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#1
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I'm reading David Rutledge's excellent "The Electronics of Radio."
In Chapter 10 -- Power Amplifiers, he discusses Class C amps and says, "In addition, if we drive the transistor clear to saturation, using the transistor as a switch, the dissipated power can be greatly reduced, because the saturation voltage is low. This is Class C amplification..." I'd always throught that in Class C, while you'd operate the device so that it was cutoff during most of the cycle, but not saturated. Is this just a different definition of Class C? I checked back with SSDRA and EMRFD, and didn't see anything about driving Class C amps into saturation? What says the group? Do we saturate in Class C or not? 73 from London Bill M0HBR N2CQR CU2JL http://www.gadgeteer.us |
#3
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On 2006-07-12, Roy Lewallen wrote:
The Class C transistor amplifiers I design and use most certainly saturate. I believe that's standard practice for solid state amplifiers. And as a followup, is the power dissipation in the final controlled by the duty cycle of the pulses (or number of degrees of conduction more generally if the input is not a square wave)? -- Ben Jackson AD7GD http://www.ben.com/ |
#4
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#5
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Ben Jackson wrote:
On 2006-07-12, Roy Lewallen wrote: The Class C transistor amplifiers I design and use most certainly saturate. I believe that's standard practice for solid state amplifiers. And as a followup, is the power dissipation in the final controlled by the duty cycle of the pulses (or number of degrees of conduction more generally if the input is not a square wave)? For operation at HF and higher, it's not always that simple. For anything but the lowest-speed PWM/class C/class D/class E operations, the transistor switching time (and the influence of the load on the switching time) can be a most important contributor to power dissipated in the device. An indication of the "bare transistor" switching time can be found on the datasheets as the delay time, rise time, and most importantly storage time and fall time. Very closely related if you are operating anywhere near max power dissipation or max current is the SOA curve. Tim. |
#6
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Ben Jackson wrote:
On 2006-07-12, Roy Lewallen wrote: The Class C transistor amplifiers I design and use most certainly saturate. I believe that's standard practice for solid state amplifiers. And as a followup, is the power dissipation in the final controlled by the duty cycle of the pulses (or number of degrees of conduction more generally if the input is not a square wave)? Not really, although it's a factor. What counts more is the transistor voltage when the current is maximum, and the relationship between V and I during switching. This in turn depends on the nature of the output network among other things. If you really want to find out more about the efficiency of a saturating transistor amplifier, look up some of the papers on "class E" operation by Sokal & Sokal and by Raab. They basically took a time-domain approach to find the optimum conditions for maximum efficiency of a saturating class C amplifier. I've never consciously designed a "class E" amplifier, but have gotten more than 85% collector efficiency from saturating Class C amplifiers at the 10 watt input level. Roy Lewallen, W7EL |
#7
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What says the group? Do we saturate in Class C or not?
----------------------- In order to get any semblence of efficiency, the drive waveform should drive the stage from near-cut-off to near-saturation, somewhat like a switch. However, depending on how hard the sinusoidal input signal drives the "Clacc C" stage, the conduction-angle and thus the duty-cycle of the output (before filtering) will increase or decrease, resulting in variable output power in the load. Joe W3JDR "Tim Shoppa" wrote in message oups.com... wrote: I'm reading David Rutledge's excellent "The Electronics of Radio." In Chapter 10 -- Power Amplifiers, he discusses Class C amps and says, "In addition, if we drive the transistor clear to saturation, using the transistor as a switch, the dissipated power can be greatly reduced, because the saturation voltage is low. This is Class C amplification..." I'd always throught that in Class C, while you'd operate the device so that it was cutoff during most of the cycle, but not saturated. Is this just a different definition of Class C? I checked back with SSDRA and EMRFD, and didn't see anything about driving Class C amps into saturation? What says the group? Do we saturate in Class C or not? Saturation is itself a somewhat mushy point. There's a V_sat specified on the datasheets but the actual definition of saturated is entirely application-sensitive. As a practical matter as you add more base current you will go further into saturation (up until you melt the base-emitter junction and then all sorts of wacky things ensue). Choice of drive level and output level and load impedance in a Class C amplifier certainly will in many cases put the device into saturation. Tim. |
#8
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The 'power dissipation' in a Cass C stage is primarily a function of what
percentage of the cycle the stage is neither in saturation nor cutoff. A perfectly rectangular output switching waveform of any duty cycle , if it could be achieved, would result in nearly 100% efficiency. Joe W3JDR "Ben Jackson" wrote in message ... On 2006-07-12, Roy Lewallen wrote: The Class C transistor amplifiers I design and use most certainly saturate. I believe that's standard practice for solid state amplifiers. And as a followup, is the power dissipation in the final controlled by the duty cycle of the pulses (or number of degrees of conduction more generally if the input is not a square wave)? -- Ben Jackson AD7GD http://www.ben.com/ |
#9
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W3JDR wrote:
In order to get any semblence of efficiency, the drive waveform should drive the stage from near-cut-off to near-saturation, somewhat like a switch. However, depending on how hard the sinusoidal input signal drives the "Clacc C" stage, the conduction-angle and thus the duty-cycle of the output (before filtering) will increase or decrease, resulting in variable output power in the load. Joe W3JDR Who said the drive had to be sinusoidal? If the final can run almost square wave, why can't the driver? |
#10
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![]() The 'power dissipation' in a Cass C stage is primarily a function of what percentage of the cycle the stage is neither in saturation nor cutoff. A perfectly rectangular output switching waveform of any duty cycle , if it could be achieved, would result in nearly 100% efficiency. Joe W3JDR =================================== Unfortunately, it makes a terrible mess of any modulation. To avoid distortion and non-linearity, saturation can be permitted only on the extreme peaks of the modulated driving waveform. Class-C is ruled out. Class-B or Class-AB prevails. ---- Reg. |
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