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Improving my best diode detector

Started by amdx April 19, 2017
On 4/23/2017 2:39 PM, rickman wrote:
> On 4/22/2017 2:01 PM, amdx wrote: >> On 4/21/2017 8:24 PM, rickman wrote: >>> On 4/21/2017 1:53 PM, amdx wrote: >>>> On 4/21/2017 12:25 PM, rickman wrote: >>>>> >>>>> BTW, if you are only interested in a given voltage range at the output >>>>> of the diode circuit, you can connect the meter ground leg to the >>>>> adjustable voltage source (a pot) and make the zero current point on >>>>> the meter correspond to a 3 Vrms input and the max reading on the >>>>> meter face correspond to a 5 Vrms input. Then you have the entire >>>>> range of the meter to measure the Q more accurately. There is no >>>>> reason why the low end has to correspond to a Q of 1. >>>>> >>>> If biasing the diode detector up by 3 volts will make it more >>>> linear on >>>> the bottom end, That's good. I can bias the other end of my meter up by >>>> three volts to remove the 3 volts. The Q meter already is setup with an >>>> adjustable bias on the negative side of the meter. >>>> I'll keep this in mind. >>>> I would just as soon keep the zero to 250 scale. >>> >>> What I am talking about doesn't affect the diode circuit, it changes >>> the range of your meter range from 0-5 to 3-5 or whatever numbers you >>> want. Because you can change your power input level to bring the >>> output of the circuit to whatever level you want there is no reason to >>> work with the low end of the output range, so no need to linearize the >>> entire output range. Besides, your circuit is pretty durn linear as >>> it is. >>> >>> If you want to keep the 0 to 250 scale, consider using an adder (+250 >>> say) before you use the multiplier. That would make the true Q scale >>> 250 to 500. >>> >> I'll keep that as a possibility. >> >> Earlier in our discussion there was a question about why have the >> termination resistor. I turns out to be important if the circuit is >> capacitively coupled. It doesn't work without a termination. I'm >> now using an 11K termination after the coupling capacitor, with a 500 >> ohm source resistance. I don't know how the termination value affects >> circuit operation. > > I can't say I follow the need for a termination resistor after the > coupling cap. What does it do for you other than raise the corner > frequency of the high pass filter? >
I'll get back to you on that, When I removed it earlier, the detector output dropped to zero. I have disassembled the detector, when I put it back together, I'll verify it. Btw, I have tried several emails to you, they all bounced. I have other gmail accounts I send to, they are fine, it is only your address that I have a problem with. ¯\_(ツ)_/¯ Mikek --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus
On 4/23/2017 3:56 PM, amdx wrote:
> On 4/23/2017 2:35 PM, rickman wrote: >> On 4/23/2017 3:05 AM, Robert Baer wrote: >>> rickman wrote: >>>> On 4/22/2017 2:59 AM, Robert Baer wrote: >>>>> rickman wrote: >>>>>> On 4/21/2017 4:33 AM, Robert Baer wrote: >>>>>>> Can't get/see that schematic; in any case, for low(er) level >>>>>>> sensitivity and "linearity", try adding a small forward DC bias to >>>>>>> move >>>>>>> the average OP up the curve. >>>>>> >>>>>> The bias would then need to be subtracted out again after >>>>>> rectification. >>>>>> >>>>> Too complicated?? >>>> >>>> More complicated than using dropbox. >>>> >>> Actually, i think the bias does not need subtraction; in the "proper" >>> circuit, it helps. >> >> If you look at the schematic he is using it is already doing a pretty >> good job of biasing the diode. He doesn't need the low end, so there >> is no reason to bias the measurement out of the low end, it's already >> doing that just by the way it is used. >> > > I'm putting a cap between the diode. > Biasing is an option, but. I'm off the detector for a while, I need to > get a flat response out of my amplifier. > Here's the amp as wired, measurements are without the diode detector > connected. > https://www.dropbox.com/s/qsu1olsvza878t6/Dagmar%27s%20complete%20tube%20replacement.png?dl=0
I'm curious, what is the AC voltage on the emitter of Q3? It should be about the same as the base voltage. If that is correct, I don't get why the gain is 4 rather than 500/220. Oh, I think it would instead be 500/(220/2) because the other 220 ohm resistor is AC coupled and in parallel with the resistor in the emitter leg. Then the gain should be over 4, about 4.5. What is the AC voltage on the point between the 220 ohm resistor and the 3.47 kohm resistor? Certainly 10 uF should be nearly invisible to 1 MHz.
> Up until recently, all my testing has been at 1MHz, yesterday I ran the > SigGen up to 10 MHz. > At 10Mhz the detector had much more output than at 1MHz. I decided the > 27PF need to be increased.
How about the detector input? Is your amp level between 1 and 10 MHz?
> After increasing it to 0.1µf it was much better, but it was still about > 35% higher at 10MHz than at 1MHZ. > I have seen circuits where they use an inductor where the termination > resistor is. (68 ohm in the article)> > Maybe this would block the high frequency and equalize the response.
The use of the inductor is to DC ground that point. I don't think you need to do that.
> I have another problem, my amplifier is down about 17% at 10MHz. > So, I'd like to make that flat before I get involved working on the > detector. > I just swapped in a 2N3866 for the BC547b, it didn't help. > > Or maybe degrade the amp further {by decreasing the 10µf cap) to oppose > the increase of the detector.
I can't follow what you are talking about, which parts are which. I would try to make the amp flat and then figure out how to deal with the detector. First, which stage of the amp is not flat? Have you measured the output of each stage to see where the response problem is? What are the AC readings at the base and emitter of Q2 and Q3 at 1 and 10 MHz? -- Rick C
On 4/23/2017 3:56 PM, amdx wrote:
> On 4/23/2017 2:35 PM, rickman wrote: >> On 4/23/2017 3:05 AM, Robert Baer wrote: >>> rickman wrote: >>>> On 4/22/2017 2:59 AM, Robert Baer wrote: >>>>> rickman wrote: >>>>>> On 4/21/2017 4:33 AM, Robert Baer wrote: >>>>>>> Can't get/see that schematic; in any case, for low(er) level >>>>>>> sensitivity and "linearity", try adding a small forward DC bias to >>>>>>> move >>>>>>> the average OP up the curve. >>>>>> >>>>>> The bias would then need to be subtracted out again after >>>>>> rectification. >>>>>> >>>>> Too complicated?? >>>> >>>> More complicated than using dropbox. >>>> >>> Actually, i think the bias does not need subtraction; in the "proper" >>> circuit, it helps. >> >> If you look at the schematic he is using it is already doing a pretty >> good job of biasing the diode. He doesn't need the low end, so there >> is no reason to bias the measurement out of the low end, it's already >> doing that just by the way it is used. >> > > I'm putting a cap between the diode. > Biasing is an option, but. I'm off the detector for a while, I need to > get a flat response out of my amplifier. > Here's the amp as wired, measurements are without the diode detector > connected. > https://www.dropbox.com/s/qsu1olsvza878t6/Dagmar%27s%20complete%20tube%20replacement.png?dl=0 > > > Up until recently, all my testing has been at 1MHz, yesterday I ran the > SigGen up to 10 MHz. > At 10Mhz the detector had much more output than at 1MHz. I decided the > 27PF need to be increased. > After increasing it to 0.1µf it was much better, but it was still about > 35% higher at 10MHz than at 1MHZ. > I have seen circuits where they use an inductor where the termination > resistor is. (68 ohm in the article)> > Maybe this would block the high frequency and equalize the response. > > I have another problem, my amplifier is down about 17% at 10MHz. > So, I'd like to make that flat before I get involved working on the > detector. > I just swapped in a 2N3866 for the BC547b, it didn't help. > > Or maybe degrade the amp further {by decreasing the 10µf cap) to oppose > the increase of the detector.
Also check for AC signal on your power rail. Why is there no cap larger than 100 nF? -- Rick C
On 4/23/2017 4:03 PM, amdx wrote:
> On 4/23/2017 2:39 PM, rickman wrote: >> On 4/22/2017 2:01 PM, amdx wrote: >>> On 4/21/2017 8:24 PM, rickman wrote: >>>> On 4/21/2017 1:53 PM, amdx wrote: >>>>> On 4/21/2017 12:25 PM, rickman wrote: >>>>>> >>>>>> BTW, if you are only interested in a given voltage range at the >>>>>> output >>>>>> of the diode circuit, you can connect the meter ground leg to the >>>>>> adjustable voltage source (a pot) and make the zero current point on >>>>>> the meter correspond to a 3 Vrms input and the max reading on the >>>>>> meter face correspond to a 5 Vrms input. Then you have the entire >>>>>> range of the meter to measure the Q more accurately. There is no >>>>>> reason why the low end has to correspond to a Q of 1. >>>>>> >>>>> If biasing the diode detector up by 3 volts will make it more >>>>> linear on >>>>> the bottom end, That's good. I can bias the other end of my meter >>>>> up by >>>>> three volts to remove the 3 volts. The Q meter already is setup >>>>> with an >>>>> adjustable bias on the negative side of the meter. >>>>> I'll keep this in mind. >>>>> I would just as soon keep the zero to 250 scale. >>>> >>>> What I am talking about doesn't affect the diode circuit, it changes >>>> the range of your meter range from 0-5 to 3-5 or whatever numbers you >>>> want. Because you can change your power input level to bring the >>>> output of the circuit to whatever level you want there is no reason to >>>> work with the low end of the output range, so no need to linearize the >>>> entire output range. Besides, your circuit is pretty durn linear as >>>> it is. >>>> >>>> If you want to keep the 0 to 250 scale, consider using an adder (+250 >>>> say) before you use the multiplier. That would make the true Q scale >>>> 250 to 500. >>>> >>> I'll keep that as a possibility. >>> >>> Earlier in our discussion there was a question about why have the >>> termination resistor. I turns out to be important if the circuit is >>> capacitively coupled. It doesn't work without a termination. I'm >>> now using an 11K termination after the coupling capacitor, with a 500 >>> ohm source resistance. I don't know how the termination value affects >>> circuit operation. >> >> I can't say I follow the need for a termination resistor after the >> coupling cap. What does it do for you other than raise the corner >> frequency of the high pass filter? >> > I'll get back to you on that, When I removed it earlier, the detector > output dropped to zero. I have disassembled the detector, when I put it > back together, I'll verify it. > Btw, I have tried several emails to you, they all bounced. I have other > gmail accounts I send to, they are fine, it is only your address that > I have a problem with. ¯\_(ツ)_/¯
The problem is not gmail, it is my arius.com server rejecting your emails because they are on spam lists. You need to deal with the spam listing problem. I've given you info about that and it is in the bounce reply you get. btw, you shouldn't be emailing me at gmail. I gave you an arius.com address to use. -- Rick C
On 4/23/2017 3:37 PM, rickman wrote:
> On 4/23/2017 3:56 PM, amdx wrote: >> On 4/23/2017 2:35 PM, rickman wrote: >>> On 4/23/2017 3:05 AM, Robert Baer wrote: >>>> rickman wrote: >>>>> On 4/22/2017 2:59 AM, Robert Baer wrote: >>>>>> rickman wrote: >>>>>>> On 4/21/2017 4:33 AM, Robert Baer wrote: >>>>>>>> Can't get/see that schematic; in any case, for low(er) level >>>>>>>> sensitivity and "linearity", try adding a small forward DC bias to >>>>>>>> move >>>>>>>> the average OP up the curve. >>>>>>> >>>>>>> The bias would then need to be subtracted out again after >>>>>>> rectification. >>>>>>> >>>>>> Too complicated?? >>>>> >>>>> More complicated than using dropbox. >>>>> >>>> Actually, i think the bias does not need subtraction; in the "proper" >>>> circuit, it helps. >>> >>> If you look at the schematic he is using it is already doing a pretty >>> good job of biasing the diode. He doesn't need the low end, so there >>> is no reason to bias the measurement out of the low end, it's already >>> doing that just by the way it is used. >>> >> >> I'm putting a cap between the diode. >> Biasing is an option, but. I'm off the detector for a while, I need to >> get a flat response out of my amplifier. >> Here's the amp as wired, measurements are without the diode detector >> connected. >> https://www.dropbox.com/s/qsu1olsvza878t6/Dagmar%27s%20complete%20tube%20replacement.png?dl=0 >> > > I'm curious, what is the AC voltage on the emitter of Q3? It should be > about the same as the base voltage. If that is correct, I don't get why > the gain is 4 rather than 500/220. Oh, I think it would instead be > 500/(220/2) because the other 220 ohm resistor is AC coupled and in > parallel with the resistor in the emitter leg. Then the gain should be > over 4, about 4.5.
Yes, you have that correct, the 10uf cap AC couples the other 220 into the circuit. What is the AC voltage on the point between the 220
> ohm resistor and the 3.47 kohm resistor? Certainly 10 uF should be > nearly invisible to 1 MHz.
with 14Vpp input that point is 3.15Vpp.
> > >> Up until recently, all my testing has been at 1MHz, yesterday I ran the >> SigGen up to 10 MHz. >> At 10Mhz the detector had much more output than at 1MHz. I decided the >> 27PF need to be increased. > > How about the detector input? Is your amp level between 1 and 10 MHz?
My amplifier is down about 17% at 10MHz.
>> After increasing it to 0.1µf it was much better, but it was still about >> 35% higher at 10MHz than at 1MHZ. >> I have seen circuits where they use an inductor where the termination >> resistor is. (68 ohm in the article)> >> Maybe this would block the high frequency and equalize the response. > > The use of the inductor is to DC ground that point. I don't think you > need to do that. > > >> I have another problem, my amplifier is down about 17% at 10MHz. >> So, I'd like to make that flat before I get involved working on the >> detector. >> I just swapped in a 2N3866 for the BC547b, it didn't help. >> >> Or maybe degrade the amp further {by decreasing the 10µf cap) to oppose >> the increase of the detector. >
> I can't follow what you are talking about, which parts are which. I > would try to make the amp flat and then figure out how to deal with the > detector. > > First, which stage of the amp is not flat?
T1 and Q2 output is flat. Q3 rolls off between 1MHz and 10MHz.
> Have you measured the output > of each stage to see where the response problem is?
Yes, Q3. > What are the AC
> readings at the base and emitter of Q2 and Q3 at 1 and 10 MHz?
The problem lies with Q3, the input is flat the output rolls off. Mikek --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus
On 4/23/2017 3:42 PM, rickman wrote:
> On 4/23/2017 4:03 PM, amdx wrote: >> On 4/23/2017 2:39 PM, rickman wrote: >>> On 4/22/2017 2:01 PM, amdx wrote: >>>> On 4/21/2017 8:24 PM, rickman wrote: >>>>> On 4/21/2017 1:53 PM, amdx wrote: >>>>>> On 4/21/2017 12:25 PM, rickman wrote: >>>>>>> >>>>>>> BTW, if you are only interested in a given voltage range at the >>>>>>> output >>>>>>> of the diode circuit, you can connect the meter ground leg to the >>>>>>> adjustable voltage source (a pot) and make the zero current point on >>>>>>> the meter correspond to a 3 Vrms input and the max reading on the >>>>>>> meter face correspond to a 5 Vrms input. Then you have the entire >>>>>>> range of the meter to measure the Q more accurately. There is no >>>>>>> reason why the low end has to correspond to a Q of 1. >>>>>>> >>>>>> If biasing the diode detector up by 3 volts will make it more >>>>>> linear on >>>>>> the bottom end, That's good. I can bias the other end of my meter >>>>>> up by >>>>>> three volts to remove the 3 volts. The Q meter already is setup >>>>>> with an >>>>>> adjustable bias on the negative side of the meter. >>>>>> I'll keep this in mind. >>>>>> I would just as soon keep the zero to 250 scale. >>>>> >>>>> What I am talking about doesn't affect the diode circuit, it changes >>>>> the range of your meter range from 0-5 to 3-5 or whatever numbers you >>>>> want. Because you can change your power input level to bring the >>>>> output of the circuit to whatever level you want there is no reason to >>>>> work with the low end of the output range, so no need to linearize the >>>>> entire output range. Besides, your circuit is pretty durn linear as >>>>> it is. >>>>> >>>>> If you want to keep the 0 to 250 scale, consider using an adder (+250 >>>>> say) before you use the multiplier. That would make the true Q scale >>>>> 250 to 500. >>>>> >>>> I'll keep that as a possibility. >>>> >>>> Earlier in our discussion there was a question about why have the >>>> termination resistor. I turns out to be important if the circuit is >>>> capacitively coupled. It doesn't work without a termination. I'm >>>> now using an 11K termination after the coupling capacitor, with a 500 >>>> ohm source resistance. I don't know how the termination value affects >>>> circuit operation. >>> >>> I can't say I follow the need for a termination resistor after the >>> coupling cap. What does it do for you other than raise the corner >>> frequency of the high pass filter? >>> >> I'll get back to you on that, When I removed it earlier, the detector >> output dropped to zero. I have disassembled the detector, when I put it >> back together, I'll verify it. >> Btw, I have tried several emails to you, they all bounced. I have other >> gmail accounts I send to, they are fine, it is only your address that >> I have a problem with. ¯\_(ツ)_/¯ > > The problem is not gmail, it is my arius.com server rejecting your > emails because they are on spam lists. You need to deal with the spam > listing problem. I've given you info about that and it is in the bounce > reply you get. btw, you shouldn't be emailing me at gmail. I gave you > an arius.com address to use. >
I tried both. I looked at what you posted, I had 5 items from 2008 and before. Do you think that is the problem? Here's what the page gave me. Summary information for me. Note: Times shown are for the latest entry only! Found 4 network entries and 0 host/domain entries. Problem Entries, (listings will cause email problems.) 5 "Escalated" entries [16:30:03 21 Nov 2008 GMT+00]. Usage classification (only important if you run your own mailserver.) 1 "DUHL" entries [13:59:07 19 Sep 2004 GMT+00]. 1 "exDUHL" entries [14:05:20 19 Sep 2004 GMT+00]. Note: Active "exDUHL" entries mean that the IP/Network has been unblocked for some or all IPs from the DUHL. Mikek --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus
On 4/23/2017 6:10 PM, amdx wrote:
> On 4/23/2017 3:37 PM, rickman wrote: >> On 4/23/2017 3:56 PM, amdx wrote: >>> On 4/23/2017 2:35 PM, rickman wrote: >>>> On 4/23/2017 3:05 AM, Robert Baer wrote: >>>>> rickman wrote: >>>>>> On 4/22/2017 2:59 AM, Robert Baer wrote: >>>>>>> rickman wrote: >>>>>>>> On 4/21/2017 4:33 AM, Robert Baer wrote: >>>>>>>>> Can't get/see that schematic; in any case, for low(er) level >>>>>>>>> sensitivity and "linearity", try adding a small forward DC bias to >>>>>>>>> move >>>>>>>>> the average OP up the curve. >>>>>>>> >>>>>>>> The bias would then need to be subtracted out again after >>>>>>>> rectification. >>>>>>>> >>>>>>> Too complicated?? >>>>>> >>>>>> More complicated than using dropbox. >>>>>> >>>>> Actually, i think the bias does not need subtraction; in the >>>>> "proper" >>>>> circuit, it helps. >>>> >>>> If you look at the schematic he is using it is already doing a pretty >>>> good job of biasing the diode. He doesn't need the low end, so there >>>> is no reason to bias the measurement out of the low end, it's already >>>> doing that just by the way it is used. >>>> >>> >>> I'm putting a cap between the diode. >>> Biasing is an option, but. I'm off the detector for a while, I need to >>> get a flat response out of my amplifier. >>> Here's the amp as wired, measurements are without the diode detector >>> connected. >>> https://www.dropbox.com/s/qsu1olsvza878t6/Dagmar%27s%20complete%20tube%20replacement.png?dl=0 >>> >> >> I'm curious, what is the AC voltage on the emitter of Q3? It should >> be about the same as the base voltage. If that is correct, I don't >> get why the gain is 4 rather than 500/220. Oh, I think it would >> instead be 500/(220/2) because the other 220 ohm resistor is AC >> coupled and in parallel with the resistor in the emitter leg. Then >> the gain should be over 4, about 4.5. > > Yes, you have that correct, the 10uf cap AC couples the other 220 into > the circuit. > > What is the AC voltage on the point between the 220 >> ohm resistor and the 3.47 kohm resistor? Certainly 10 uF should be >> nearly invisible to 1 MHz. > > with 14Vpp input that point is 3.15Vpp.
That shows your AC coupling is not very good. This is the bootstrap to reduce the input capacitance of this stage. You need to up the value of either the capacitor or the resistor or both. You didn't say what the AC reading is on the emitter, but it should be pretty close to 14 Vpp and both sides of the caps should be about the same AC voltage for the bootstrap to work properly. The 220 ohm resistor on the base side of the caps can be increased if you reduce the 3.47 kohm resistor by the same amount. The limiting case is two resistors of 1.75 kohms giving an AC load to the caps of 8.75 kohm which is about 40 times the 220 they now see. This also increases the input impedance and makes the capacitance reduction work properly which may be part of your frequency response problem.
>>> Up until recently, all my testing has been at 1MHz, yesterday I ran the >>> SigGen up to 10 MHz. >>> At 10Mhz the detector had much more output than at 1MHz. I decided the >>> 27PF need to be increased. >> >> How about the detector input? Is your amp level between 1 and 10 MHz? > > > My amplifier is down about 17% at 10MHz.
I don't know what "my amplifier" includes. Is that from the 14.14 volt input to the collector of Q3? Ok, I see later you say the first two stages are flat to 10 MHz. Where do you measure the response of Q2? Is this while connected to Q3?
>>> After increasing it to 0.1µf it was much better, but it was still about >>> 35% higher at 10MHz than at 1MHZ. >>> I have seen circuits where they use an inductor where the termination >>> resistor is. (68 ohm in the article)> >>> Maybe this would block the high frequency and equalize the response. >> >> The use of the inductor is to DC ground that point. I don't think you >> need to do that. >> >> >>> I have another problem, my amplifier is down about 17% at 10MHz. >>> So, I'd like to make that flat before I get involved working on the >>> detector. >>> I just swapped in a 2N3866 for the BC547b, it didn't help. >>> >>> Or maybe degrade the amp further {by decreasing the 10µf cap) to oppose >>> the increase of the detector. >> > >> I can't follow what you are talking about, which parts are which. I >> would try to make the amp flat and then figure out how to deal with >> the detector. >> >> First, which stage of the amp is not flat? > > T1 and Q2 output is flat. Q3 rolls off between 1MHz and 10MHz. > >> Have you measured the output of each stage to see where the response >> problem is? > > Yes, Q3. > > > What are the AC >> readings at the base and emitter of Q2 and Q3 at 1 and 10 MHz? > > The problem lies with Q3, the input is flat the output rolls off.
Try the changes to the Q3 bootstrap and see if that helps the frequency response. Otherwise try proportionally increasing the values of all the resistors. If you double them all and the response increases, you are fighting something fundamental with the circuit. To isolate the circuits, try driving the Q3 input directly without the rest of the amp connected. The Q3 bootstrap can impact the loading on the Q2 stage output. -- Rick C
On 4/23/2017 6:28 PM, amdx wrote:
> On 4/23/2017 3:42 PM, rickman wrote: >> On 4/23/2017 4:03 PM, amdx wrote: >>> On 4/23/2017 2:39 PM, rickman wrote: >>>> On 4/22/2017 2:01 PM, amdx wrote: >>>>> On 4/21/2017 8:24 PM, rickman wrote: >>>>>> On 4/21/2017 1:53 PM, amdx wrote: >>>>>>> On 4/21/2017 12:25 PM, rickman wrote: >>>>>>>> >>>>>>>> BTW, if you are only interested in a given voltage range at the >>>>>>>> output >>>>>>>> of the diode circuit, you can connect the meter ground leg to the >>>>>>>> adjustable voltage source (a pot) and make the zero current >>>>>>>> point on >>>>>>>> the meter correspond to a 3 Vrms input and the max reading on the >>>>>>>> meter face correspond to a 5 Vrms input. Then you have the entire >>>>>>>> range of the meter to measure the Q more accurately. There is no >>>>>>>> reason why the low end has to correspond to a Q of 1. >>>>>>>> >>>>>>> If biasing the diode detector up by 3 volts will make it more >>>>>>> linear on >>>>>>> the bottom end, That's good. I can bias the other end of my meter >>>>>>> up by >>>>>>> three volts to remove the 3 volts. The Q meter already is setup >>>>>>> with an >>>>>>> adjustable bias on the negative side of the meter. >>>>>>> I'll keep this in mind. >>>>>>> I would just as soon keep the zero to 250 scale. >>>>>> >>>>>> What I am talking about doesn't affect the diode circuit, it changes >>>>>> the range of your meter range from 0-5 to 3-5 or whatever numbers you >>>>>> want. Because you can change your power input level to bring the >>>>>> output of the circuit to whatever level you want there is no >>>>>> reason to >>>>>> work with the low end of the output range, so no need to linearize >>>>>> the >>>>>> entire output range. Besides, your circuit is pretty durn linear as >>>>>> it is. >>>>>> >>>>>> If you want to keep the 0 to 250 scale, consider using an adder (+250 >>>>>> say) before you use the multiplier. That would make the true Q scale >>>>>> 250 to 500. >>>>>> >>>>> I'll keep that as a possibility. >>>>> >>>>> Earlier in our discussion there was a question about why have the >>>>> termination resistor. I turns out to be important if the circuit is >>>>> capacitively coupled. It doesn't work without a termination. I'm >>>>> now using an 11K termination after the coupling capacitor, with a 500 >>>>> ohm source resistance. I don't know how the termination value affects >>>>> circuit operation. >>>> >>>> I can't say I follow the need for a termination resistor after the >>>> coupling cap. What does it do for you other than raise the corner >>>> frequency of the high pass filter? >>>> >>> I'll get back to you on that, When I removed it earlier, the detector >>> output dropped to zero. I have disassembled the detector, when I put it >>> back together, I'll verify it. >>> Btw, I have tried several emails to you, they all bounced. I have other >>> gmail accounts I send to, they are fine, it is only your address that >>> I have a problem with. ¯\_(ツ)_/¯ >> >> The problem is not gmail, it is my arius.com server rejecting your >> emails because they are on spam lists. You need to deal with the spam >> listing problem. I've given you info about that and it is in the >> bounce reply you get. btw, you shouldn't be emailing me at gmail. I >> gave you an arius.com address to use. >> > I tried both. > I looked at what you posted, I had 5 items from 2008 and before. Do you > think that is the problem? > > Here's what the page gave me. > > Summary information for me. > Note: Times shown are for the latest entry only! > Found 4 network entries and 0 host/domain entries. > > Problem Entries, (listings will cause email problems.) > 5 "Escalated" entries [16:30:03 21 Nov 2008 GMT+00]. > > Usage classification (only important if you run your own mailserver.) > 1 "DUHL" entries [13:59:07 19 Sep 2004 GMT+00]. > 1 "exDUHL" entries [14:05:20 19 Sep 2004 GMT+00]. > Note: Active "exDUHL" entries mean that the IP/Network has been > unblocked for some or all IPs from the DUHL.
I don't know, but doesn't the page indicate a way to get taken off the blacklist? Either way you should complain to your hosting provider. You are sharing a server and someone is using it to send spam, although that might be long ago. This is what I see when I check the email you forwarded to me... Problem Entries, (listings will cause email problems.) 85 "Spam" entries [02:08:44 20 Apr 2017 GMT+00]. 1 "Virus" entries [20:38:42 21 Nov 2016 GMT+00]. http://www.sorbs.net/lookup.shtml?64.8.71.112 You ask for your IP to be delisted, but if the spam continues it will be put back on. Clearly there is an ongoing problem with your shared server hosting. Open a ticket with your hosting support and send them the bounce messages. -- Rick C
On 4/23/2017 7:17 PM, rickman wrote:
> On 4/23/2017 6:10 PM, amdx wrote: >> On 4/23/2017 3:37 PM, rickman wrote: >>> On 4/23/2017 3:56 PM, amdx wrote: >>>> On 4/23/2017 2:35 PM, rickman wrote: >>>>> On 4/23/2017 3:05 AM, Robert Baer wrote: >>>>>> rickman wrote: >>>>>>> On 4/22/2017 2:59 AM, Robert Baer wrote: >>>>>>>> rickman wrote: >>>>>>>>> On 4/21/2017 4:33 AM, Robert Baer wrote: >>>>>>>>>> Can't get/see that schematic; in any case, for low(er) level >>>>>>>>>> sensitivity and "linearity", try adding a small forward DC >>>>>>>>>> bias to >>>>>>>>>> move >>>>>>>>>> the average OP up the curve. >>>>>>>>> >>>>>>>>> The bias would then need to be subtracted out again after >>>>>>>>> rectification. >>>>>>>>> >>>>>>>> Too complicated?? >>>>>>> >>>>>>> More complicated than using dropbox. >>>>>>> >>>>>> Actually, i think the bias does not need subtraction; in the >>>>>> "proper" >>>>>> circuit, it helps. >>>>> >>>>> If you look at the schematic he is using it is already doing a pretty >>>>> good job of biasing the diode. He doesn't need the low end, so there >>>>> is no reason to bias the measurement out of the low end, it's already >>>>> doing that just by the way it is used. >>>>> >>>> >>>> I'm putting a cap between the diode. >>>> Biasing is an option, but. I'm off the detector for a while, I need to >>>> get a flat response out of my amplifier. >>>> Here's the amp as wired, measurements are without the diode detector >>>> connected. >>>> https://www.dropbox.com/s/qsu1olsvza878t6/Dagmar%27s%20complete%20tube%20replacement.png?dl=0 >>>> >>>> >>> >>> I'm curious, what is the AC voltage on the emitter of Q3? It should >>> be about the same as the base voltage. If that is correct, I don't >>> get why the gain is 4 rather than 500/220. Oh, I think it would >>> instead be 500/(220/2) because the other 220 ohm resistor is AC >>> coupled and in parallel with the resistor in the emitter leg. Then >>> the gain should be over 4, about 4.5. >> >> Yes, you have that correct, the 10uf cap AC couples the other 220 into >> the circuit. >> >> What is the AC voltage on the point between the 220 >>> ohm resistor and the 3.47 kohm resistor? Certainly 10 uF should be >>> nearly invisible to 1 MHz. >> >> with 14Vpp input that point is 3.15Vpp. >
> That shows your AC coupling is not very good. This is the bootstrap to > reduce the input capacitance of this stage. You need to up the value of > either the capacitor or the resistor or both. You didn't say what the > AC reading is on the emitter, but it should be pretty close to 14 Vpp > and both sides of the caps should be about the same AC voltage for the > bootstrap to work properly. >
I might have confused you, I should have wrote, with 14Vpp on the input to the circuit (before T1) I have 3.15Vpp on the 220 ohm 3.47k junction. The AC voltage is the same on both sides of the 10uf cap,
> The 220 ohm resistor on the base side of the caps can be increased if > you reduce the 3.47 kohm resistor by the same amount. The limiting case > is two resistors of 1.75 kohms giving an AC load to the caps of 8.75 > kohm which is about 40 times the 220 they now see. This also increases > the input impedance and makes the capacitance reduction work properly > which may be part of your frequency response problem. > > >>>> Up until recently, all my testing has been at 1MHz, yesterday I ran >>>> the >>>> SigGen up to 10 MHz. >>>> At 10Mhz the detector had much more output than at 1MHz. I decided >>>> the >>>> 27PF need to be increased. >>> >>> How about the detector input? Is your amp level between 1 and 10 MHz? >> >> >> My amplifier is down about 17% at 10MHz. > > I don't know what "my amplifier" includes. Is that from the 14.14 volt > input to the collector of Q3? Ok, I see later you say the first two > stages are flat to 10 MHz. Where do you measure the response of Q2?
At the emitter of Q2.
> Is this while connected to Q3? >
Yes.
> >>>> After increasing it to 0.1µf it was much better, but it was still >>>> about >>>> 35% higher at 10MHz than at 1MHZ. >>>> I have seen circuits where they use an inductor where the termination >>>> resistor is. (68 ohm in the article)> >>>> Maybe this would block the high frequency and equalize the response. >>> >>> The use of the inductor is to DC ground that point. I don't think you >>> need to do that. >>> >>> >>>> I have another problem, my amplifier is down about 17% at 10MHz. >>>> So, I'd like to make that flat before I get involved working on the >>>> detector. >>>> I just swapped in a 2N3866 for the BC547b, it didn't help. >>>> >>>> Or maybe degrade the amp further {by decreasing the 10µf cap) to oppose >>>> the increase of the detector. >>> >> >>> I can't follow what you are talking about, which parts are which. I >>> would try to make the amp flat and then figure out how to deal with >>> the detector. >>> >>> First, which stage of the amp is not flat? >> >> T1 and Q2 output is flat. Q3 rolls off between 1MHz and 10MHz. >> >>> Have you measured the output of each stage to see where the response >>> problem is? >> >> Yes, Q3. >> >> > What are the AC >>> readings at the base and emitter of Q2 and Q3 at 1 and 10 MHz? >> >> The problem lies with Q3, the input is flat the output rolls off. > > Try the changes to the Q3 bootstrap and see if that helps the frequency > response. Otherwise try proportionally increasing the values of all the > resistors. If you double them all and the response increases, you are > fighting something fundamental with the circuit. > > To isolate the circuits, try driving the Q3 input directly without the > rest of the amp connected. The Q3 bootstrap can impact the loading on > the Q2 stage output. >
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On 4/23/2017 7:27 PM, rickman wrote:
> On 4/23/2017 6:28 PM, amdx wrote: >> On 4/23/2017 3:42 PM, rickman wrote: >>> On 4/23/2017 4:03 PM, amdx wrote: >>>> On 4/23/2017 2:39 PM, rickman wrote: >>>>> On 4/22/2017 2:01 PM, amdx wrote: >>>>>> On 4/21/2017 8:24 PM, rickman wrote: >>>>>>> On 4/21/2017 1:53 PM, amdx wrote: >>>>>>>> On 4/21/2017 12:25 PM, rickman wrote: >>>>>>>>> >>>>>>>>> BTW, if you are only interested in a given voltage range at the >>>>>>>>> output >>>>>>>>> of the diode circuit, you can connect the meter ground leg to the >>>>>>>>> adjustable voltage source (a pot) and make the zero current >>>>>>>>> point on >>>>>>>>> the meter correspond to a 3 Vrms input and the max reading on the >>>>>>>>> meter face correspond to a 5 Vrms input. Then you have the entire >>>>>>>>> range of the meter to measure the Q more accurately. There is no >>>>>>>>> reason why the low end has to correspond to a Q of 1. >>>>>>>>> >>>>>>>> If biasing the diode detector up by 3 volts will make it more >>>>>>>> linear on >>>>>>>> the bottom end, That's good. I can bias the other end of my meter >>>>>>>> up by >>>>>>>> three volts to remove the 3 volts. The Q meter already is setup >>>>>>>> with an >>>>>>>> adjustable bias on the negative side of the meter. >>>>>>>> I'll keep this in mind. >>>>>>>> I would just as soon keep the zero to 250 scale. >>>>>>> >>>>>>> What I am talking about doesn't affect the diode circuit, it changes >>>>>>> the range of your meter range from 0-5 to 3-5 or whatever numbers >>>>>>> you >>>>>>> want. Because you can change your power input level to bring the >>>>>>> output of the circuit to whatever level you want there is no >>>>>>> reason to >>>>>>> work with the low end of the output range, so no need to linearize >>>>>>> the >>>>>>> entire output range. Besides, your circuit is pretty durn linear as >>>>>>> it is. >>>>>>> >>>>>>> If you want to keep the 0 to 250 scale, consider using an adder >>>>>>> (+250 >>>>>>> say) before you use the multiplier. That would make the true Q >>>>>>> scale >>>>>>> 250 to 500. >>>>>>> >>>>>> I'll keep that as a possibility. >>>>>> >>>>>> Earlier in our discussion there was a question about why have the >>>>>> termination resistor. I turns out to be important if the circuit is >>>>>> capacitively coupled. It doesn't work without a termination. I'm >>>>>> now using an 11K termination after the coupling capacitor, with a 500 >>>>>> ohm source resistance. I don't know how the termination value affects >>>>>> circuit operation. >>>>> >>>>> I can't say I follow the need for a termination resistor after the >>>>> coupling cap. What does it do for you other than raise the corner >>>>> frequency of the high pass filter? >>>>> >>>> I'll get back to you on that, When I removed it earlier, the detector >>>> output dropped to zero. I have disassembled the detector, when I put it >>>> back together, I'll verify it. >>>> Btw, I have tried several emails to you, they all bounced. I have >>>> other >>>> gmail accounts I send to, they are fine, it is only your address that >>>> I have a problem with. ¯\_(ツ)_/¯ >>> >>> The problem is not gmail, it is my arius.com server rejecting your >>> emails because they are on spam lists. You need to deal with the spam >>> listing problem. I've given you info about that and it is in the >>> bounce reply you get. btw, you shouldn't be emailing me at gmail. I >>> gave you an arius.com address to use. >>> >> I tried both. >> I looked at what you posted, I had 5 items from 2008 and before. Do you >> think that is the problem? >> >> Here's what the page gave me. >> >> Summary information for me. >> Note: Times shown are for the latest entry only! >> Found 4 network entries and 0 host/domain entries. >> >> Problem Entries, (listings will cause email problems.) >> 5 "Escalated" entries [16:30:03 21 Nov 2008 GMT+00]. >> >> Usage classification (only important if you run your own mailserver.) >> 1 "DUHL" entries [13:59:07 19 Sep 2004 GMT+00]. >> 1 "exDUHL" entries [14:05:20 19 Sep 2004 GMT+00]. >> Note: Active "exDUHL" entries mean that the IP/Network has been >> unblocked for some or all IPs from the DUHL. > > I don't know, but doesn't the page indicate a way to get taken off the > blacklist? Either way you should complain to your hosting provider. You > are sharing a server and someone is using it to send spam, although that > might be long ago. > > This is what I see when I check the email you forwarded to me... > > Problem Entries, (listings will cause email problems.) > 85 "Spam" entries [02:08:44 20 Apr 2017 GMT+00]. > 1 "Virus" entries [20:38:42 21 Nov 2016 GMT+00]. > > http://www.sorbs.net/lookup.shtml? 64.8.71.112 >
That is not my IP, that is the example they use on their site.
> You ask for your IP to be delisted, but if the spam continues it will be > put back on. > > Clearly there is an ongoing problem with your shared server hosting.
You're using info from the wrong IP.
> Open a ticket with your hosting support and send them the bounce messages. >
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