06-28-2024, 01:03 AM
(This post was last modified: 06-28-2024, 06:38 AM by physics.
Edit Reason: grammar
)
Hi!
In TUT2, figure 2-62 we're shown a "typical bias supply":
It's stated that the impedance of the supply is about 1k || 15k at balance, or about 1k. I need to work out the supply impedance for the bias supply I'm building in an amp, so I want to make sure that I understand how this calculation was done in the book.
First, is the diode assumed to be a short? Otherwise the 1k resistor is isolated. Is this because in thermal runaway, (conventional) current would flow out of the grid into the bias network, so from that perspective the diode would be forward-biased if there is enough current flowing to develop a diode-drop's worth of voltage across it?
Second, how come the resistance of the centertapped pot is ignored? At balance, the current from the grid has to flow through 5k to the centertap and wiper, then out to the 1k and 15k assuming the diode is forward biased, so the supply impedance would be (1k || 15k) + 5k ~= 6k. Admittedly the answer isn't that different, so maybe that was why it wasn't accounted for?
Also, how come the 70u capacitor doesn't factor in? Elsewhere in the TUTs, it's stated that if we have to increase the resistances to get adequate bias voltages, we can drop caps from the bias wipers to ground to get a good AC ground for the grids. Wouldn't the 70u cap act similarly, meaning that the supply impedance is just half of the center-tapped pot at most, so about 5k?
Sorry if it seems like I'm being nitpicky, I just want to make sure I understand the thought process behind the calculation so I have a better idea of how to go about the calculations for my own projects.
Thanks!
In TUT2, figure 2-62 we're shown a "typical bias supply":
It's stated that the impedance of the supply is about 1k || 15k at balance, or about 1k. I need to work out the supply impedance for the bias supply I'm building in an amp, so I want to make sure that I understand how this calculation was done in the book.
First, is the diode assumed to be a short? Otherwise the 1k resistor is isolated. Is this because in thermal runaway, (conventional) current would flow out of the grid into the bias network, so from that perspective the diode would be forward-biased if there is enough current flowing to develop a diode-drop's worth of voltage across it?
Second, how come the resistance of the centertapped pot is ignored? At balance, the current from the grid has to flow through 5k to the centertap and wiper, then out to the 1k and 15k assuming the diode is forward biased, so the supply impedance would be (1k || 15k) + 5k ~= 6k. Admittedly the answer isn't that different, so maybe that was why it wasn't accounted for?
Also, how come the 70u capacitor doesn't factor in? Elsewhere in the TUTs, it's stated that if we have to increase the resistances to get adequate bias voltages, we can drop caps from the bias wipers to ground to get a good AC ground for the grids. Wouldn't the 70u cap act similarly, meaning that the supply impedance is just half of the center-tapped pot at most, so about 5k?
Sorry if it seems like I'm being nitpicky, I just want to make sure I understand the thought process behind the calculation so I have a better idea of how to go about the calculations for my own projects.
Thanks!