Message	Date	From	Subject
12006	2015-12-14 13:13:08	Claudio Girardi	R: [emrfd] IRF510 : how high?
In practice I'll say that you can't get 100W from a pair of IRF510. They are rated for 100 V and 5.6 A max and these are the DC values and the current max at 25 C, so should probably be derated at high frequency, but let's be optimistic and assume we can push the devices to these values also at HF. For a push-pull class (A)B, the peak drain voltage Vd_max is 2Vdd (of course in practice it could be worse, hi), so you can use a Vdd_max of 50 V. The peak drain current Id_max is Vdd/R, where R is the load seen by each drain, so to have Id_max = 5.6 A you have to choose R = 8.9 ohm. With this load resistance the power output is Vdd^2/(2R) = 140 W. Yes, it's higher than 100 W but we are driving the devices to their maximum limits and in practice you have other factors that will reduce the maximum output power. The classic book "Solid State Radio Engineering" by Krauss, Bostian and Raab (and probably others) has all the formulas for the "normalized power output capability" of each amplifier class, i.e. what is the theoretical maximum output power you could obtain by driving the active devices to reach their Vdd_max and Id_max . For a class-B amplifier it is 0.25 so the calculation above could also be simply expressed as Pout_max = 0. 25 * 100 * 5.6 = 140 W 73 de Claudio, IN3OTD farhanbox@gmail.com wrote: > whats the maximum drain voltage one can use with the IRF510? > >i have pushed them to 48v. some blew up on the load test at 20 watts (single ended). i am wondering if i can get 100 watts from a pair at 14 Mhz. at 30v I am clocking 30 watts. > > - f

In practice I'll say that you can't get 100W from a pair of IRF510.
They are rated for 100 V and 5.6 A max and these are the DC values and the current max at 25 C, so should probably be derated at high frequency, but let's be optimistic and assume we can push the devices to these values also at HF.

For a push-pull class (A)B, the peak drain voltage Vd_max is 2*Vdd (of course in practice it could be worse, hi), so you can use a Vdd_max of 50 V. The peak drain current Id_max is Vdd/R, where R is the load seen by each drain, so to have Id_max = 5.6 A you have to choose R = 8.9 ohm. With this load resistance the power output is Vdd^2/(2*R) = 140 W.
Yes, it's higher than 100 W but we are driving the devices to their maximum limits and in practice you have other factors that will reduce the maximum output power.

The classic book "Solid State Radio Engineering" by Krauss, Bostian and Raab (and probably others) has all the formulas for the "normalized power output capability" of each amplifier class, i.e. what is the theoretical maximum output power you could obtain by driving the active devices to reach their Vdd_max and Id_max . For a class-B amplifier it is 0.25 so the calculation above could also be simply expressed as Pout_max = 0. 25 * 100 * 5.6 = 140 W

73 de Claudio, IN3OTD

farhanbox@gmail.com wrote:

> whats the maximum drain voltage one can use with the IRF510?
>
>i have pushed them to 48v. some blew up on the load test at 20 watts (single ended). i am wondering if i can get 100 watts from a pair at 14 Mhz. at 30v I am clocking 30 watts.

> - f