Tech help needed

Brick · October 12th, 2006, 09:43 PM

I have an old 50s champ like Valco that has a replacement speaker in it. I put my multimeter across the speaker terminals and it read 18 ohms, which I'm assuming makes it a 16ohm speaker. I want to solder in a 1/4 inch jack and plug so I can use other speaker cabs with the amp. I was just thinking that with the champ tube configuration,( aren't Champs 8 ohm?), who's to say that the person who replaced the speaker had a clue as to what they were doing? Is there a way to tell for sure what ohm speaker rating the amp wants to see?

Tremo · October 12th, 2006, 10:51 PM

Champs are 4 ohms. The proper output impedance of your old Valco is a function of the OT. It may be 4 or 8 ohms, who knows?

Dacious · October 13th, 2006, 02:26 AM

You could unsolder the two primary leads from the tube socket, and then connect an external AC source, like 120vac. Measure what you get in AC volts at the secondaries. Maybe safer with a low-voltage wall wart, rather than AC line voltage. You need to very careful if you do this.

Lets say you get 0.15 volts AC with 120 volts in. 120/0.15=800. As V=R You have a 800:1 winding ratio - that direct translates into relative impedance. That means 800 winds on the primary side to 1 wind on the secondary. A single 6V6 wants to see about a 5kohm load. So with 800:1, the reflected load into the secondary is about 6.25 ohms, close enough to 8 to be intended for 8 ohms.

If you got .08vac, the winding ratio would be 1500 or 1600:1, meaning the output would be intended to be 4ohms. OTOH, if it was 0.3 or 0.4 vac, it would be closer to 16 ohms.

Just knowing the voltage at the secondary pole, you can determine what the output was likely meant to be. Bear in mind, they may have mixed speakers and output trannies depending on what was available day-day. With the limited frequency response required for electric guitar, the same output tranny may have been used for 4 and 8 ohm speakers because impedance varies with frequency and power input load almost infinitely.

That's a rough way to do it, but it will give roughly the minimum impedance the transformer is designed for.

Tremo · October 13th, 2006, 02:53 AM

The impedance ratio is the square of the turns ratio. If the turns ratio is 10:1, the impedance ratio is 100:1.

If you can determine the turns ratio by putting an AC voltage on one winding and measuring the resulting voltage on the other winding, the voltage ratio is the same as the turns ratio.

I would NOT hook an OT up to 120 volts coming out of the wall, too dangerous. I'd put 6.3 volts from a heater winding on it instead.

Once you get the impedance ratio, assume the 6V6 wants to see about 5k reflected to it's plate, so you can then calculate what secondary load would do that.

October 12th, 2006, 09:43 PM	#1 (permalink)
Brick Tele-Holic Join Date: Mar 2003 Location: San Jose Age: 58 Posts: 787	Tech help needed I have an old 50s champ like Valco that has a replacement speaker in it. I put my multimeter across the speaker terminals and it read 18 ohms, which I'm assuming makes it a 16ohm speaker. I want to solder in a 1/4 inch jack and plug so I can use other speaker cabs with the amp. I was just thinking that with the champ tube configuration,( aren't Champs 8 ohm?), who's to say that the person who replaced the speaker had a clue as to what they were doing? Is there a way to tell for sure what ohm speaker rating the amp wants to see?

October 13th, 2006, 02:26 AM	#3 (permalink)
Dacious Friend of Leo's Join Date: Mar 2003 Location: Godzone Posts: 2,612 35 Photos	You could unsolder the two primary leads from the tube socket, and then connect an external AC source, like 120vac. Measure what you get in AC volts at the secondaries. Maybe safer with a low-voltage wall wart, rather than AC line voltage. You need to very careful if you do this. Lets say you get 0.15 volts AC with 120 volts in. 120/0.15=800. As V=R You have a 800:1 winding ratio - that direct translates into relative impedance. That means 800 winds on the primary side to 1 wind on the secondary. A single 6V6 wants to see about a 5kohm load. So with 800:1, the reflected load into the secondary is about 6.25 ohms, close enough to 8 to be intended for 8 ohms. If you got .08vac, the winding ratio would be 1500 or 1600:1, meaning the output would be intended to be 4ohms. OTOH, if it was 0.3 or 0.4 vac, it would be closer to 16 ohms. Just knowing the voltage at the secondary pole, you can determine what the output was likely meant to be. Bear in mind, they may have mixed speakers and output trannies depending on what was available day-day. With the limited frequency response required for electric guitar, the same output tranny may have been used for 4 and 8 ohm speakers because impedance varies with frequency and power input load almost infinitely. That's a rough way to do it, but it will give roughly the minimum impedance the transformer is designed for. __________________ My other Telecaster is a Thinline The Tele Bible, Ch 1, v 10 Love thy Telecaster, covet not thy neighbour's Strat!

October 12th, 2006, 10:51 PM	#2 (permalink)
Tremo Friend of Leo's Join Date: Mar 2003 Location: Silicon Valley, CA, USA Posts: 3,767	Champs are 4 ohms. The proper output impedance of your old Valco is a function of the OT. It may be 4 or 8 ohms, who knows?

October 13th, 2006, 02:53 AM	#4 (permalink)
Tremo Friend of Leo's Join Date: Mar 2003 Location: Silicon Valley, CA, USA Posts: 3,767	The impedance ratio is the square of the turns ratio. If the turns ratio is 10:1, the impedance ratio is 100:1. If you can determine the turns ratio by putting an AC voltage on one winding and measuring the resulting voltage on the other winding, the voltage ratio is the same as the turns ratio. I would NOT hook an OT up to 120 volts coming out of the wall, too dangerous. I'd put 6.3 volts from a heater winding on it instead. Once you get the impedance ratio, assume the 6V6 wants to see about 5k reflected to it's plate, so you can then calculate what secondary load would do that.