Saturday, November 25, 2006

What I Bought at the Kempton Park Hamfest (Rally)

The blue box is the power supply. The yellow box (tobacco tin) houses a broadband amp and seems to foreshadow the use of Altoid tins. The Black box is an old signal generator. I bought it because the model is "Windsor" (I like the town of that name). It was manufactured in Slough, Bucks, UK. You can also see the Hi-Z headphones, the PC boards, 73 magazine, and the RSGB's Handbook (a very good book!)

Thursday, November 23, 2006

Dipole visible from space!

I was surprised to be able to see my fishing pole dipole on Google Earth. I realize that this is probably a shot taken from an aircraft (not from space).

Friday, November 10, 2006

Should Class C Amps Saturate?

I have been trying to figure this out and have been posting the following questions on If anyone out there can help, please send me some comments or e-mail. Thanks, Bill

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? -------------------------------------------------------- SECOND POST
I've been thinking about this some more. The 1980 ARRL handbook points out that "Solid State power amplifiers should be operated just below their saturation points for best efficiency and stability." Also, the formula that we use to determine load resistance (Rl=Vcc^2/2Po) implies that we are looking for a combination of Vcc, Load resistance and power out that will prevent saturation. And wouldn't we end up with far lower harmonic content if we only clip one side of the wave form (at cutoff) instead of both sides (cutoff and saturation)?
I know there are more exotic modes beyond C, but for plain old ordinary ham radio applications, don't we normally avoid saturation in Class C amps?
Also, what about this business of having the efficiency improve through saturation "because the saturation voltage is low" Could that be right? If you put a voltage across a conductor and generate a large current, you can't sit back and say "Great! Power consumption across the conductor is low because the voltage drop across it is now minimal!"
Thanks to all who responded. I think I'm starting to understand this. LTSpice helps a lot. I set up a class C amp and looked out power dissipated in the transistor vs. power dissipated in the load. The big efficiency gains that come with saturation were very apparent.
To better understand WHY this happens, I set up a spreadsheet that looked at power dissipated in a variable resistor as it swept from .1 ohms to 10 ohms. It had a fixed 10 ohm resistor in series and 10 volts DC across both of them. Yes indeed, the power dissipated in the variable resistor drops off dramatically when the resistance (and hence the voltage across it) gets very low. I guess is why this happens in the saturating Class C amp, Right?
But I still have some questions. When we design a Class A amp, the familiar formula Rload = (Vcc-Ve)^2/2Pout allows us to come up with a load value that will prevent the amplifier from saturating. A load of this value will cause the voltage across the transistor (collector to emitter) to vary from zero to twice Vcc. But it won't go into saturation.
Why then do so many of the books (EMRFD, SSDRA, the W1FB books) call for the use of essentially the same formula for the load when selecting a load for Class C amps? We're no longer worried about staying out of saturation, correct? In fact, we want to saturate. So why the same formula? In fact, it seems to me that if you have a Class C amplifier that is designed with this formula and is operating just below saturation, you can get it to saturate just by increasing the value of the load presented to the collector. Power out and efficiency immediately improves. Linearity, of course, does not.
Thanks, 73 Bill M0HBR

Sunday, October 29, 2006

My amp design efforts

24 OCTOBER 2006


You have to keep in mind the differences between input (DC) power and output (AC) power. And throughout you have to keep in mind the differences between peak, average, and rms values. Remember that when we talk about power, we are talking about AVERAGE power: Vrms*Irms. Or (Vpeak*Ipeak)/2

I have an oscillator/amplifier circuit that produces .050 watts in a 50 ohm load. I want to build an amplifier stage that will increase power out to .5 Watts.


Assume 20 percent efficiency. So I will need 2.5 Watts DC input.
Assume 11 volts on the collector and 3 volts bias on the emitter.
So there is 8 volts across the transistor.
For 2.5 watts DC input I’ll need 312 ma of idle current. (8*.3125=2.5 watts)
With 3 volts on the emitter, that corresponds to an emitter resistor of 9.6 ohms (bypassed)
To get the 3 volts on the emitter, you need approximately 3.6 volts on the base. (Because the silicon diode that forms the base-emitter junction drops .6 volts.) Use a simple resistor voltage divider network to get this.


The 312 milliamps of idle current results in an emitter resistance of .0833 ohms (from the formula 26/312ma.)
To get the input impedance, you take this value and multiply it by (Beta+1) Beta= ft/f
250/14 = 17.8
So 18.8*.0833=1.566 ohms input impedance.

A transformer with a turns ratio of 5.6:1 will yield a 50 ohm input.

Using the 4X rule of thumb, I took the input impedance 1.6ohms * 4 and got 6.4 ohms. At 14Mhz that is .0727 uH for the secondary, .4072 uH for the primary.


Remember that rms volts X rms amperes = AVERAGE power.
And Peak volts X Peak current = 2 X AVERAGE POWER

To determine necessary load: Rload= (Vcc-Ve)^2/2Pout

This formula derives from the standard formulas linking power calculations to ohms law: P= E^2/R. Vcc-Ve represents the highest possible peak voltage. If you have only 8 volts between the collector and the emitter, to avoid voltage flat-topping, you can’t have a load that will, at peak signal current, drop more than 8 volts. But the power equations need rms values. In the numerator, you are squaring the peak voltage value. Assume that the peak voltage was 1. 1^2=1 Now, take the peak voltage value down to rms: 1*.707 = .707 I you square .707 you get .5 volts. So, putting the 2 in the denominator of this equation is just simple way of converting the peak voltage value to an rms value needed by the P=E^2/R formula.

So: .5 = (11-3)^2/2(x) x= 64 ohms

Play with this a bit to see what it means.
8Vpeak * Ipeak = 2*(.5) (What peak current multiplied by 8 volts peak voltage will give you 1 watt?) Remember: Peak volts * Peak current = 2*Average Power out.
So Ipeak = .125 amps.

At peak, what resistance would drop 8V at .125 amps? Ohms law: 8/.125 = 64 ohms.

Now, drop the peak voltage and current values down to rms.
8*.707=5.56V rms
.125*.707= .088Amps rms

5.56V*.088A= .5 watts

.5 watts in 64 ohms implies .088 amps (rms) (P=I^2R)
.5 watts in 64 ohms yields 5.64 volts (rms) (P=E^2/R)

Now, convert these current and voltage figures from rms back up to peak to see how it all fits together:

5.64 V * 1.414 = 8 volts. That is the max maximum possible voltage drop (that would drop collector voltage from 11 down to 3).

.088 * 1.414 = .125 amps. .125 amps flowing through 64 ohms drops… 8 volts.

Another way of looking at this would be this. You have 8 volts peak to work with. You are saying you want .5 watts out. Since P=IE and I = E/R, this means you are calling for a certain impedance and a corresponding value for current. Take the volts down to rms.
.5 = 5.656I I=.088 amps rms. 125 ma peak

????Now, let’s take a look at the efficiency of this stage. Assume we put the idle current at 125ma. You have 8 volts across the transistor. That’s 1 watt DC input. 50 percent efficiency. This also works if you look at input power of one complete cycle of the signal. Voltage (collector to emitter) is varying from 8, up to 16, down to 0. Current is varying from 125 ma, up to 250, down to zero. So average current is .125 Average Voltage is 8 power input is 1 watt. ????/

So, to make my 50ohm antenna system look like 64 ohms. I use the formula
Np/Ns = square root Zp/Zs. That yields a turns ratio of 1.131:1.

Using the rule of thumb of making the Z = 4*Z of smaller coil, Z=200
At 14 Mhz this is 2.27 uH for the 50 ohm secondary, 2.567 uH on the primary.

Saturday, August 19, 2006

Steve Weber's COOL 25 Watt Linear Amplifier

I was looking at Steve "Melt Solder" Weber's (fairly) simple SSB rig. I really liked the 25 Watt linear he built for it. Take a look at his technique for mounting the MOSFETS on the big finned heat sink: he just cut holes in the PC board and, through these holes mounted the MOSFETS on the heatsink. Then the rest of the circuit is mounted on the board using our traditional ugly/Manhattan techniques.

Friday, August 11, 2006

The Very First Homebrewers -- Hominid HB!

I'm reading "A Short History of Nearly Everything" by Bill Bryson. In his description of early toolmaking, I came across some passages that I think will resonate with modern electronic homebrewers:

"Sometime about a million and a half years ago, some forgotten genius of the hominid world did an unexpected thing. He (or very possibly she) took one stone and carefully used it to shape another. The result was a simple teardrop-shaped hand axe, but it was the worlds first piece of advanced technology.

It was so superior to existing tools that soon others were following the inventor's lead and making hand axes of their own. Eventually whole societies existed that seemed to do nothing else. 'They made them in their thousands,' says Ian Tattersal... It's strange because they are quite intensive objects to make. It was as if they made them for the sheer pleasure of it.'

...The axes became known as Acheulean tools.... These early Homo sapiens loved their Acheulean tools... They carried them vast distances. Sometimes they even took unshaped rocks with them to make into tools later on. They were, in a word, devoted to the technology."

Sound familiar?

Saturday, August 05, 2006

Watch Richard Feynman talk about radio waves!

I'm a big fan of Richard Feynman. I've been reading his books for years, but until I came across this YouTube video clip, I'd never heard his voice or seen a film of him in action.

I was struck by his accent. Brooklyn comes through very clearly. It is a great thing that humble, working class origins didn't prevent Feynman from achieving what he did.

Check it out: