Pre-Amplifier Biasing... Missing voltages at base and additional on output

Thread Starter

Kain6622

Joined Apr 10, 2021
8
Hello everyone,

I come to you today with an interesting problem (at least I think it is myself), I'm designing a pre-amplifier and gone through all the necessary steps, I get the desired current on the collector and most the voltages are correct but one thing is bothering me and that is that I have 400mV missing from the base bias and have an additional 1v on my output (note in my setup at present I do not have capacitors as they seem to affect the DC bias), but not matter where i take my multimeter I cannot figure where the missing voltage has gone too nor the additional on the output, I hope you might be able to help me here, here is my setup:

1627089755632.png
I am using a desktop power supply (measuring 9v on both display and multimeter), my transistors are both 2n3904, Q1 with gain of 384.6, Q2 344.8, Diode drops from both BE is roughtly 654mV. When I measure the voltage drops across the power rail I get 9v, Across R1 = 6.59v, R2 = 2.01v... if I am correct these should add up to 9v using KVL but as we can see, 400mV is missing, I first thought it might be partially carried across the base-emitter of the transistor but I have no idea where this could be hiding? Also when checking the voltages on the other side I get 3.02v across Rc and 1.007v across Re and (drum role)... 5.97v from the output from collector to ground, giving me ~10 total giving me an extra 600mV (if I am to consider 400mV to be from the other side... but I don't think this is true, so 1v extra), When I run this through Microcap, I get different values to on the actual breadboard, resistor values have been chosen to match the above values (some additions in series and measure at the values needed).

Can anyone advise what could be wrong here? or if I am missing something/ any addition debugging tests I might be missing?

The circuit works but I don't like the fact of unaccountable voltages missing/present. I also measured the current coming from the output to ground and it measured 2.98mA, if that has anything to do with it, both RC to collector and RE to ground measure 1mA as expected, so the additional 1.98mA on the output is confusing me also.

Kind Regards
David
 

Audioguru again

Joined Oct 21, 2019
3,516
You have very high value base bias resistors then the input resistance of your multimeter (10M?) is reducing the voltage across the bias resistors. What is your signal source that needs a preamp with such a high input impedance?

You never need to measure output current to ground because the current meter shorts the output to ground. It is simply 9V/Rc= 9V/3k= 3.0mA.

There is no "extra" 1V. 9V supply to collector= 3V. Collector to ground= 6V. Total is 9V.
The voltage gain of the amplifier is less than 3 times so it barely amplifies.
 

Thread Starter

Kain6622

Joined Apr 10, 2021
8
Good Morning, I'll checked the input resistance of my multi-meter and does compute to roughly 10M, so if I'm thinking correctly that creates a second voltage divider circuit to ground working out at 9v*617000/10617000 = 0.523? (which is close to the missing voltage, given that I don't know the input resistance of my multimeter to 100% accuracy).

The preamp is for a guitar + cable, hence the requirement for high input impedance (was aiming for >500k but just short of that i believe... zin = R1 || R2 || B(re + RE) where B = Q1HFe * Q2HFe + (Q1HFe + Q2HFe), which with the above values equals about 470K.

Thanks for clearing the other side up of the 1v, a sillymistake on my part but that makes sense.

Yes the gain at present is only 3 as I wanted to get the DC biasing correct first before adding bypass capacitors to the circuit, I test my peak of my guitar output at about 700mV (when played hard) so will tune it to a gain of ~11x for a clean amplification but still need to do the math for that to recalculate the value of RE once I've places the feedback resistor in series with the emitter capacitor, from what I've found from experimenting with the class A config is that the maximum swing I can get from this is Vcc - Ve = 9v - 1v = 8v (with no distortion) hence the 11x gain calculated.
 

Thread Starter

Kain6622

Joined Apr 10, 2021
8
Thanks for the reply, I've not have much time to experiment with JFETS or depletion mode mosfets yet but might look into them and how to bias them (I Like the math part more than I probably should), I was aware of the 1Meg+ rule of thumb with regards to guitars, but was working with what I had on hand (found today 4.7M resistors at 1/4W, I usuallly use 1W resistors).

At present I only have BJT's and a few mosfets (enhancement mode), I'm currently wandering if it would be worth me using the above circuit in my original post with higher Resistances like the 4.7M with a 1.62M giving just over 1.2M Zin or to using a Common collector voltage follower as a buffer input stage with roughly a Zin of 2.3M before the preamp stage, Obviously high impedance is better but I'm still working to figure out the effects of using multistages with a single supply are, from what I've seen up to now it seem to lower the maximum swing on the preamp section when we place a Buffer in, but would need to do further experiments to be sure of what is happening.

The schematic you gave will serve as a good reference information when designing, I will look up some tutorials on jfets so I can enjoy the math myself :). Thank you for clearing up some of my confusion.
 
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