Hi all, I'm sure there are at least a few who are familiar with using a BiasProbe.
According to its instruction manual, the method and formula for setting the bias correctly is:
I(p) = P(D) X 500 / E(p) X Qty.
I(p) = normal bias current, P(D) = rated RMS output power in watts, E(p) = B+ voltage.
I was recently going to bias a 100 watt Marshall head containing only 2 of the power tubes. Also I don't have a load - I realize I need to get one fast - my speaker cab is offsite at my band's rehearsal site. Given this info:
According to my calculations, my I(p) = 50, E(p) = 485V, Qty = 2. Using the bias probe, 50 x 500 = 25,000 / 485 X 2 = 970, or 25,000 / 970 = 25.77 milliamps. So my I(p) should be around 25.77 milliamps. My measurment was about twice that.
My questions are:
I read in Aspen Pittman's book that less negative bias equals more current. Likewise, more negative bias would reduce the current. In my example above it sounds like I need to increase the resistance on my bias pot in order to reduce the bias voltage (?) I assume it's a matter of simply using a screwdriver to turn the bias pot until the current measures what the formula says it should be.
I always hear about bias described using the adjectives of "hot" and "cold". In my situation where the current is twice what it should be, is this considered "hot" or "cold"?
Last, with using only 2 tubes instead of 4, and having no load, is my whole measurement process completely flawed and basically mute?
Thank you.
According to its instruction manual, the method and formula for setting the bias correctly is:
I(p) = P(D) X 500 / E(p) X Qty.
I(p) = normal bias current, P(D) = rated RMS output power in watts, E(p) = B+ voltage.
I was recently going to bias a 100 watt Marshall head containing only 2 of the power tubes. Also I don't have a load - I realize I need to get one fast - my speaker cab is offsite at my band's rehearsal site. Given this info:
According to my calculations, my I(p) = 50, E(p) = 485V, Qty = 2. Using the bias probe, 50 x 500 = 25,000 / 485 X 2 = 970, or 25,000 / 970 = 25.77 milliamps. So my I(p) should be around 25.77 milliamps. My measurment was about twice that.
My questions are:
I read in Aspen Pittman's book that less negative bias equals more current. Likewise, more negative bias would reduce the current. In my example above it sounds like I need to increase the resistance on my bias pot in order to reduce the bias voltage (?) I assume it's a matter of simply using a screwdriver to turn the bias pot until the current measures what the formula says it should be.
I always hear about bias described using the adjectives of "hot" and "cold". In my situation where the current is twice what it should be, is this considered "hot" or "cold"?
Last, with using only 2 tubes instead of 4, and having no load, is my whole measurement process completely flawed and basically mute?
Thank you.
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