Impedance bridging question (I Guess...)...
Due to physical layout, I consider connecting modules inside an amp via 10-15cm shielded cable.
Here are some options attached...
Can I leave the opamp input without a shunt resistor to ground?
Do I miss something in the network at the opamp's input?
Which is best value for the trimpot - 10k or 100k?
Which is the resulting Zout before the cable?
(is it the resistance of the trimpot to ground??)
Regards - Emil
Due to physical layout, I consider connecting modules inside an amp via 10-15cm shielded cable.
Here are some options attached...
Can I leave the opamp input without a shunt resistor to ground?
Do I miss something in the network at the opamp's input?
Which is best value for the trimpot - 10k or 100k?
Which is the resulting Zout before the cable?
(is it the resistance of the trimpot to ground??)
Regards - Emil
Emil,
The Zout varies with pot setting, it is Zout but in fact is the Z when you look back into it. Assuming that the externally connected source is close to zero, which normally is the case, when the pot is halfway you see 2 x 5k in parallel which of course is 2.5k. This is the max. Turning the pot up or down it decreases to zero in either limit position.
If the external source is 1k, the total becomes 11k and the midpoint is at 5.5k, and Zout is at 2,75k max.
You show the interconnect cable as grounded on both sides. Be careful not to have multiple grounds to avoid ground loops. If both stages have their own ground it normally is better to ground the cable screen only at the receiving end which normally is most sensitive. Alternatively you can use the screen to carry the ground (the signal reference really) to the next stage and don't interconnect the two stages additionally.
The 100k at the 2nd stage input is superfluous except as a safety if the cable should come disconnected while in operation.
Jan
PS Nice to see someone using Proteus. I normally set the grid to 50th before placing parts to avoid the slight wire jumps ;-)
The Zout varies with pot setting, it is Zout but in fact is the Z when you look back into it. Assuming that the externally connected source is close to zero, which normally is the case, when the pot is halfway you see 2 x 5k in parallel which of course is 2.5k. This is the max. Turning the pot up or down it decreases to zero in either limit position.
If the external source is 1k, the total becomes 11k and the midpoint is at 5.5k, and Zout is at 2,75k max.
You show the interconnect cable as grounded on both sides. Be careful not to have multiple grounds to avoid ground loops. If both stages have their own ground it normally is better to ground the cable screen only at the receiving end which normally is most sensitive. Alternatively you can use the screen to carry the ground (the signal reference really) to the next stage and don't interconnect the two stages additionally.
The 100k at the 2nd stage input is superfluous except as a safety if the cable should come disconnected while in operation.
Jan
PS Nice to see someone using Proteus. I normally set the grid to 50th before placing parts to avoid the slight wire jumps ;-)
Last edited:
Because your coaxial cable is short you can get away with bad or poor practice and still get reasonable results.
Option 2 works, but option 4 should be better because the vol pot output impedance is lower.
Options 1 & 2 have a 1uF into 100k high pass filter. This allows all the audio to get through.
Options 3 & 4 have 1uF into 10k and this attenuates the low bass. I recommend you increase C7 & C8 to 10uF MKT to give the same passing ability as options 1 & 2.
10uF MKT is quite big and expensive, so look at using a 22uF bi-polar electrolytic, or a pair of 47uF polar electrolytics in back to back formation.
All options have the +IN dc voltage applied to the vol pot wiper. Insert a DC blocking capacitor to protect the wiper contact, or move the earlier DC blocker to AFTER the wiper.
The +IN and -IN pins have a current output/input.
You must provide a route for this current. It is specified as input bias current in the datasheet. Input offset current is the maximum DIFFERENCE in the two bias currents.
BJT input opamps need a lowish value resistor for this offset current and for minimim output offset the two input pins need to see the same resistance. Typically 10k to 100k does. The Iof spec may require the two resistances to be slightly different for minimum output offset.
jFET input opamps have MUCH LOWER input offset current and can manage with >1M and they don't need to be balanced/matched.
Using 1M sets the input impedance to the same 1M and allows much lower value DC blocking capacitors to be used.
I suggest you add pads for a resistor in the feedback route. Some opamps require this as protection, most don't specify it. Try 100r
The TL0xx series are notorious for poor performance into capacitive loads. Add an output resistor, after the feedback tapping, so that the opamp sees the output resistor as the dominant load. Try 51r to 100r.
Why did you choose 08x? You may find that the 07x performs a bit better.
Add some RF attenuation. RC somewhere between 300us and 1500us
Option 2 works, but option 4 should be better because the vol pot output impedance is lower.
Options 1 & 2 have a 1uF into 100k high pass filter. This allows all the audio to get through.
Options 3 & 4 have 1uF into 10k and this attenuates the low bass. I recommend you increase C7 & C8 to 10uF MKT to give the same passing ability as options 1 & 2.
10uF MKT is quite big and expensive, so look at using a 22uF bi-polar electrolytic, or a pair of 47uF polar electrolytics in back to back formation.
All options have the +IN dc voltage applied to the vol pot wiper. Insert a DC blocking capacitor to protect the wiper contact, or move the earlier DC blocker to AFTER the wiper.
The +IN and -IN pins have a current output/input.
You must provide a route for this current. It is specified as input bias current in the datasheet. Input offset current is the maximum DIFFERENCE in the two bias currents.
BJT input opamps need a lowish value resistor for this offset current and for minimim output offset the two input pins need to see the same resistance. Typically 10k to 100k does. The Iof spec may require the two resistances to be slightly different for minimum output offset.
jFET input opamps have MUCH LOWER input offset current and can manage with >1M and they don't need to be balanced/matched.
Using 1M sets the input impedance to the same 1M and allows much lower value DC blocking capacitors to be used.
I suggest you add pads for a resistor in the feedback route. Some opamps require this as protection, most don't specify it. Try 100r
The TL0xx series are notorious for poor performance into capacitive loads. Add an output resistor, after the feedback tapping, so that the opamp sees the output resistor as the dominant load. Try 51r to 100r.
Why did you choose 08x? You may find that the 07x performs a bit better.
Add some RF attenuation. RC somewhere between 300us and 1500us
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.