Embedded Files
The LTPI is used in many tube guitar and bass amps. The circuit topology can produce attractive features such as high gain, high headroom, good balance, and ease of accommodating negative feedback (NFB). Despite its widespread use, there are many misunderstandings about its operation, imperfections, and variations. Firstly, this article is about understanding the circuit. This is not a cookbook recipe for modifications or miracles. Once the circuit is understood, changes or adjustments can be made based on reasoning, rather than guesswork or hearsay. Secondly, the circuit is well know and any changes to an existing amp LTPI should be carefully considered before implementation. Hopefully this article can help with that. A simple LTPI circuit (without NFB) is shown below.
The LTPI is used in many tube guitar and bass amps. The circuit topology can produce attractive features such as high gain, high headroom, good balance, and ease of accommodating negative feedback (NFB). Despite its widespread use, there are many misunderstandings about its operation, imperfections, and variations. Firstly, this article is about understanding the circuit. This is not a cookbook recipe for modifications or miracles. Once the circuit is understood, changes or adjustments can be made based on reasoning, rather than guesswork or hearsay. Secondly, the circuit is well know and any changes to an existing amp LTPI should be carefully considered before implementation. Hopefully this article can help with that. A simple LTPI circuit (without NFB) is shown below.
In this circuit there is no NFB and the input signal (A) drives the V1 grid while the V2 grid is held at signal ground by C2. The two cathodes are connected to a shared bias resistor, R5, and a tail resistor, R6. The tail resistor elevates the voltage at the cathodes and grids since all the tube currents pass through it. Typically, the elevation is around 50-60 volts. The tail resistor's purpose is to act as a current source and there will be more discussion about that later. (Similar circuits are used in many amps, particularly the JMP 18 watt Marshall variants.) First let's consider the circuit without any applied signal. With B+ applied (around 325V), the symmetry of the circuit leads both tubes to similar conditions. They both pass current through their plate resistors and the shared bias and tail resistors. The currents will settle to a value that creates a bias voltage across R5 that produces a stable current. Assuming identical tubes (which is OK for our purpose), the tube currents will be identical. For the circuit shown, each 12at7 will pass about 1.5ma so the R5 and R6 current will be about 3ma. The grid voltages will be slightly lower than the cathode voltage ( by about 1.5 volts) due to the voltage drop across the bias resistor R5. Both the grids and cathodes are elevated by the voltage created by the tail resistor. With 3ma passing through 22K, 66 volts are created to elevate the grids and cathodes. The elevation has little effect on the discussion here. (It does reduce the headroom of the PI circuit, but that is not an issue for this discussion.) Both grids have DC blocking caps. C1 connects to a ground-referenced signal source (A), while C2 is directly connected to ground. We'll assume the caps pass all AC signals, but block DC, so the stable bias conditions described above are not affected by the cap connections.
In this circuit there is no NFB and the input signal (A) drives the V1 grid while the V2 grid is held at signal ground by C2. The two cathodes are connected to a shared bias resistor, R5, and a tail resistor, R6. The tail resistor elevates the voltage at the cathodes and grids since all the tube currents pass through it. Typically, the elevation is around 50-60 volts. The tail resistor's purpose is to act as a current source and there will be more discussion about that later. (Similar circuits are used in many amps, particularly the JMP 18 watt Marshall variants.) First let's consider the circuit without any applied signal. With B+ applied (around 325V), the symmetry of the circuit leads both tubes to similar conditions. They both pass current through their plate resistors and the shared bias and tail resistors. The currents will settle to a value that creates a bias voltage across R5 that produces a stable current. Assuming identical tubes (which is OK for our purpose), the tube currents will be identical. For the circuit shown, each 12at7 will pass about 1.5ma so the R5 and R6 current will be about 3ma. The grid voltages will be slightly lower than the cathode voltage ( by about 1.5 volts) due to the voltage drop across the bias resistor R5. Both the grids and cathodes are elevated by the voltage created by the tail resistor. With 3ma passing through 22K, 66 volts are created to elevate the grids and cathodes. The elevation has little effect on the discussion here. (It does reduce the headroom of the PI circuit, but that is not an issue for this discussion.) Both grids have DC blocking caps. C1 connects to a ground-referenced signal source (A), while C2 is directly connected to ground. We'll assume the caps pass all AC signals, but block DC, so the stable bias conditions described above are not affected by the cap connections.
Now, let's apply an input sine wave signal (a) of 1V peak to C1. As the signal swings positive, the V1 grid will rise by 1 volt at the signal crest. Let's freeze time at that point and examine what's happened in the rest of the circuit. The cathodes will try to follow the V1 grid (like a cathode follower circuit). Increasing the cathode voltage draws more current through the bias and tail resistors. Also, as the cathode voltage increases, the V2 grid remains fixed at its bias voltage. It's held fixed by C2, so the V2 grid-to-cathode voltage decreases, which decreases the current through V2.
Now, let's apply an input sine wave signal (a) of 1V peak to C1. As the signal swings positive, the V1 grid will rise by 1 volt at the signal crest. Let's freeze time at that point and examine what's happened in the rest of the circuit. The cathodes will try to follow the V1 grid (like a cathode follower circuit). Increasing the cathode voltage draws more current through the bias and tail resistors. Also, as the cathode voltage increases, the V2 grid remains fixed at its bias voltage. It's held fixed by C2, so the V2 grid-to-cathode voltage decreases, which decreases the current through V2.
So, let's walk through that again. V1 has its grid voltage increased by a volt. In response to that, the current in V1 increases and the cathode voltage increases as some of that additional current passes through R5 and R6. At the same time, the V2 grid voltage stays unchanged, and the current in V2 decreases.
So, let's walk through that again. V1 has its grid voltage increased by a volt. In response to that, the current in V1 increases and the cathode voltage increases as some of that additional current passes through R5 and R6. At the same time, the V2 grid voltage stays unchanged, and the current in V2 decreases.
This is a key point. Additional current through V1 can only come from two sources. Additional current can pass through R5 and R6 (raising the cathode voltage), or current in V2 is reduced and shifted to V1. Actually, both of these are happening at the same time. The cathode voltage goes up as additional current flows through R5 and R6, and V2 current decreases as it shifts to V1.
This is a key point. Additional current through V1 can only come from two sources. Additional current can pass through R5 and R6 (raising the cathode voltage), or current in V2 is reduced and shifted to V1. Actually, both of these are happening at the same time. The cathode voltage goes up as additional current flows through R5 and R6, and V2 current decreases as it shifts to V1.
A reasonably accurate approximation is that the cathode voltage increases by about 1/2 the input voltage. So if the signal (A) increases the V1 grid voltage by 1 volt, the cathode voltage (C) increases by 1/2 volt. That means that the V1 grid-to-cathode voltage (Vgk1) increases by 1/2 volt and the V2 grid-to-cathode voltage (Vgk2) decreases by 1/2 volt. Since tube current changes are roughly proportional to grid-cathode voltage changes (for small signals), the changes in both tube currents are equal, but opposite. V1 current will increase by some amount and the V2 current will decrease by roughly the same amount. The result is a shift of current from V2 to V1. However, there is a slight wrinkle to these approximations. Recall that the cathode voltage has increased by 1/2 of the input voltage (or 1/2 volt), so there is a slight increase in the bias and tail resistor current since the cathode voltage increased. This additional current adds to the shift of current from V2 to V1, making the V1 current increase slightly larger than the V2 current decrease.
A reasonably accurate approximation is that the cathode voltage increases by about 1/2 the input voltage. So if the signal (A) increases the V1 grid voltage by 1 volt, the cathode voltage (C) increases by 1/2 volt. That means that the V1 grid-to-cathode voltage (Vgk1) increases by 1/2 volt and the V2 grid-to-cathode voltage (Vgk2) decreases by 1/2 volt. Since tube current changes are roughly proportional to grid-cathode voltage changes (for small signals), the changes in both tube currents are equal, but opposite. V1 current will increase by some amount and the V2 current will decrease by roughly the same amount. The result is a shift of current from V2 to V1. However, there is a slight wrinkle to these approximations. Recall that the cathode voltage has increased by 1/2 of the input voltage (or 1/2 volt), so there is a slight increase in the bias and tail resistor current since the cathode voltage increased. This additional current adds to the shift of current from V2 to V1, making the V1 current increase slightly larger than the V2 current decrease.
This is a key point. The increased R5 and R6 current adds to the current shifting from V2 to V1. This additional bias and tail resistor current causes imbalance in the tube current changes. V1 has a greater current increase than the decrease in V2. That translates into a greater voltage swing on the plate resistor (R1) of V1 than on the plate resistor (R2) of V2. This creates an imbalance in the output signal amplitudes. The V1 output signal is greater than that of V2, and we say the phase inverter is "unbalanced".
This is a key point. The increased R5 and R6 current adds to the current shifting from V2 to V1. This additional bias and tail resistor current causes imbalance in the tube current changes. V1 has a greater current increase than the decrease in V2. That translates into a greater voltage swing on the plate resistor (R1) of V1 than on the plate resistor (R2) of V2. This creates an imbalance in the output signal amplitudes. The V1 output signal is greater than that of V2, and we say the phase inverter is "unbalanced".
Taking a small tangent at this point, let's clarify the use of the term "balanced". In this context, the PI circuit is said to be "unbalanced" only because of the fact that the current changes in the tubes are unequal and therefore the output signals at the plates are also unequal. The PI circuit balance is quite different than the usage of "balance" as it refers to tubes. Balanced tubes refer to matching of the intrinsic tube characteristics, not the circuit they are used in. PI circuit balance refers to a circuit behavior. Of course, these are related in that the above PI circuit behavior is analyzed with the assumption that the V1 and V2 tubes are the same or "balanced". So the tube balance will impact circuit behavior somewhat, but tube balance will not create a balanced PI circuit because of the reasons stated above. In reality, the PI circuit behavior discussed here is only slightly affected by tube balance. It can be considered a minor issue, as one would consider any component tolerances.
Taking a small tangent at this point, let's clarify the use of the term "balanced". In this context, the PI circuit is said to be "unbalanced" only because of the fact that the current changes in the tubes are unequal and therefore the output signals at the plates are also unequal. The PI circuit balance is quite different than the usage of "balance" as it refers to tubes. Balanced tubes refer to matching of the intrinsic tube characteristics, not the circuit they are used in. PI circuit balance refers to a circuit behavior. Of course, these are related in that the above PI circuit behavior is analyzed with the assumption that the V1 and V2 tubes are the same or "balanced". So the tube balance will impact circuit behavior somewhat, but tube balance will not create a balanced PI circuit because of the reasons stated above. In reality, the PI circuit behavior discussed here is only slightly affected by tube balance. It can be considered a minor issue, as one would consider any component tolerances.
A common solution to PI circuit imbalance is to simply make the V2 load resistor greater than the V1 load resistor, leading to the common 82k value for R1 and the 100k value for R2. The imbalance is caused by the change in bias and tail resistor current, so lower resistance tail resistors (R6) usually lead to greater imbalance. For any reasonable tail resistor value and its contribution to imbalance, there are load resistor values that will cause the outputs to have the same amplitude. The selection of different load resistors is the common way to "balance" the outputs of the PI. Note that balancing the PI output by varying the load resistors does not change the fact that the tube currents are still unbalanced, it only compensates for the current imbalance.
A common solution to PI circuit imbalance is to simply make the V2 load resistor greater than the V1 load resistor, leading to the common 82k value for R1 and the 100k value for R2. The imbalance is caused by the change in bias and tail resistor current, so lower resistance tail resistors (R6) usually lead to greater imbalance. For any reasonable tail resistor value and its contribution to imbalance, there are load resistor values that will cause the outputs to have the same amplitude. The selection of different load resistors is the common way to "balance" the outputs of the PI. Note that balancing the PI output by varying the load resistors does not change the fact that the tube currents are still unbalanced, it only compensates for the current imbalance.
Most all amps have some imbalance in their PI circuits. Other phase inverter circuits also have imbalance. Part tolerances simply make perfect balance very elusive. Although imbalance can impact an amp's tone, only a listener can decide whether it is good or bad. It's a subjective quality. There is no right or wrong. Many people argue that some imbalance is good for an amp's tone. Hopefully, now you know how imbalance happens and where it comes from.
Most all amps have some imbalance in their PI circuits. Other phase inverter circuits also have imbalance. Part tolerances simply make perfect balance very elusive. Although imbalance can impact an amp's tone, only a listener can decide whether it is good or bad. It's a subjective quality. There is no right or wrong. Many people argue that some imbalance is good for an amp's tone. Hopefully, now you know how imbalance happens and where it comes from.
There are more details about balancing the LTPI circuit and how feedback factors into all of this found in Guitar Amplifier Design: Tubes and Semiconductors Play Together.
There are more details about balancing the LTPI circuit and how feedback factors into all of this found in Guitar Amplifier Design: Tubes and Semiconductors Play Together.
Page updated
Google Sites
Report abuse