Hello all, this might be a bit of ridiculous question, but here goes...
I am building an amplifier circuit for a class project. As part of the project I need to determine the power consumed by my circuit under normal operation. I'm using a pre-amp/tone control into an LM3875. I have 25V rails.
Here's what I'm thinking: I know the voltage (25V) so I just need to find the current. I want to place an ammeter in the ground path and measure the current. In my mind, this should be the total current of the circuit.
Is this the way to do it?
Thanks for any help.
I am building an amplifier circuit for a class project. As part of the project I need to determine the power consumed by my circuit under normal operation. I'm using a pre-amp/tone control into an LM3875. I have 25V rails.
Here's what I'm thinking: I know the voltage (25V) so I just need to find the current. I want to place an ammeter in the ground path and measure the current. In my mind, this should be the total current of the circuit.
Is this the way to do it?
Thanks for any help.
Hi,
the most precise way to determine the total current draw of the circuit would probably be to insert a current (measurement) transformer into the supply line(s), preferrably the transformer primary. I guess you're using a laboratory supply, so you would require two of them for it's dual output rails. Probably a bit expensive (except you have them in class).
Inserting a multimeter into the ground line would work too, though. But if you do it for a class project, lifting the ground line probably won't be what the teacher wants to see.
There's also the possibility that current flows between either supply line. This wouldn't be reflected in the ground line measurement.
By multiplying your measurement value with your supply voltage, you get an approximation of the power consumed. If you don't measure both at the same time using an appropriate technique, you only approximate. But measuring voltage and current at the same time introduces a systematic error that would - in theory - have to be corrected for: the ammeter has a resistance (that drops voltage) and the voltmeter has a resistance, too (which drains current).
Cheers,
Sebastian.
the most precise way to determine the total current draw of the circuit would probably be to insert a current (measurement) transformer into the supply line(s), preferrably the transformer primary. I guess you're using a laboratory supply, so you would require two of them for it's dual output rails. Probably a bit expensive (except you have them in class).
Inserting a multimeter into the ground line would work too, though. But if you do it for a class project, lifting the ground line probably won't be what the teacher wants to see.
There's also the possibility that current flows between either supply line. This wouldn't be reflected in the ground line measurement.
By multiplying your measurement value with your supply voltage, you get an approximation of the power consumed. If you don't measure both at the same time using an appropriate technique, you only approximate. But measuring voltage and current at the same time introduces a systematic error that would - in theory - have to be corrected for: the ammeter has a resistance (that drops voltage) and the voltmeter has a resistance, too (which drains current).
Cheers,
Sebastian.
You'll find that the ground line won't show much current draw; most goes directly from the positive rail to the negative rail. To measure it properly you will need 2 current meters, or move it between the two measurement points. If yours is an electronics class, maybe you've heard of Kirchoff's Current Law. By knowing two of the values, you can calculate the third. You only need the positive rail current and the negative rail current, however.
You can also estimate it from the National datasheets for the device. Assume a certain power output (maybe 5 watts?) and look up the value on the graph for your voltage supply and speaker impedance. See page 10 of
http://cache.national.com/ds/LM/LM3875.pdf
You must add the power to the loudspeaker (5W) to that of the device (which is shown in the graph).
5 watts into a speaker is actually very loud so your measurements will probably be quite a bit less than measured values at reasonable volume levels. There is also a minimum current draw for the amplifier; you can see it on the graph as power output is close to 0.
You also have to add the power to the preamp section. I think you can assume without too much error that this will negligible compared to the power amp section.
You can also estimate it from the National datasheets for the device. Assume a certain power output (maybe 5 watts?) and look up the value on the graph for your voltage supply and speaker impedance. See page 10 of
http://cache.national.com/ds/LM/LM3875.pdf
You must add the power to the loudspeaker (5W) to that of the device (which is shown in the graph).
5 watts into a speaker is actually very loud so your measurements will probably be quite a bit less than measured values at reasonable volume levels. There is also a minimum current draw for the amplifier; you can see it on the graph as power output is close to 0.
You also have to add the power to the preamp section. I think you can assume without too much error that this will negligible compared to the power amp section.
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