ablaze 0 Report post Posted March 25, 2003 What are 1. DC offset? 2. DC leakage? I've measured them, but I've no idea what the hell they mean. someone care to explain what significance these hold? Share this post Link to post Share on other sites
jtfoo 0 Report post Posted March 25, 2003 I don't know how to explain DC offset. Experts pls correct me if I'm wrong. Ideally the signal should have a DC offset of 0 volts. so say a AC 2V pk-pk signal will be a +1V and -1V wave. If say there's an DC offset of positive 100mV (0.1V) offset in one of the channel. the resultant wave will have high peak of +1.1V and low peak 0f -0.9V Let's say there's a gross DC offset of 3V, then the same 2Vpk-pk AC signal will have maximum peak at 4V and minimum peak at 2V. This peak at 4V could be potentially damaging. Share this post Link to post Share on other sites
Audio 0 Report post Posted March 25, 2003 When there is audio signal, your amp output MUST be zero volt DC!!! If it is not, and say, it is 1.5V DC appearing across the output, it just like coonect the wire of your headphone directly to a battery. The membrance of your transducer will be pushed to the extreme, inward of outward will depend on the polarity. Most modern amp will have a zero dc voltage output due to good feedback characteristic but sometimes, when certain components drift, it is likely you will get DC drift. Normally, it can range froma few millavolts and upwards. Offset voltage is commonly used in Op amp circuitry where DC voltage appearing at the output can be removed by feedback loop or trimming the offset compensation control (available on some op amp IC). DC leakage is a loose term sometimes used for offset voltage. This term is used quite often in Tube amp to describe voltage that is not design or not suppose to be there...like the output. When you are playing music, what appears at the output should be AC voltage...and if you can measure DC, you're in trouble, the bigger the voltage, the higher heat is being dissipated at your headphone coils. (Audio) Share this post Link to post Share on other sites
ablaze 0 Report post Posted March 25, 2003 ok i just posted this in the other thread, DC leakage test, from my rudimentary understanding the multimeter has to be connected in SERIES to measure the current. so, lets say if i wanna test the right channel, i place one lead of the multimeter touching the right channel of the amp's output, the other lead touching the right channel of the headphone's plug, then you'll need another wire connecting the ground of the amp output to the ground on the headphone (to complete the circuit). repeat for other channel. the "leaky" channel will read out a much higher current. correct me if I'm wrong, Northern Oak whats this current I'm measuring? a lil lost Share this post Link to post Share on other sites
Audio 0 Report post Posted March 25, 2003 ok i just posted this in the other thread, DC leakage test, from my rudimentary understanding the multimeter has to be connected in SERIES to measure the current. so, lets say if i wanna test the right channel, i place one lead of the multimeter touching the right channel of the amp's output, the other lead touching the right channel of the headphone's plug, then you'll need another wire connecting the ground of the amp output to the ground on the headphone (to complete the circuit). repeat for other channel. the "leaky" channel will read out a much higher current. correct me if I'm wrong, Northern Oak whats this current I'm measuring? a lil lost No, you don't use the ampmeter part of your multimeter, you measure the voltage, not the current. It is unwise to place a current meter across your output as you will be shorting it!!! To measure DC voltage...set the voltage selectror to DC volt, place your probe across the output. You may like to unscrew the headphone jacks to get at the connections. (Audio) Share this post Link to post Share on other sites
