I purchased a few LME49810's for an experiment as tube drivers. I purchased the LME49810 because it can provide up to 60ma of current. In comparing the test circuits from the LM4702 and the LME49810 there is a difference. On the LM4702 the source and sink pins can be connected directly.
The LME49810 test circuit has 10ohm resistors in series with the pins connected to a common point.
Is it possible to also connect the two pins together on the LME49810 without sacrificing its performance?
I want to drive a single ended tube for a tube curve tracer. The output will be connected to the tubes grid input and I will be sending a reference stepped signal. I am afraid that when driving tubes into class A2 (drawing driver's current) the 10ohm resistor will have a voltage drop that will affect the voltage between steps.
Any suggestions are appreciated.
Alfredo
An externally hosted image should be here but it was not working when we last tested it.
The LME49810 test circuit has 10ohm resistors in series with the pins connected to a common point.
An externally hosted image should be here but it was not working when we last tested it.
Is it possible to also connect the two pins together on the LME49810 without sacrificing its performance?
I want to drive a single ended tube for a tube curve tracer. The output will be connected to the tubes grid input and I will be sending a reference stepped signal. I am afraid that when driving tubes into class A2 (drawing driver's current) the 10ohm resistor will have a voltage drop that will affect the voltage between steps.
Any suggestions are appreciated.
Alfredo
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I guess there are not a lot of LME49810 users out there. Any ideas if I may get an answer by posting in another area?
> connect the two pins together on the LME49810
Read the wretched sales-sheet. '49810 has internal bias diodes and emitter followers. It is not expected these are bias-matched. Therefore the emitter follower idle current "could" be infinite. You need resistors.
They do no harm.
> driving tubes into class A2 ...the 10ohm resistor will have a voltage drop
1) 10/2= 5r, 60mA in 5r is 0.3V. In general, 5 ohms or 0.3V is "nothing" in power grid circuits. At least less error than the parameter-spread from one tube to another.
2) That's why you have negative feedback. The claimed audio THD is point-zero-zero-something. With correct connections (their "DC coupled" suggestion is not DC-coupled at input or NFB) the DC error will likewise be point-zero-zero and far-far better than any tube-work needs.
Read the wretched sales-sheet. '49810 has internal bias diodes and emitter followers. It is not expected these are bias-matched. Therefore the emitter follower idle current "could" be infinite. You need resistors.
They do no harm.
> driving tubes into class A2 ...the 10ohm resistor will have a voltage drop
1) 10/2= 5r, 60mA in 5r is 0.3V. In general, 5 ohms or 0.3V is "nothing" in power grid circuits. At least less error than the parameter-spread from one tube to another.
2) That's why you have negative feedback. The claimed audio THD is point-zero-zero-something. With correct connections (their "DC coupled" suggestion is not DC-coupled at input or NFB) the DC error will likewise be point-zero-zero and far-far better than any tube-work needs.
Thanks PRR
The voltage drop concern is more about the accuracy of the resulting curves. I actually put the chip on my protoboard and tested without the resistors. It actually looks pretty good. I will play with different values to see if i can detect any differences. Since all i am amplifying is a staircase signal the crossover distortion may not play a big role.
In a real A2 tube amp the voltage drop may be a benefit as i think i will compensate for the larger current spreads at higher positive grid voltages. Kind of self adjusting.
Alfredo
Alfredo,
Maybe you missed what PRR was getting at. There is a feedback loop so the amplifier will correct for the voltage drop across those resistors. You don't need to worry about the voltage drop across those resistors since they are inside the loop and will be corrected for. You shouldn't be able to discern any difference in performance of this chip with or without them within reasonable use. Keep them there to keep the LME49810 safe.
I'm interested in how your experiments turn out with the tracer so keep posting updates in the Tubes forum.
Maybe you missed what PRR was getting at. There is a feedback loop so the amplifier will correct for the voltage drop across those resistors. You don't need to worry about the voltage drop across those resistors since they are inside the loop and will be corrected for. You shouldn't be able to discern any difference in performance of this chip with or without them within reasonable use. Keep them there to keep the LME49810 safe.
I'm interested in how your experiments turn out with the tracer so keep posting updates in the Tubes forum.
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