# A qestion about class C ampl. design?

Discussion in 'General Electronics Chat' started by samy555, Aug 28, 2015.

1. ### samy555 Thread Starter Active Member

May 24, 2010
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Hi

I have read many of the pages on the Web about class C amplifier, and I have benefited and learned a lot, but unfortunately non of these sites spoke about how to choose the values of C1 and R1 (look at the circuit below).

I think that C1 & R1 form a HPF, if the input signal is of 2V peak and 10MHz frequency, I think that if I choose the value of C1 such that it has very low reactance at 10MHz then using the high pass RC filter law to calculate R1 when cutoff freq = 10MHz or lower.

Thanks

2. ### nsaspook AAC Fanatic!

Aug 27, 2009
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3. ### Colin55 Member

Aug 27, 2015
174
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Remove the resistor.
The transistor is turned on when the input signal is above 0.7v. During this time the capacitor charges a small amount and when the signal reduces or "goes negative" the base will go negative too up to a value of about -5v to -7v and put NO effective resistance on the capacitor to discharge it. Thus the capacitor will charge a small amount.
In the following cycles the capacitor gets charged more and more and eventually it will not deliver the signal to the transistor.
Fit the resistor.
Each time the signal goes low or negative, the resistor discharges the capacitor.
The value of the resistor is selected so it discharges the capacitor fully on each cycle so it does not gradually get charged.

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4. ### #12 Expert

Nov 30, 2010
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You can attach an o-scope to the base of the transistor and inject lower and lower frequency while watching DC build up on C1 until the transistor is locked, "on". The emitter to base path is equal to a rectifier. Failing to dump the charge on C1 is a guaranteed "fail".

This is true in vacuum tubes, too. No matter what resistor you use, you can find a frequency that will load a charge into the input coupling capacitor. That's why the answer isn't obvious. As usual, "It depends".

5. ### nsaspook AAC Fanatic!

Aug 27, 2009
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Don't forget this is a class C amplifier so we want a level of charge left on the capacitor after the negative swing to set the conduction angle. (transistor on time during the positive swing of the signal)

6. ### Colin55 Member

Aug 27, 2015
174
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Don't forget this is a class C amplifier so we want a level of charge left on the capacitor

You cannot adjust the value of resistance to get a "partial charge" on the capacitor.
It will get either fully charged or not charged AT ALL.

Nov 30, 2010
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May 24, 2010
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9. ### samy555 Thread Starter Active Member

May 24, 2010
116
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There is no R1 in the circuit above and C1 transmits the signal from mic to the base witout any problems regard to charge or discharge!
thank you

10. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I did not understand
Thank you

11. ### Colin55 Member

Aug 27, 2015
174
18
The transistor in the circuit above is CLASS A - commonly called SELF BIAS.

Class C is not turned on AT ALL by the transistor stage. IT is turned ON via the signal entering the stage.

12. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Yes, I'm sorry, you are right

I read in a book (Electronic Devices and Circuit Theory 7th Edition) page 83:

Now, if the operating frequency of my class C amp. = 10MHz, then T = 1/f = 0.1 usec and 0.5T = 0.05 usec

5RC (5 * time constants) should be much greater than 0.05 usec that is 5*RC = at least 10 * 0.05 usec

RC >= (10 * 0.05 usec)/5 = 0.1 usec

In general: RC >= 10*(1/2f)/5

RC>= 1/f

Design Steps:

Choose C1 so that Xc1=0.1 ohm at the operating freq

C1=160 nF

R1 >= 1/(C1 * f)

R1 >= 0.625 ohm nooooooooooooooooooooo,,, there is something wrong

13. ### Colin55 Member

Aug 27, 2015
174
18
You are talking about a clamping network.
One minute you are confused about class A and class C and now you are talking about a clamping network, just because it looks like the resistor and capacitor on the front end of the class C amplifier.

I already explained the purpose of the resistor.
Class C tries to get as close as possible to 180 degrees to get the max energy transfer.
Class C has lots of problems and these could cover a whole new topic.
You don't specify what you want to do, but at 10MHz it sounds like you want to amplify an FM signal.
Class C takes an enormous amount to drive it and you need to have a reason why you select this class.

14. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I did notchoosethis class, the designer of the following (toy car) TX circuit ischosen it.

thank you Colin55

15. ### Colin55 Member

Aug 27, 2015
174
18
This is class C because the base is not being turned on from a voltage-source (that is via a resistor from the supply rail). The transistor is turned ON from the energy coming via the 47p and some will be lost in the 33k but since the 33k is very high you may get some voltage appearing on the base and you may get some voltage being developed across the 100R.

16. ### nsaspook AAC Fanatic!

Aug 27, 2009
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Class C usually has conduction angle significantly less than 180 degrees, typical might be 120 with reverse bias of the base-emitter junction with a grounded choke. Exactly where is not exactly critical in his circuit but it's a trade-off between efficiency and output power.
The reason it's hard to find information about what value of RC to use it that it's a cheap hack for a BJT that's normally used at low power levels (the high negative input voltage from high drive levels at the same time as high peak collector voltage in a resonant output circuit makes it eat transistors). A proper bias method for stable operation at significant power would use a choke or transformer to ground and maybe a bias supply feedback system to set the class C conduction angle if efficiency over a range of power settings is a design consideration.

17. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Why we want a level of charge left on the capacitor???