Lrl from Italy

[FrancoItaly]

I have 3 volts in output with antenna and without quarts
So the sensor is self oscillation to avoid the self oscillation i need to zero the voltage i should decrease the c13 and c14 (according to help file)
But with decreasing c13 and c14 there is no change in output voltage

Try R10 = R12 = 1.5K for less gain.

[edward]

Try R10 = R12 = 1.5K for less gain.

I replaced r10 and r12 with 1.5 kohm but the output voltage is yet 3 volts

[FrancoItaly]

I replaced r10 and r12 with 1.5 kohm but the output voltage is yet 3 volts

Maybe the transistors have a too high amplification, try to use BC183B or BC549B for TR3 and TR4. Did you use the double-sided pcb with the lower face connected to the negative? If the sensor stage self oscillates too easily the fault can be of the pcb.

[edward]

Maybe the transistors have a too high amplification, try to use BC183B or BC549B for TR3 and TR4. Did you use the double-sided pcb with the lower face connected to the negative? If the sensor stage self oscillates too easily the fault can be of the pcb.

I used bc183c for tr1 to tr5

[FrancoItaly]

I used bc183c for tr1 to tr5

As I said try to use BC183B or BC549B for TR3 and TR4.

[FrancoItaly]

As I said try to use BC183B or BC549B for TR3 and TR4.

Also try C13/C14 = 220pF or 100pF

[edward]

Also try C13/C14 = 220pF or 100pF

I used 470 pf instead of 560 pf the voltage falls about 0.3 volts

[FrancoItaly]

I used 470 pf instead of 560 pf the voltage falls about 0.3 volts

The voltage must be about 0 volt, 0.3 V at the output means that there is at least 1 V of alternating voltage on the collector of TR4, the voltage drop of the diodes must be taken into account. As I said, put C13 / C14 = 220 pF or 100 pF.

[edward]

I used 470 pf instead of 560 pf the voltage falls about 0.3 volts

[edward]

i tested 100 pf the voltage is about 1 volts when i put 20 pf for c13 and c14 the output is 0 volt but there is another problem when i connect the quartz the output is 0.6 volts

[FrancoItaly]

i tested 100 pf the voltage is about 1 volts when i put 20 pf for c13 and c14 the output is 0 volt but there is another problem when i connect the quartz the output is 0.6 volts

You have to increase the total value of C2 / C3 / C4, ie put only C2 = 1pF, if not enough it increases to 2pF until the output voltage is in the range 3 - 6V (non-critical).
Remember that with two or three capacitors in series the total capacity is less.

[erfan]

appreciation

Hello every body
And hello dear franco italy
And a special thanks to franco italy.
Because he shared what ever he knows with no financial out look.I wish best for franco italy.

[FrancoItaly]

Hello every body
And hello dear franco italy
And a special thanks to franco italy.
Because he shared what ever he knows with no financial out look.I wish best for franco italy.

Thanks for the compliments, there can be real progress only with the collaboration of everyone. The "phenomenon" that allows the operation of the lrls is still largely unknown, but it is as real as many members have already been able to verify. As far as possible I help everyone in the realization of my lrl and take this opportunity to summarize the useful advice for the construction:
- the sensor stage has a very high gain and therefore there is the risk of self oscillations, it is advisable to use a double-sided pcb with the lower face connected to the negative and on the upper one the components will be soldered, no holes are necessary , only that, if necessary, to connect the upper mass to the screen constituted by the lower face of the pcb.
- the transistors are of the BC183C type or equivalent (for example BC549C or others), also the BC183B can be used as long as the beta (gain) is higher than 500.
- the total value of C2 / C3 / C4 establishes how much the quartz oscillator signal will be amplified by the sensor stage and therefore
determines the output value.
- C13 / C14 determine the gain of the sensor stage, in fact, increasing the value of C13 / C14 lowers the actual value of R10 / R12 (for the frequency of 8Mhz) and therefore increases the gain of the stage.
- On the TR2 base there is the mixing between the 8Mhz signal and the "phenomenon" captured by the stylus antenna.

[edward]

Thanks for the compliments, there can be real progress only with the collaboration of everyone. The "phenomenon" that allows the operation of the lrls is still largely unknown, but it is as real as many members have already been able to verify. As far as possible I help everyone in the realization of my lrl and take this opportunity to summarize the useful advice for the construction:
- the sensor stage has a very high gain and therefore there is the risk of self oscillations, it is advisable to use a double-sided pcb with the lower face connected to the negative and on the upper one the components will be soldered, no holes are necessary , only that, if necessary, to connect the upper mass to the screen constituted by the lower face of the pcb.
- the transistors are of the BC183C type or equivalent (for example BC549C or others), also the BC183B can be used as long as the beta (gain) is higher than 500.
- the total value of C2 / C3 / C4 establishes how much the quartz oscillator signal will be amplified by the sensor stage and therefore
determines the output value.
- C13 / C14 determine the gain of the sensor stage, in fact, increasing the value of C13 / C14 lowers the actual value of R10 / R12 (for the frequency of 8Mhz) and therefore increases the gain of the stage.
- On the TR2 base there is the mixing between the 8Mhz signal and the "phenomenon" captured by the stylus antenna.

Hi mr franco and thank you for your help

I tested bc548, bc547 b , bc 238b , bc 547c , and the results was same, my pcb is self oscillation, i have to use 20 pf capacitors for c13 , c14 to avoid from self oscilation
I think when i increase total value of c2 , c3 , c4 for increasing output voltage the lrl sensivity decreases

Now i have 2 question mr franco
1- how much is maximum amount that can be used for r10 and r12
2- does your lrl dont work with low voltage in output? ( for example 0.5 volts or 0.6 volts)

[FrancoItaly]

Hi mr franco and thank you for your help

I tested bc548, bc547 b , bc 238b , bc 547c , and the results was same, my pcb is self oscillation, i have to use 20 pf capacitors for c13 , c14 to avoid from self oscilation
I think when i increase total value of c2 , c3 , c4 for increasing output voltage the lrl sensivity decreases

Now i have 2 question mr franco
1- how much is maximum amount that can be used for r10 and r12
2- does your lrl dont work with low voltage in output? ( for example 0.5 volts or 0.6 volts)

1 -Usually it is chosen for R10 / R12 a value lower than that of R9 / R13. I made several copies of my lrl and I always managed to eliminate self oscillation. Try to remove C13 / C14, in this way it should stop oscillating, if so, decrease R10 / R12 until it oscillates again, then increase again R10 / R12 until the oscillations cease. For example R10 / R12 = 1 K no oscillations, R10 / R12 = 680 ohms no oscillations ... R10 / R12 = 220 ohms no oscillations, R10 / R12 = 100 ohms yes oscillations, choose R10 / R12 = 220 ohms. This difficulty in eliminating self oscillations makes me think of a defect in the pcb, there may be some parasitic coupling. I always recommend using a double-sided pcb.
2 - my lrl works also with low voltage output but for me it's better at least 2V, in fact it could happen when the installation was completed or in the field that the tension lowered further, in this way it would also greatly reduce the sensitivity, since it could not regulate the threshold effectively.