My amp design efforts
24 OCTOBER 2006
HOW TO DESIGN AN AMPLIFIER
(CLASS A, COMMON EMITTER, NO FEEDBACK)
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.
FOR DC BIAS:
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.
INPUT (BASE) CIRCUIT
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.
OUTPUT (COLLECTOR) CIRCUIT
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.
HOW TO DESIGN AN AMPLIFIER
(CLASS A, COMMON EMITTER, NO FEEDBACK)
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.
FOR DC BIAS:
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.
INPUT (BASE) CIRCUIT
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.
OUTPUT (COLLECTOR) CIRCUIT
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.
posted by Bill Meara at 6:59 AM
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