# Harry's Basic linear amplifiers

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## Re: Harry's Basic linear amplifiers

Second attempt:

My circuits are written to be simple and easy to reproduce, but I used to quote Harry's laws of coils:

1 - No two people can wind the same coil given the same data

2 - coil-winding data is a constant that varies from person to person

Now take another known fact, AF and HF circuits are quite different to VHF circuits. A typical VHF transistor input circuit will have two capacitors forming a voltage divider (and current multiplier) to transform the 50Ω input impedance to (say) 5Ω. But at VHF you cannot be nonchalant about capacitance, to the input of a transinstor may be 5Ω-J3Ω which is another way of saying that there are two input impedances, one of which is capacitive. The two capacitor trick works but you need to add an inductor to cancel the input capacitance.

At AF and HF we can use the battery and bulb matching technique: If you have a battery and a bulb both having the same internal impedance, then when you connect the bulb the battery terminal voltage will be 50% but you will get maximum power transfer. If your battery internal impedance is only 5% of the load impedance then you will have almost the full battery voltage across the load. In other words, drive with a low impedance, load with a high impedance, and simply ignore the small capacitances.

The input (B-E) impedance of the single BC547 is about 26Ω (plus the emitter resistor), which just happens to be 1/4 of 50Ω. So a 4:1 transformer (turns ratio = 2:1) is just perfect. The lower input drive impedance leads to improved stability and flexibility at the expense of a couple of dB.

When driving four or more) transistors in parallel then the transistor input impedance is going to be typically 3Ω to 5Ω so designing for 1Ω works fine. A 50:1 transformer (turns ratio = 7:1) is perfect. It will also cover most variations tried by readers of my homepages. I sacrificed a couple of dB for a simplified circuit that will work every time.

If you want to milk out the last dB then take the datasheet of the devices you are using and determine the base-emitter impedance. Add the emitter-Gnd impedance and design your transformer accordingly. Do not forget to balance out any capacitance.

Have I explained this well enough for you?

Very best regards from Harry - SM0VPO

My circuits are written to be simple and easy to reproduce, but I used to quote Harry's laws of coils:

1 - No two people can wind the same coil given the same data

2 - coil-winding data is a constant that varies from person to person

Now take another known fact, AF and HF circuits are quite different to VHF circuits. A typical VHF transistor input circuit will have two capacitors forming a voltage divider (and current multiplier) to transform the 50Ω input impedance to (say) 5Ω. But at VHF you cannot be nonchalant about capacitance, to the input of a transinstor may be 5Ω-J3Ω which is another way of saying that there are two input impedances, one of which is capacitive. The two capacitor trick works but you need to add an inductor to cancel the input capacitance.

At AF and HF we can use the battery and bulb matching technique: If you have a battery and a bulb both having the same internal impedance, then when you connect the bulb the battery terminal voltage will be 50% but you will get maximum power transfer. If your battery internal impedance is only 5% of the load impedance then you will have almost the full battery voltage across the load. In other words, drive with a low impedance, load with a high impedance, and simply ignore the small capacitances.

The input (B-E) impedance of the single BC547 is about 26Ω (plus the emitter resistor), which just happens to be 1/4 of 50Ω. So a 4:1 transformer (turns ratio = 2:1) is just perfect. The lower input drive impedance leads to improved stability and flexibility at the expense of a couple of dB.

When driving four or more) transistors in parallel then the transistor input impedance is going to be typically 3Ω to 5Ω so designing for 1Ω works fine. A 50:1 transformer (turns ratio = 7:1) is perfect. It will also cover most variations tried by readers of my homepages. I sacrificed a couple of dB for a simplified circuit that will work every time.

If you want to milk out the last dB then take the datasheet of the devices you are using and determine the base-emitter impedance. Add the emitter-Gnd impedance and design your transformer accordingly. Do not forget to balance out any capacitance.

Have I explained this well enough for you?

Very best regards from Harry - SM0VPO

_________________

*(Thanks to those who tried and couldn't make the QSO. Back on 14.200MHz Sundays from the 24th*

*Sept.)*

**Admin**- Admin
- Posts : 416

Join date : 2012-11-24

Age : 65

Location : Märsta, Sweden

## Re: Harry's Basic linear amplifiers

I just spent 30 minutes replying to this but the server was busy and the whole lot was lost. I will try to replay again to this later. Sorry for the delay.

BR Harry

BR Harry

_________________

*(Thanks to those who tried and couldn't make the QSO. Back on 14.200MHz Sundays from the 24th*

*Sept.)*

**Admin**- Admin
- Posts : 416

Join date : 2012-11-24

Age : 65

Location : Märsta, Sweden

## Harry's Basic linear amplifiers

Hello,

I have questions regarding Harry's design approach. The article is found at the following: http://www.sm0vpo.com/tx/rf_pa_cct_00.htm

Harry calculates the input transformer but I am lost at what assumptions were made to come to the conclusion. Also, can someone hammer out a math explanation of how these values are derrived.

Best Regards,

Nick

I have questions regarding Harry's design approach. The article is found at the following: http://www.sm0vpo.com/tx/rf_pa_cct_00.htm

Harry calculates the input transformer but I am lost at what assumptions were made to come to the conclusion. Also, can someone hammer out a math explanation of how these values are derrived.

Best Regards,

Nick

**macgman2000**- Posts : 1

Join date : 2013-02-19

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