahead and use the 100K. How bad can it be? If you can tell the difference,
then it's too much.
😎
calvert73 said:
Go ahead and use the 100K. How bad can it be? If you can tell the difference,
then it's too much.
😎
High R low C
Hi, the reason I want to use a higher value for the pot is I use some expensive caps, hence the higher the pot R the lower the cap C (in the source output stage) and the lower the $$. I'll try the 100K, thanks for the input guys.
RC
Hi, the reason I want to use a higher value for the pot is I use some expensive caps, hence the higher the pot R the lower the cap C (in the source output stage) and the lower the $$. I'll try the 100K, thanks for the input guys.
RC
Member
Joined 2002
Hey nelson, Do you think that the B1 Buffer would be good with the Mini A ? I'm currently running the mini a with a passive pot, but a little more gain would be nice 😛
Why are the R102, (and maybe R104 ) not close to the FETs instead of being on the other side of C200 and C201?