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I am trying to design a pre-amplifier before going into a power amplifier. I essentially just want to signal condition a little more before the power amplifier.

Whats really confusing me is the input/output impedance of the OP-AMP (LM833N) and what it can potentially drive. The datasheet doesn't say its maximum current capability so I dont know where to start.

I am trying to create 1st order passive filters just to make a simple, however I know that I have to take into account of its impedance as well. I was wondering if anyone can tell me if I am on the right track and some design tips in creating filters for this case.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ If you don't think the current is enough, you could always add a current buffer to the output. \$\endgroup\$ – Hearth Nov 20 '18 at 4:20
  • \$\begingroup\$ First time hearing about it. I believe the current isn't enough, however in the datasheet of the LM833N it doesnt tell you otherwise. \$\endgroup\$ – Pllsz Nov 20 '18 at 4:29
  • \$\begingroup\$ What input bias current is needed? What is the RIn of the LM833? \$\endgroup\$ – analogsystemsrf Nov 20 '18 at 4:44
  • \$\begingroup\$ What I meant by Circuit bias, Its running a Single Supply Topology meaning, the input is being centered around VSS(5V) - VEE(0V) So the V+ will have a DC voltage of 2.5V. There's no RIN into the LM833N doesnt OP-AMP usually have built in HIGH impedance? \$\endgroup\$ – Pllsz Nov 20 '18 at 4:46
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The product page for the LM833N on ti.com lists a typical output current of 40mA. Such an output current in the tens of milliamps is a typical output current maximum for most general purpose op amps. Note that this is a typical value so don't assume every LM833N can supply 40mA. There's no specification on the minimum output current the LM833N can supply but as long as its load doesn't draw more than ~20mA you should be fine.

The input impedance of the LM833N isn't specified (and I don't see it on the product page, either) but the datasheet does say that the input bias current is 1000nA maximum. This is the information you need to set the voltage divider resistors \$R_1\$ and \$R_2\$ which bias the LM833N's non-inverting input to \$V_{\text{DD}}/2\$. In order to make sure that the voltage divider produces the correct voltage at its output you should make sure that the bias current through those resistors is ten times the input bias current on the LM833N -- i.e. 10uA or more. You don't want that bias current to be much higher than that, though, otherwise it will draw more power than necessary. \$237\Omega\$ is a small value for that purpose -- with \$V_{\text{DD}} = 5\text {V}\$ the bias current is approximately 10mA. You can set \$R_1\$ and \$R_2\$ a thousand times higher (i.e. \$237\text{k}\Omega\$) to achieve a 10uA bias current.

Regarding the filters at the output of the LM833N you might want to use active filters and/or buffer the passive filters to simplify the design. With those passive filters loading each other like you've shown you'll have to take into account the impedances of each passive filter. If you put an op amp buffer in between each passive filter (or use an op amp as part of the filter to make it an active filter) then you gain the high input impedance and low output impedance of the op amps and you can design the filters independently. The tradeoff is that you'll need more op amps (though you might have an unused op amp anyway since you're only showing one LM833N op amp but the LM833N is a dual op amp package).

Another thing you might consider is buffering the output of the voltage divider formed by \$R_1\$ and \$R_2\$ since you need a 2.5V bias voltage for both of the op amps (and any additional op amps used in the filters). With a buffer for this voltage divider you can connect the 2.5V bias voltage directly to the non-inverting inputs of the op amps rather than using \$R_{11}\$ and \$R_{12}\$ to produce another 2.5V bias. Instead of using \$R_{11}\$, \$R_{12}\$, and \$C_5\$ as a passive 1st order HPF you can use an active 2nd order HPF and get rid of the passive 1st order HPF formed by \$C_3\$ and \$R_7\$, thus removing a stage from the pre-amplifier. Figure 33 of the LM833N datasheet shows a Butterworth 2nd order HPF you can use, complete with the equations necessary to set the cutoff frequency.

This is what your circuit would look like with a buffered 2.5V reference:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that \$C_1\$ is still required to block DC, though it is no longer filtering.

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  • \$\begingroup\$ 1. "237Ω is a small value for that purpose" - The purpose for the resistors where to set the Fc of the High pass filter that it makes with C1 that 20Hz has a dB ~= 0 so it doesn't attenuate the bass frequencies at all. 2. Is setting the input bias that important? 3. you have to further explain the last bit. \$\endgroup\$ – Pllsz Nov 20 '18 at 17:59
  • \$\begingroup\$ @Pllsz If you add an op amp non-inverting amplifier with a gain of 1 (i.e. a voltage buffer) to the output of the voltage divider formed by \$R_1\$ and \$R_2\$ you get a stable 2.5V reference which can be used to bias all the op amps in the pre-amplifier -- the op amp buffer has a very low impedance so it can drive multiple stages and you don't have to worry about the input impedance of those stages. This allows you to get rid of \$R_{11}\$ and \$R_{12}\$. For your input HPF and the HPF at the input of the LM1875 you can replace them with active, higher-order filter(s) to do the same job... \$\endgroup\$ – Null Nov 20 '18 at 18:10
  • \$\begingroup\$ Fundamentally, you are trying to do two very different jobs with the same resistors (\$R_1\$/\$R_2\$ and \$R_{11}\$/\$R_{12}\$) -- biasing and filtering. Instead, use one pair of resistors for biasing and add resistor(s) and/or op amps for the dedicated purpose of filtering. \$\endgroup\$ – Null Nov 20 '18 at 18:12
  • \$\begingroup\$ I am sure the restive network can drive two buffer inputs no? \$\endgroup\$ – Pllsz Nov 20 '18 at 18:15
  • \$\begingroup\$ @Pllsz They can but (a) it's difficult to choose the right values when the voltage divider is connected to other impedances (as is the case for \$R_{11}\$/\$R_{12}\$) and (b) you need multiple voltage dividers to produce the same 2.5V reference. \$\endgroup\$ – Null Nov 20 '18 at 18:19

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