Video Polarity and Direct Injection

[vts1134]

Can anyone verify that the circuit I posted should work? I can't get anything out of it.

[old_coot88]

Originally Posted by vts1134

Can anyone verify that the circuit I posted should work? I can't get anything out of it.

You mean you've built it already? If it's not working, the transistor may not be biased correctly. First measure the emitter/base voltage, which should be about 0.7 V. If the voltage is too low or missing, you need to increase value of the base-to-ground resistor.

Or, if the emitter/base voltage is too high, the transistor will overconduct and saturate. So lower the voltage by increasing value of the base-to-12V resistor.

This is all assuming the transistor is NPN silicon type and supply is +12V.

BTW, the circuit shown will deliver only half the output level that it would if the non-inverting feature (which you don't need) were eliminated. The mod would be easy to do if the output level turns out to be inadequate. Here's a neat little tutorial on transistor biasing..
http://www.electronics-tutorials.ws/...r-biasing.html

[vts1134]

Originally Posted by old_coot88

...the circuit shown will deliver only half the output level that it would if the non-inverting feature (which you don't need) were eliminated. The mod would be easy to do if the output level turns out to be inadequate...

How would I accomplish that?

[vts1134]

By the way, I'm using a C1815. I got the emitter/base voltage to .7VDC. I have output now, but just a minuscule amount.

[old_tv_nut]

Originally Posted by vts1134

Can anyone verify that the circuit I posted should work? I can't get anything out of it.

The problem is that the base is biased at about 1/2 of B+, which makes the emitter that (6 v) minus 0.7 volts = 5.3 v, and the collector drops to 12-5.3 = 6.7 v, so the transistor is almost saturated at one end of the circuit's output range. You need to reduce the lower base resistor or increase the top one to get about 3.7 volts at the base. Then the emitter will be at 3 volts and be able to swing almost from ground to 6 volts, while the collector will be at 9 volts and be able to swing almost from 6 to 12 volts.

You could try leaving the lower base resistor as 4.7k and increasing the upper one to 15k or 18k.

[kf4rca]

say 100K or so and adjust it till I get what I want with a scope on the output. Then measure the pot and substitute fixed resistors.

[old_coot88]

Since the non-inverting output feature is not needed, it could be eliminated. This would allow the output level to take advantage of the full supply voltage (rather than half as it is).

It's always useful to think of a transistor as a triode analog - with the emitter the cathode, the base the grid, and the collector the plate.

In the sketch shown, start out with R2 as 4.7K, and R3 as 1K. Then for R1, first try 100K and work down in value till you get 0.7 V on the base. Once a usable output is obtained, then play with values for R2 and R3 and see if the output level can be improved any, all the while tweaking R1 to maintain 0.7V on the base.

Edit.. Error in sketch, 1K resistor "R2" should be R3. Dumb, dumb.

[old_coot88]

As a follow-on to the triode analog thingy, the analogy breaks down a bit since the base is biased positive to the emitter to set the operating point, while in a vacuum triode, the grid is biased negative to the cathode.

[old_tv_nut]

Originally Posted by old_coot88

Since the non-inverting output feature is not needed, it could be eliminated. This would allow the output level to take advantage of the full supply voltage (rather than half as it is).

It's always useful to think of a transistor as a triode analog - with the emitter the cathode, the base the grid, and the collector the plate.

In the sketch shown, start out with R2 as 4.7K, and R3 as 1K. Then for R1, first try 100K and work down in value till you get 0.7 V on the base. Once a usable output is obtained, then play with values for R2 and R3 and see if the output level can be improved any, all the while tweaking R1 to maintain 0.7V on the base.

Edit.. Error in sketch, 1K resistor "R2" should be R3. Dumb, dumb.

This circuit will not work as a decent video amplifier because there is no emitter resistor. The gain will be very large and very dependent on bias - in other words, without an emitter resistor, you have a switching circuit rather than a linear amplifier. EDIT: this has sometimes been used as a sync separator circuit - a very non-linear switch that changes state on sync pulses.

If you want to have both inverted and non-inverted outputs (although with gain = 1), go back to the circuit with equal emitter and collector resistors and make the upper base resistor about 15k or 18k as I suggested. If you want non-inverting with gain, use the feedback circuit suggested by kf4rca. If you need inverting with gain, you can used the single transistor circuit, but the emitter resistor needs to be smaller and the base bias voltage needs to be reduced so the collector voltage is at the middle of its possible range.

Second edit: I forgot to add that the gain for the inverted signal on the collector is close to the ratio of the collector resistor divided by the emitter resistor. The gain of the non-inverted signal on the emitter is always (very) slightly less than 1.

[vts1134]

Originally Posted by old_coot88

Since the non-inverting output feature is not needed, it could be eliminated. This would allow the output level to take advantage of the full supply voltage (rather than half as it is).

It's always useful to think of a transistor as a triode analog - with the emitter the cathode, the base the grid, and the collector the plate.

In the sketch shown, start out with R2 as 4.7K, and R3 as 1K. Then for R1, first try 100K and work down in value till you get 0.7 V on the base. Once a usable output is obtained, then play with values for R2 and R3 and see if the output level can be improved any, all the while tweaking R1 to maintain 0.7V on the base.

Edit.. Error in sketch, 1K resistor "R2" should be R3. Dumb, dumb.

What is the 75ohm "dotted" connection to ground at the input?

[Electronic M]

Originally Posted by old_coot88

It's always useful to think of a transistor as a triode analog - with the emitter the cathode, the base the grid, and the collector the plate.

Thinking of a transistor as a tube* is a really conceptually poor design approach. A BJT is completely different in behavior/operation from a tube. A tube is a voltage amplifier and a transistor is a current amplifier. Assuming DC biases are correct and can be ignored and AC analysis of a transistor looks like a relatively low resistance resistor that measures input current and controls a current source based on the current through the resistor. A tube looks like a near infinite resistance the voltage across which controls a variable voltage source.

*Except for FET transistors which are functionally and behaviorally identical to triode tubes.

[vts1134]

Originally Posted by old_coot88

Since the non-inverting output feature is not needed, it could be eliminated. This would allow the output level to take advantage of the full supply voltage (rather than half as it is).

It's always useful to think of a transistor as a triode analog - with the emitter the cathode, the base the grid, and the collector the plate.

In the sketch shown, start out with R2 as 4.7K, and R3 as 1K. Then for R1, first try 100K and work down in value till you get 0.7 V on the base. Once a usable output is obtained, then play with values for R2 and R3 and see if the output level can be improved any, all the while tweaking R1 to maintain 0.7V on the base.

Edit.. Error in sketch, 1K resistor "R2" should be R3. Dumb, dumb.

Progress! I've got output with the correct polarity using that circuit. I'll have to do some tweaking, but it's a victory for now. One question, is there a way to adjust gain in that circuit?

[old_tv_nut]

Originally Posted by vts1134

Progress! I've got output with the correct polarity using that circuit. I'll have to do some tweaking, but it's a victory for now. One question, is there a way to adjust gain in that circuit?

1) Please tell me you are not using the circuit with the emitter tied directly to ground instead of through a resistor.
2)Are you using the collector output (which is inverted)? In that case, the gain is equal to the ratio of the collector resistor divided by the emitter resistor. Without the emitter resistor, the gain is very non-linear and depends on the amount of current in the transistor, which varies the effective emitter impedance. So, to get the gain you want, divide the collector resistor by the gain needed and use that value for the emitter resistor.
3) a fine point to add: if the following circuit (connected to the collector) has a high impedance (like a tube grid), all is well, but if it has a lower impedance, closer to the collector resistor, it will reduce the gain because it reduces the total effective resistance at the collector.

[vts1134]

Originally Posted by old_tv_nut

1) Please tell me you are not using the circuit with the emitter tied directly to ground instead of through a resistor.
2)Are you using the collector output (which is inverted)? In that case, the gain is equal to the ratio of the collector resistor divided by the emitter resistor. Without the emitter resistor, the gain is very non-linear and depends on the amount of current in the transistor, which varies the effective emitter impedance. So, to get the gain you want, divide the collector resistor by the gain needed and use that value for the emitter resistor.
3) a fine point to add: if the following circuit (connected to the collector) has a high impedance (like a tube grid), all is well, but if it has a lower impedance, closer to the collector resistor, it will reduce the gain because it reduces the total effective resistance at the collector.

OK, I remade the inverter following this schematic, with the exception of the 2.2 K resistor which is a potentiometer/parallel resistor of the same overall value. The output is greatly improved. One problem I am having is what looks to be vertical interference as you can see from this video. The vertical bars look like a horizontal problem in the television, but it's in the video itself. When I adjust the potentiometer you can see at some point the vertical bars change direction. Thoughts?

https://www.youtube.com/watch?v=G7Xc...&feature=share

[old_coot88]

Originally Posted by vts1134

Progress! I've got output with the correct polarity using that circuit. I'll have to do some tweaking, but it's a victory for now. One question, is there a way to adjust gain in that circuit?

Well to tell you the truth, I built a number of projects back in the day (late 50s to early 60s) using the common emitter configuration with no resistor in the emitter-to-ground leg. They worked fine with no nonlinearity problems etc. But these were all done with germanium transistors. Apparently that does not transfer well to later practice with silicon transistors. So I was 'waay behind the curve and gave crappy advice and should be duly reprimanded.

So by all means sub some resistances in the emitter leg per the more experienced advice given.

(In the early days of transistors the "triode" analogy was indeed used as a visualization aid for newcomers. For the purpose at hand, it disregarded the input-impedance disparity. Then when FETs finally came along, the input impedance truly mimicked a vacuum triode.)