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4th November 2005, 12:33 AM  #1 
diyAudio Member
Join Date: Jul 2004
Location: Berlin

power source resistance vs. current needs
Hello,
I'm reworking my simple mic preamp circuit which isn't more than a basic noninverting opamp circuit. Now I am thinking about the powersupply. I use batteries for a dual supply, + 3 to 4.5V In my previous version (which worked fine), I left away those buffercaps often recommended in parallel to the batteries, which are meant to have about 4.7uF to 10uF. In the power supply path, I only used 0.1uF bypass caps near the opamp pins. My total supply current is 6mA (at max, I'm still swapping opamps), and if I understand the whole thing right, there should be about 0.2mA of alternating current in addition on the powersupply rails, depending on the audio signal to be amplified. What I would like to understand better is: What are the mathematical relations between the current need and the inner resistance of the power source, so that I could determine, if a battery would be fine (regarding it's inner resistance) or how big a buffer capacitor I should use! I would be very thankful if someone could enlighten me, I guess that the physical relations are rather simple... hopefully Cheers, Dominique 
4th November 2005, 01:21 AM  #2 
diyAudio Member
Join Date: Sep 2005

Howdy! I can help you out with a little bit of your question:
Take the source resistance of your voltage source (in this case, your 9V battery), divided your source voltage by your source resistancee and this is the MAXIMUM current your source will be able to produce. For instance, if you have a 9V battery with an internal resistance of 1k, ==> 9V / 1k = .009A = 9mA maximum current output by the battery. You ask a good question: If the battery has, let's say a 10uF cap across it, what is the maximum current that the battery + capacitor can supply together to a load? This I am not too sure of... 
4th November 2005, 03:56 PM  #3 
diyAudio Member
Join Date: Jul 2004
Location: Berlin

Thank you for your reply!
That's already a part of the solution! Is there a rule on how many times the "suppliable" current should be bigger than the needed current by the circuit? Because performance would certainly not be great if the two were equal! I think that the problem is similar to poweramplifiers and their power supply. There's also the relation between current needs, the transformer inner impedance and those large filter (and at the same time buffer) caps which supply the amp. Unfortunately, looking for info on websites regarding power amp power supplies, I only found rules of thumb, but not the mathematical relations. Cheers Dominique 
4th November 2005, 05:03 PM  #4  
diyAudio Member
Join Date: Jul 2005
Location: 65N 25E

Quote:
If you want to calculate it in the hard way considering all the complexities I would recommend some good electronics textbook. dV=dt*dI/C would be one possible starting point. 

6th November 2005, 08:59 PM  #5 
diyAudio Member
Join Date: Jul 2004
Location: Berlin

thanks for answering!
Yes, probably I'll have to go through it. Good thing is afterwards I'll understand the whole thing better. I admit I'm a bit lazy sometimes and  'cause I'd like my circuit to be rather good, in front of all the problems I'd like to solve I resign and come here asking But the other day I took another look into the noise calculation of my mic preamp and found it wouldn't make much sense to use opamps with lower noise than 7nV/Hz^1/2, so I have more liberty now to look for other parameters, like quiescent current. It's nice to fix some points in a design, but well, there are still too many things I'm not aware of and just copy from other designs. That, I don't like. Best wishes, Dominique 
10th November 2005, 09:13 AM  #6  
diyAudio Member
Join Date: Sep 2005

Quote:
On a short term basis though, a capacitor will have a MUCH higher current availability than a 9V battery with 1K (!!!) source resistance. Obviously, the recharge time will be limited by the battery though. 

10th November 2005, 09:30 AM  #7  
diyAudio Member

Quote:
This is technically correct but not very interesting because in this situation, at 9 mA, the DC output from the battery will be zero... What I would do is calculate the DC drop from the DC load (say 6mA) and the battery impedance (say 20 ohms, just an example I don't know how realistically this is, but you get the point). The final DC will then be 9V  (6mA * 20ohms) = 8.880V. Then the AC load (say 0.2mA) will cause an AC ripple of 0.2mA * 20ohms = 4mV ripple. If you now parallel the batt with a cap that has  for AC  say only 2 ohms impedance the ripple will be 0.4mV only. Calculate the cap for 2 ohms impedance at 20Hz, for higher frequencies it will be even better. So, select C = 1/(2*pi*20Hz*2ohms) = about 5000uF. But be aware that during the lifetime of the battery both the DC voltage and the impedance can vary considerably! Jan Didden
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11th November 2005, 01:42 AM  #8 
diyAudio Member
Join Date: Jul 2004
Location: Berlin

Yes, that's a good starting point, you've helped me get a bit further in the design!
Thank you Janneman! Best wishes, Dominique 
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