I'm building a balanced preamp using TI DRV134 receivers, OPA604 opamps, and OPA134 line drivers. I don't know what value of dual pot to use for the volume control bridging the DRV134's to the OPA604's? Thanks!
Hey there ds9kicks, welcome!
I've unpacked your 'volume control bridging the dvr134's to the 604s' to imply that you're trying to go FULLY BALANCED through the entire volume chain? Is that right?
Cheers,
Jeff
I've unpacked your 'volume control bridging the dvr134's to the 604s' to imply that you're trying to go FULLY BALANCED through the entire volume chain? Is that right?
Cheers,
Jeff
because I could not get these IC's in PDIP form, I ordered opa2134's and am going single-ended all the way through. please see the thread I just started because it is working and sounding fine but I am not getting any level attenuation
For information some general rules :
1. Choose the less value as possible because of Johnson noise
2. A low value could be a big load for the previous stage so a minimum value shall be chosen to avoid overcurrent.
3. When you transmit a voltage signal, the input impedance of second stage shall be >> the output impedance of previous stage. If it's note the case, a resistor divider is done. BUT you can choose a small value to have low Johnson noise, loose a little bit voltage level, then add more gain in the second stage to have the correct voltage level at the output 😉 --> simulations shall be done to find best way (noise of R vs noise of AOP with bigger gain).
Examples:
You have an AOP to transmit an audio signal of 1V. Iout max of AOP = 10mA. Then the minimal load is R = U/I = 1/10mA = 10k. To have low noise, you want to use a P of 1K or 4.7K but you can't. You have to choose a 10k min P or better 22K to reduced THD+N (more the load is important more you have THD+N so reduced load is very important in audio).
1. Choose the less value as possible because of Johnson noise
2. A low value could be a big load for the previous stage so a minimum value shall be chosen to avoid overcurrent.
3. When you transmit a voltage signal, the input impedance of second stage shall be >> the output impedance of previous stage. If it's note the case, a resistor divider is done. BUT you can choose a small value to have low Johnson noise, loose a little bit voltage level, then add more gain in the second stage to have the correct voltage level at the output 😉 --> simulations shall be done to find best way (noise of R vs noise of AOP with bigger gain).
Examples:
You have an AOP to transmit an audio signal of 1V. Iout max of AOP = 10mA. Then the minimal load is R = U/I = 1/10mA = 10k. To have low noise, you want to use a P of 1K or 4.7K but you can't. You have to choose a 10k min P or better 22K to reduced THD+N (more the load is important more you have THD+N so reduced load is very important in audio).