Hey All,
Can anyone point me to a good explanation of 'drive'? I know it has to do with voltage swing, gm and impedance but I haven't found a really good explanation in the literature. I understand how to design a single stage. But how to design one stage into another escapes me.
Kevin
Can anyone point me to a good explanation of 'drive'? I know it has to do with voltage swing, gm and impedance but I haven't found a really good explanation in the literature. I understand how to design a single stage. But how to design one stage into another escapes me.
Kevin
There are at least two issues here, which people often confuse.
The first relates to impedance. The output impedance of the first stage and the input impedance of the second stage form a potential divider. This will attenuate the voltage. If either impedance is non-linear it will also introduce distortion. Note that the output impedance of most valve stages is non-linear; the issue is 'by how much?'. The normal way of coupling audio stages together it to have the output impedance much smaller than the input impedance of the next stage so that there is little attenuation and little added distortion.
The second issue is current. How much signal current can the first stage produce without clipping or distorting too much? This is related to, but usually smaller than, the quiescent current of the first stage. The problem may arise if the second stage has high input capacitance (perhaps from Miller effect).
These two issues are separate, in that it is possible to design stages which are good for one and bad for the other. People who don't understand this sometimes lump them together as 'drive', then wonder why their designs don't work too well or, alternatively, massively over-engineer things and insist that everyone else does the same.
The first relates to impedance. The output impedance of the first stage and the input impedance of the second stage form a potential divider. This will attenuate the voltage. If either impedance is non-linear it will also introduce distortion. Note that the output impedance of most valve stages is non-linear; the issue is 'by how much?'. The normal way of coupling audio stages together it to have the output impedance much smaller than the input impedance of the next stage so that there is little attenuation and little added distortion.
The second issue is current. How much signal current can the first stage produce without clipping or distorting too much? This is related to, but usually smaller than, the quiescent current of the first stage. The problem may arise if the second stage has high input capacitance (perhaps from Miller effect).
These two issues are separate, in that it is possible to design stages which are good for one and bad for the other. People who don't understand this sometimes lump them together as 'drive', then wonder why their designs don't work too well or, alternatively, massively over-engineer things and insist that everyone else does the same.
DF has it, but just to try it another way...In my simple mind I think of swing and drive as the same b/c when I design something to swing a certain amount of voltage, it is into the intended load and not just over it's self. But technically swing is just the amount of voltage you find at the output of a stage without a load I guess.
Lets make an example: Say you have a simple high gain triode (12AX7 perhaps) with a large ohmic plate resistor (200kohms or CCS even). Say b+ is 500V and the gainstage (12AX7 triode) swings +-200V peak and idles a 1.25mA. (I'm just using numbers here for sake of the example, not a circuit well thought out).
Now you want to drive a low mu tube that has a gain of 5 with the intended O.T. and speaker load. Say this tube's capacitance from grid to plate is 10pF. Cmiller is then equal 50pF (gain times Cgp). You have some more capacitance as well, but miller dominates.
You wish to impress and want to have a 50kHz top end and with this low mu tube you want the 200Vp swing at it's grid.
There are two ways to see how the 12AX7 gain stage wont drive the output tube properly.
1:
Find the impedance of 50pF at 50kHz, which is 63.7kohms. Will the +-1.25mA of the 12AX7 drive this to 200V?
2:
The slew rate of 200V at 50kHz is (SR(sinusoid)=2*pi*Vp*freq) giving SR=62.8V/us. How much current is needed to slew a 50pF capacitance 62.8V/us? (icap=cap*dv/dt), (btw dv/dt=SR), this gives 3.14mA.
Using a CCS the driver really doesn't need anymore steady state idle current than just a little more than the total load requires. And in the given example 200V swing was giving plenty overdrive ability, so settling at 4mA may be just as beefy as required, but the output impedance of the driving stage needs to be low enough to not introduce your 3dB knee untill 50kHz, which we know from point 1 would be 62.7kohm.
So basically the driver needs to be able to swing it's plate resistance and the load connected and that's your drive.
Lets make an example: Say you have a simple high gain triode (12AX7 perhaps) with a large ohmic plate resistor (200kohms or CCS even). Say b+ is 500V and the gainstage (12AX7 triode) swings +-200V peak and idles a 1.25mA. (I'm just using numbers here for sake of the example, not a circuit well thought out).
Now you want to drive a low mu tube that has a gain of 5 with the intended O.T. and speaker load. Say this tube's capacitance from grid to plate is 10pF. Cmiller is then equal 50pF (gain times Cgp). You have some more capacitance as well, but miller dominates.
You wish to impress and want to have a 50kHz top end and with this low mu tube you want the 200Vp swing at it's grid.
There are two ways to see how the 12AX7 gain stage wont drive the output tube properly.
1:
Find the impedance of 50pF at 50kHz, which is 63.7kohms. Will the +-1.25mA of the 12AX7 drive this to 200V?
2:
The slew rate of 200V at 50kHz is (SR(sinusoid)=2*pi*Vp*freq) giving SR=62.8V/us. How much current is needed to slew a 50pF capacitance 62.8V/us? (icap=cap*dv/dt), (btw dv/dt=SR), this gives 3.14mA.
Using a CCS the driver really doesn't need anymore steady state idle current than just a little more than the total load requires. And in the given example 200V swing was giving plenty overdrive ability, so settling at 4mA may be just as beefy as required, but the output impedance of the driving stage needs to be low enough to not introduce your 3dB knee untill 50kHz, which we know from point 1 would be 62.7kohm.
So basically the driver needs to be able to swing it's plate resistance and the load connected and that's your drive.
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