Build Your Own Clone Message Board

It is currently Fri Mar 29, 2024 12:22 am

All times are UTC - 6 hours




Post new topic Reply to topic  [ 4 posts ] 
Author Message
PostPosted: Sat Oct 28, 2017 12:15 pm 
Offline

Joined: Sat Oct 28, 2017 12:06 pm
Posts: 2
For practice I just finished breadboarding the confidence boost kit and it's working but the gain is very high. The pot at 50% is no where near unity. I ran through the trouble shooting steps and everything checked out except for "The emitter of the transistor (E pad) should have about 1 volt or so. (my build has 0.98 volts)". I'm reading only 0.15v on the emitter. Not sure if that's related. Anyway, it works and it was fun but I wanted to check if there was something simple I screwed up that would lead to the low voltage on the emitter and the extra gain. If I take the signal from right after the buffer it's perfect.

https://www.dropbox.com/s/zzu6cnabz4rlcsu/IMG_6998%202.jpg?dl=0


Last edited by jdriscoll on Sat Oct 28, 2017 5:16 pm, edited 2 times in total.

Top
 Profile  
 
PostPosted: Sat Oct 28, 2017 3:53 pm 
Offline

Joined: Tue May 24, 2016 11:16 am
Posts: 104
The resistors connected to the base and the emitter of the transistor are mostly what affect the emitter DC voltage. The two resistors connected to the base should try to make the base voltage right around 1/11 of the battery voltage. Then current flowing between the emitter and collector should cause a voltage drop across the emitter resistor.

The exact voltages will depend on the current gain of the particular transistor. If very high gain, the emitter voltage should rise to limit the base current, in which case the two resistors connected to the base will have the most influence on the voltages. If lower gain, the base and emitter voltages will be less than if the transistor were higher gain.

The gain of the transistor stage, if there is enough base current available, should about be equal to the the ratio of collector resistor divided by the emitter resistor. That is 10,000/360 = 1000/36 which would be a gain of around 28. If the gain is a lot higher, probably wise to disconnect the resistors and measure their resistance to make sure they are the correct values.


Top
 Profile  
 
PostPosted: Sat Oct 28, 2017 5:28 pm 
Offline

Joined: Sat Oct 28, 2017 12:06 pm
Posts: 2
Thank you for the info. I checked all the resistors and they were within tolerances. The voltage at the base is just a smidge below 1/11 or 9v and matches the trouble shooting instructions. The only discrepancy from the troubleshooting docs is I'm showing 1/10th the voltage on the emitter. Is it possible that's a typo?

I'm fairly new to this. How should I measure the gain. Is 28 a ratio between the voltage at the input and the output? Thanks again for the help. Breadboarding something simple like this has been a good way to learn and experiment.


Top
 Profile  
 
PostPosted: Sat Oct 28, 2017 5:47 pm 
Offline

Joined: Tue May 24, 2016 11:16 am
Posts: 104
With an NPN silicon transistor like you have there, the emitter voltage should be about 0.7 volts less than the base voltage. Maybe slightly less than that.

Knowing the emitter voltage and the emitter resistor, you can calculate the emitter current using Ohm's Law. I=E/R which says "I" the current equals the voltage "E" divided by the resistance "R"

The current in the collector is slightly less than the emitter current, but close enough to assume they are about equal. Using Ohm's Law again, "I" for the collector will be about the same as the "I" for the emitter, so we will use the same value of "I", and use the value of the collector resistor, our new "R", to find "E" the voltage across the collector resistor. Doing a little algebra Ohm's Law becomes E=IR.

Once we know E for the collector resistor, we can subtract that voltage from the voltage of our 9 volt battery and that should give the expected collector voltage.

If everything checks out okay, then the circuit is probably working properly.

All the the measurements, of course, are carried out in the absence of any guitar input. We just want to find the DC bias voltages in the absence of any signal.


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC - 6 hours


Who is online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group