Home |
Search |
Today's Posts |
|
#1
|
|||
|
|||
Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A amplifier)." The no-signal loss of a Class A amplifier is 100%. It equals volts x amps and appears in the amplifier. Now feed a signal to the ideal amplifier set just below the clipping level. Average d-c power is unchanged from the unloaded and no-signal conditions. Connect a matched load resistor to the amplifier output. If physically small, the resistor may become warm with heat that were it not for the load would be otherwise dissipated in the amplifier. Input power to the Class A amplifier is unchanging. Finding the internal resistance theoretically is simple. It is simply the open-circuit output voltage divided by the short-circuit current. Open-circuit voltage at full output and short-circuit current may be severe. Internal resistance can be found under less stressful conditions. Internal resistance will drop the voltage to any load reasistance. Use the voltage-divider formula to calculate the internal resistance. With pure resistances, half the open-circuit volts are dropped by the internal resistance when the load is a match. A power amplifier`s internal impedance can be determined. Power output from a Class A amplifier cools it. Best regards, Richard Harrison, KB5WZI |
#2
|
|||
|
|||
Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current) At present I have no facilities to make such measurements. Perhaps a few curious readers, who do have facilities, might make crude measurements on their HF rigs and report approximate impedance values on one or two bands on this newsgroup. Some sort of average could be obtained. ---- Reg, G4FGQ =================================== "Richard Harrison" wrote in message ... Asimov wrote: "Let`s look at it from the dynamic point of view (loss in a Class A amplifier)." The no-signal loss of a Class A amplifier is 100%. It equals volts x amps and appears in the amplifier. Now feed a signal to the ideal amplifier set just below the clipping level. Average d-c power is unchanged from the unloaded and no-signal conditions. Connect a matched load resistor to the amplifier output. If physically small, the resistor may become warm with heat that were it not for the load would be otherwise dissipated in the amplifier. Input power to the Class A amplifier is unchanging. Finding the internal resistance theoretically is simple. It is simply the open-circuit output voltage divided by the short-circuit current. Open-circuit voltage at full output and short-circuit current may be severe. Internal resistance can be found under less stressful conditions. Internal resistance will drop the voltage to any load reasistance. Use the voltage-divider formula to calculate the internal resistance. With pure resistances, half the open-circuit volts are dropped by the internal resistance when the load is a match. A power amplifier`s internal impedance can be determined. Power output from a Class A amplifier cools it. Best regards, Richard Harrison, KB5WZI |
#3
|
|||
|
|||
Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#4
|
|||
|
|||
Cecil Moore wrote:
Reg Edwards wrote: Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. Some of Motorola's devices might actually be able to stand Reg's little test. The MRF150, for instance is advertised as being able to withstand a 30:1 VSWR at all phase angles. I think Reg should offer to pay for any damage sustained as the result of this test. 73, Tom Donaly, KA6RUH |
#5
|
|||
|
|||
Reg Edwards wrote:
Richard, your method of measuring internal impedance of an amplifier sounds interesting. (o/c volts divided by s/c current) Reg, I tried that one time in college. I got as far as torching two ARC-5 1625's driving an open circuit so I don't know what would have happened with a short circuit. -- 73, Cecil ============================== Cecil, Presumably you hadn't heard of Ohms Law. The internal resistance of a generator is independent of its internal voltage. Just reduce the drive level to some small, no particular value, and stop making excuses. ( Somebody will say it IS dependent. But we are interested only in ball-park accuracy. And in any case the operating point will be within the normal range of operation between no-drive and full drive, or between no modulation and full modulation.) ---- Reg, G4FGQ |
#6
|
|||
|
|||
Reg Edwards wrote:
The internal resistance of a generator is independent of its internal voltage. Just reduce the drive level to some small, no particular value, and stop making excuses. ( Somebody will say it IS dependent. But we are interested only in ball-park accuracy. And in any case the operating point will be within the normal range of operation between no-drive and full drive, or between no modulation and full modulation.) That might work better for Class-AB than Class-C. If one reduces drive level on a Class-C, one is reducing the 'on' time of the device, thus changing the internal impedance probably somewhat inversely proportional to the 'on' time. If I remember correctly, my IC-706 would not fold back under any load condition when run at minimum power. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#8
|
|||
|
|||
Asimov wrote:
"I was confusing it with Class B which does warm up the heatsink" True. Class B amplifiers can be single-ended for R-F if used with the right tank circuit to supply the missing 1/2-cycle. More often they are push-pull to reduce harmonics. SSB final amplifiers are biased to near current cut-off. During modulation gaps, current falls to nearly zero. When current is zero, so is the dissipation. Reg mentioned the blistering heat of the 6N7. Thai`s my recollection too when getting nearly 10 watts of audio from the small twin-triode octal-based metal tube. Reduce the audio volume and the heat is reduced simultaneously. Best regards, Richard Harrison, KB5WZI |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Forum | |||
Discone antenna plans | Antenna | |||
The "TRICK" to TV 'type' Coax Cable [Shielded] SWL Loop Antennas {RHF} | Antenna | |||
Poor quality low + High TV channels? How much dB in Preamp? | Shortwave | |||
X-terminator antenna | CB | |||
Outdoor Antenna and lack of intermod | Scanner |