another FM transmitter project

[Willen]

Ok than I will understand Reversed Parallel diode at suppy as a protection diode is bad way. And will connect series diode (forward at ve line) for polarity protection.

It should be obvious that your circuit "A" hasthe diode always reverse-biased so it will never do anything.

Then I found a faulty circuit here- https://solomonsmusic.net/FM_CrystalRadio.html Here used reversed diode as a detector.

I am thinking that reversed diode will rectify negative half cycles of RF and modulates audio from that half negative cycles, but why not? Have you found exact reason why reversed diode will not work on radio reception?

 

[audioguru]

The Techlib circuit has the detector diode always forward biased by the positive output voltage from the opamp. The diode has a very high load resistance so the diode becomes a peak detector of the envelope of the signal.
If the diode is reverse-biased then it needs a very high signal level to overcome the reverse bias voltage.

The Soloman circuit and all other simple AM crystal diode circuits have the diode biased at 0V DC so it can easily rectify the envelope of the radio signal. The polarity of the diode makes no difference.

 

[Willen]

Hi,
I found these transistor on very old transistor radio. Transistors has unusual packaging- BC148A, BC148B, BF195D, BF194B. Are they 'Vintage'? How old they are?

(This photo is downloaded from google. My actual same type of transistors are marked as 'BC148A' at top and 'BEL-9' at bottom)

 

[audioguru]

My very first job was repairing then designing car radios on the Philips production line in about 1968. The circuit used Philips BC148A transistors like you found except yours were made by another manufacturer and might be fakes.
They are detailed in the 1968 Philips Semiconductors Databook that I still have. The same chip is used in modern BC548 transistors that are in a modern TO92 case.

 

[Willen]

Actually why they marked as '148' or '547' or '3904'? Are there any specific reason?

 

[audioguru]

Actually why they marked as '148' or '547' or '3904'? Are there any specific reason?

I do not know why or how the manufacturer selected the part numbers.

I forgot to tell you that the BC148 in the unusual case and pins had a previous BC108 version which had the same chip but it had a TO-18 metal case.
Some datasheets for the BC108 are dated 1980.

 

[Willen]

I forgot a simpler calculation so fall in problem.

I have a 12V liquid (fast cycle) acid battery (of motor bike) almost 9Amp. And I have a 15V AC transformer. I thought If I added rectifier on it, it would be a ideal charger for 12V battery. So I added bridge rectifier with hi value capacitor. I measured the output but it was almost 20V DC. I think 20V DC on 12V will damage the battery. I need almost 15V DC.

-Did I miss few RMS and voltage drop calculation? How it happened?

-How can I design a transformer which produce 15V DC after rectifier? (need 12V AC output for 15V DC?)

I can reduce its output (secondary) turns, so asking.

 

[audioguru]

AC is measured in RMS volts so that 15VAC produces the same amount of heating or same brightness for an incandescent light bulb as 15VDC.
But a full-wave bridge rectifier charges a filter capacitor or battery to the peak voltage (1.414 times the RMS voltage) minus two diode voltage drops (1.3V to 2V). So 15VAC will try to overcharge a battery to 20VDC. 12VAC will be a little too high to charge a 12V lead-acid battery.
A battery needs a current limiter for charging. It does not need a filter capacitor.

 

[Willen]

Wow! Very useful in daily life to me, but I did not think about it. I thought 15AC gives 15V DC

-And if there is a device which needs 12V DC, then I have to use 10V AC transformer with bridge rectifier, isn't it? (10x1.414)-2V= around 12.1V DC.

-Did you mean if output of transformer is marked as '15V AC' and multimeter also shows- 15V AC, this measured voltage is RMS?

- Then what happens if I used half-wave rectifier (two diodes), or only single diode?

 

[audioguru]

A full-wave rectifier produces ripple at 100Hz (when the mains is 50Hz) and has two diodes in series producing a voltage drop.
A half-wave rectifier produces ripple at 50Hz (which needs a filter capacitor double the capacitance) and it has a single diode in series producing a voltage drop.
Both types of rectifiers use the "1.414 times" (the peak voltage, not the RMS voltage) in the calculation for output voltage.

 

[ericgibbs]

-And if there is a device which needs 12V DC, then I have to use 10V AC transformer with bridge rectifier, isn't it? (10x1.414)-2V= around 12.1V DC.

Hi Willen,

This Tool will give you a close approximation for a transformer/rectifier/smoothing cap/regulator PSU.

https://www.electro-tech-online.com/tools/PSUCapV1.php

E

 

[Willen]

OK, and about current limiter:
I have a transformer from old Black and White TV which is 18V and few amps (?). I have 12V and almost 9 amp battery.

Should I have to use current limiter if transformer has only few amp (3 or 4?) output?
What is the average charging current for the battery? (so I have to limit?)

 

[audioguru]

Nobody makes a "9A" battery. Batteries are rated in Amp-Hours so maybe it is a 9Ah battery. Then it can supply 450mA for 20 hours.
An 18V transformer produces rectified 23.5V which is way too high. The transformer and battery might catch on fire from the extremely high unlimited charging current. You also need to limit the voltage.

 

[Willen]

Are there any suitable current limiter? (lets say I have 14.5V DC source to charge 12V battery.) I didn't search on google because I know nothing about current limiter for this purpose.

Simply can I say- 9Ah= can supply 9 Amp current till one hour?

 

[audioguru]

You cannot charge a dead lead-acid battery in 1 hour. Its datasheet will list its maximum allowed charging current and it might take a few hours.
Modern Ni-MH and Li-Po batteries can be charged in 10 or 15 minutes.

You need a charging circuit that limits the current and voltage then detects a full charge then turns off.

 

[audioguru]

Quote from Willen: "Digital multimeter shows just hFE only. It shows 195 - 210 for 2N3904. How can I explain? I mean at how much 'V' and how much 'Ic'?"
The datasheet for the 2M3904 shows that its hFE is somewhere from 100 to 300 when its collector current is 10mA and its collector to emitter voltage is 1V or more.
We do not know the current used in your multimeter (it is probably less than 10mA) and we do not know its voltage (which is probably high enough).

 

[Willen]

Dear AG, I know 100pf cap will eliminate AC from audio line (you added base to emitter on mod 4). Can I get same effect if I added it with Base to Ground?

 

[audioguru]

Dear AG, I know 100pf cap will eliminate AC from audio line (you added base to emitter on mod 4). Can I get same effect if I added it with Base to Ground?

The 100pF capacitor prevents the preamp transistor from amplifying (and clipping) the very strong 100MHz radio signal.
It works perfectly from the base to the emitter.
I have not tried it from base to ground, you try it.

 

[Nepaliman]

Hi again dear audioguru,
(still playing with fundamental of electronics)

I found a post here where you were calculating Vb, Ib and bias resistor with Vc, Ic and hFE. That was so clear to me. But I saw a circuit where used a base resistor is extremely high 10 Meg. It is a crystal radio amplifier.

(also i found a VLF receiver with a single jEET which has 44 Meg resistor for base bias and this base was antenna too. And collector of the device was connected to Mic in. I can attach it if you want.)

You were calculating with 1 Meg base resistor with almost 110 to 800 hFE. Here this transistor has almost same hFE around 200-300 BUT what is the purpose to use such large value resistor here (like 10 Meg here or 44 Meg on jFET)? Should we have to use larger R if base input has smaller signal?

Are there any interesting Vb, Ib, hFE, Ic, Vc etc. calculation related with the circuit? Thank you.

 

[audioguru]

Nepaliman, the high resistor values used in the schematic you posted are used to keep the power consumption VERY low.
I calculated its voltages and currents when the hFE of the transistor is only 100 but you should re-calculate with the hFE at the minimum of 20 at a low current or with the hFE of typically 200 at a low current shown on the graph. The hFE might actually be 300 which will cause the transistor to be almost saturated and therefore work poorly as an amplifier.