Hi all. Was playing with mosfets over the weekend when I noticed something a bit strange going on with the information on CBtricks with regards the EN-369FN.
Whilst the information below does contain some maths and Ohms law it does clearly prove that the EN-369FN drawing and in particular the value of R2 is infact not just incorrect, but it's waaaay off!
The example being used is a bog standard Magnum 257 that uses the EN369DR and EN369FN companion parts on the driver and final mosfets. Below is the schematic of the driver and finals mosfets and their biasing resistors and below that the EN-369FN and DR nabbed from CB Tricks .com
Now for the calculations:
Worked driver and final PA bias examples using a Magnum 257 as our example:
All examples are taken with RV11 and RV13 at their midpoint which is 10k and 250k ohm respectively.
Keep in mind that we would expect to bias the driver mosfet at about 3.1 - 3.3 volts and the finals at 3.7 volts.
Driver biasing example - Please refer to red box 1 on the above schematic.
Transmit line voltage is the usual 8 volts DC.
First we calculate the current:
current total = transmit line voltage / (RV13 midpoint + R233 + [R2 of the EN369DR]) =
So current total = 8 / (250000 + 270000 + 320000) = 8 / 840000 = 0.000009523A or 9.523 uA (Micro amps)
Now we can calculate the bias voltage across the EN-369DR
bias voltage across the EN36DR = current total x [R2 of the EN369DR]
So bias voltage = 0.000009523 x 320000 = 3.04 volts.
3.04 volts is a nice driver bias voltage to start with and we can tweak it from there.
Final biasing example - Please refer to red box 2 on the above schematic..
Please note before we start the transmit line voltage is now 7.4 volts as we drop 0.6 volts across transistor Q5
First we calculate the current:
current total = transmit line voltage / (RV11 midpoint + R239 + [R2 of the EN369FN])
So current total = 7.4 / (10000 + 10000 + 230000) = 8 / 250000 = 0.0000296A or 29.6 uA (Micro amps)
Now we can calculate the voltage across the EN-369FN
bias voltage across the EN369FN = current total x [R2 of the EN369FN]
So our bias voltage = 0.0000296 x 230000 = 6.808 volts
Hang on a moment! That’s not good at all, actually this is very BAD. 6.808 volts is almost double what we should be biasing the mosfet at and will destroy it very quickly!
So what’s wrong here? We know full well that the Magnum was never designed to run at 6.8 volts bias, or else they’d be popping finals left right and centre.
The answer to that question is R2 in the EN369FN in the drawing is incorrect.
If we were to decrease value of R2 in the EN369FN to 20k and then redo the maths.
Final biasing example with R2 in the EN369FN at 20k ohms
First we calculate the current:
current total = transmit line voltage / (RV11 midpoint + R239 + [our new R2 of the EN369FN]) =
So current total = 7.4 / (10000 + 10000 + 20000) = 8 / 40000 = 0.000185A or 185 uA (Micro amps)
Now we can calculate the voltage across the EN-369FN
bias voltage across the EN36FN = current total x [our new R2 of the EN369FN]
So bias voltage = 0.000185 x 20000 = 3.7 volts
3.7 volts = A perfect bias!
As you can clearly see from the above simple maths the EN369FN should not be 230k ohms as per the drawing, but should be 20k ohms.
If you use the following drawing edit it will not only work, but work as intended.
Psi
Whilst the information below does contain some maths and Ohms law it does clearly prove that the EN-369FN drawing and in particular the value of R2 is infact not just incorrect, but it's waaaay off!
The example being used is a bog standard Magnum 257 that uses the EN369DR and EN369FN companion parts on the driver and final mosfets. Below is the schematic of the driver and finals mosfets and their biasing resistors and below that the EN-369FN and DR nabbed from CB Tricks .com
Now for the calculations:
Worked driver and final PA bias examples using a Magnum 257 as our example:
All examples are taken with RV11 and RV13 at their midpoint which is 10k and 250k ohm respectively.
Keep in mind that we would expect to bias the driver mosfet at about 3.1 - 3.3 volts and the finals at 3.7 volts.
Driver biasing example - Please refer to red box 1 on the above schematic.
Transmit line voltage is the usual 8 volts DC.
First we calculate the current:
current total = transmit line voltage / (RV13 midpoint + R233 + [R2 of the EN369DR]) =
So current total = 8 / (250000 + 270000 + 320000) = 8 / 840000 = 0.000009523A or 9.523 uA (Micro amps)
Now we can calculate the bias voltage across the EN-369DR
bias voltage across the EN36DR = current total x [R2 of the EN369DR]
So bias voltage = 0.000009523 x 320000 = 3.04 volts.
3.04 volts is a nice driver bias voltage to start with and we can tweak it from there.
Final biasing example - Please refer to red box 2 on the above schematic..
Please note before we start the transmit line voltage is now 7.4 volts as we drop 0.6 volts across transistor Q5
First we calculate the current:
current total = transmit line voltage / (RV11 midpoint + R239 + [R2 of the EN369FN])
So current total = 7.4 / (10000 + 10000 + 230000) = 8 / 250000 = 0.0000296A or 29.6 uA (Micro amps)
Now we can calculate the voltage across the EN-369FN
bias voltage across the EN369FN = current total x [R2 of the EN369FN]
So our bias voltage = 0.0000296 x 230000 = 6.808 volts
Hang on a moment! That’s not good at all, actually this is very BAD. 6.808 volts is almost double what we should be biasing the mosfet at and will destroy it very quickly!
So what’s wrong here? We know full well that the Magnum was never designed to run at 6.8 volts bias, or else they’d be popping finals left right and centre.
The answer to that question is R2 in the EN369FN in the drawing is incorrect.
If we were to decrease value of R2 in the EN369FN to 20k and then redo the maths.
Final biasing example with R2 in the EN369FN at 20k ohms
First we calculate the current:
current total = transmit line voltage / (RV11 midpoint + R239 + [our new R2 of the EN369FN]) =
So current total = 7.4 / (10000 + 10000 + 20000) = 8 / 40000 = 0.000185A or 185 uA (Micro amps)
Now we can calculate the voltage across the EN-369FN
bias voltage across the EN36FN = current total x [our new R2 of the EN369FN]
So bias voltage = 0.000185 x 20000 = 3.7 volts
3.7 volts = A perfect bias!
As you can clearly see from the above simple maths the EN369FN should not be 230k ohms as per the drawing, but should be 20k ohms.
If you use the following drawing edit it will not only work, but work as intended.
Psi
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